Arpa's blog

[Tutorial] Sack (dsu on tree)
By Arpa, history, 9 years ago, In English

UPD: I found a blog describing this blog: Link.

Hi!

Most of the people know about dsu but what is the "dsu on tree"?

In Iran, we call this technique "Guni" (the word means "sack" in English), instead of "dsu on tree".

I will explain it and post ends with several problems in CF that can be solved by this technique.

What is the dsu on tree?

With dsu on tree we can answer queries of this type:

How many vertices in the subtree of vertex v has some property in time (for all of the queries)?

For example:

Given a tree, every vertex has color. Query is how many vertices in subtree of vertex v are colored with color c?

Let's see how we can solve this problem and similar problems.

First, we have to calculate the size of the subtree of every vertice. It can be done with simple dfs:

int sz[maxn];
void getsz(int v, int p){
    sz[v] = 1;  // every vertex has itself in its subtree
    for(auto u : g[v])
        if(u != p){
            getsz(u, v);
            sz[v] += sz[u]; // add size of child u to its parent(v)
        }
}

Now we have the size of the subtree of vertex v in sz[v].

The naive method for solving that problem is this code(that works in O(N ^ 2) time)

int cnt[maxn];
void add(int v, int p, int x){
    cnt[ col[v] ] += x;
    for(auto u: g[v])
        if(u != p)
            add(u, v, x)
}
void dfs(int v, int p){
    add(v, p, 1);
    //now cnt[c] is the number of vertices in subtree of vertex v that has color c. You can answer the queries easily.
    add(v, p, -1);
    for(auto u : g[v])
        if(u != p)
            dfs(u, v);
}

Now, how to improve it? There are several styles of coding for this technique.

1. easy to code but .

map<int, int> *cnt[maxn];
void dfs(int v, int p){
    int mx = -1, bigChild = -1;
    for(auto u : g[v])
       if(u != p){
           dfs(u, v);
           if(sz[u] > mx)
               mx = sz[u], bigChild = u;
       }
    if(bigChild != -1)
        cnt[v] = cnt[bigChild];
    else
        cnt[v] = new map<int, int> ();
    (*cnt[v])[ col[v] ] ++;
    for(auto u : g[v])
       if(u != p && u != bigChild){
           for(auto x : *cnt[u])
               (*cnt[v])[x.first] += x.second;
       }
    //now (*cnt[v])[c] is the number of vertices in subtree of vertex v that has color c. You can answer the queries easily.

}

2. easy to code and .

vector<int> *vec[maxn];
int cnt[maxn];
void dfs(int v, int p, bool keep){
    int mx = -1, bigChild = -1;
    for(auto u : g[v])
       if(u != p && sz[u] > mx)
           mx = sz[u], bigChild = u;
    for(auto u : g[v])
       if(u != p && u != bigChild)
           dfs(u, v, 0);
    if(bigChild != -1)
        dfs(bigChild, v, 1), vec[v] = vec[bigChild];
    else
        vec[v] = new vector<int> ();
    vec[v]->push_back(v);
    cnt[ col[v] ]++;
    for(auto u : g[v])
       if(u != p && u != bigChild)
           for(auto x : *vec[u]){
               cnt[ col[x] ]++;
               vec[v] -> push_back(x);
           }
    //now cnt[c] is the number of vertices in subtree of vertex v that has color c.
    // note that in this step *vec[v] contains all of the subtree of vertex v.
    if(keep == 0)
        for(auto u : *vec[v])
            cnt[ col[u] ]--;
}

3. heavy-light decomposition style .

int cnt[maxn];
bool big[maxn];
void add(int v, int p, int x){
    cnt[ col[v] ] += x;
    for(auto u: g[v])
        if(u != p && !big[u])
            add(u, v, x)
}
void dfs(int v, int p, bool keep){
    int mx = -1, bigChild = -1;
    for(auto u : g[v])
       if(u != p && sz[u] > mx)
          mx = sz[u], bigChild = u;
    for(auto u : g[v])
        if(u != p && u != bigChild)
            dfs(u, v, 0);  // run a dfs on small childs and clear them from cnt
    if(bigChild != -1)
        dfs(bigChild, v, 1), big[bigChild] = 1;  // bigChild marked as big and not cleared from cnt
    add(v, p, 1);
    //now cnt[c] is the number of vertices in subtree of vertex v that has color c. You can answer the queries easily.
    if(bigChild != -1)
        big[bigChild] = 0;
    if(keep == 0)
        add(v, p, -1);
}

4. My invented style .

This implementation for "Dsu on tree" technique is new and invented by me. This implementation is easier to code than others. Let st[v] dfs starting time of vertex v, ft[v] be it's finishing time and ver[time] is the vertex which it's starting time is equal to time.

int cnt[maxn];
void dfs(int v, int p, bool keep){
    int mx = -1, bigChild = -1;
    for(auto u : g[v])
       if(u != p && sz[u] > mx)
          mx = sz[u], bigChild = u;
    for(auto u : g[v])
        if(u != p && u != bigChild)
            dfs(u, v, 0);  // run a dfs on small childs and clear them from cnt
    if(bigChild != -1)
        dfs(bigChild, v, 1);  // bigChild marked as big and not cleared from cnt
    for(auto u : g[v])
	if(u != p && u != bigChild)
	    for(int p = st[u]; p < ft[u]; p++)
		cnt[ col[ ver[p] ] ]++;
    cnt[ col[v] ]++;
    //now cnt[c] is the number of vertices in subtree of vertex v that has color c. You can answer the queries easily.
    if(keep == 0)
        for(int p = st[v]; p < ft[v]; p++)
	    cnt[ col[ ver[p] ] ]--;
}

But why it is ? You know that why dsu has time (for q queries); the code uses the same method. Merge smaller to greater.

If you have heard heavy-light decomposition you will see that function add will go light edges only, because of this, code works in time.

Any problems of this type can be solved with same dfs function and just differs in add function.

Hmmm, this is what you want, problems that can be solved with this technique:

(List is sorted by difficulty and my code for each problem is given, my codes has heavy-light style)

600E - Lomsat gelral : heavy-light decomposition style : Link, easy style : Link. I think this is the easiest problem of this technique in CF and it's good to start coding with this problem.

570D - Tree Requests : 17961189 Thanks to Sora233; this problem is also good for start coding.

Sgu507 (SGU is unavailable, read the problem statements here) This problem is also good for the start.

HackerEarth, The Grass Type This problem is also good for start (See bhishma's comment below).

246E - Blood Cousins Return : 15409328

208E - Blood Cousins : 16897324

IOI 2011, Race (See SaYami's comment below).

291E - Tree-String Problem : See bhargav104's comment below.

1009F - Dominant Indices : 40332812 Arpa-Style. Thanks to Tanmoy_Datta.

343D - Water Tree : 15063078 Note that problem is not easy and my code doesn't use this technique (dsu on tree), but AmirAz 's solution to this problem uses this technique : 14904379.

375D - Tree and Queries : 15449102 Again note that problem is not easy :)).

716E - Digit Tree : 20776957 A hard problem. Also can be solved with centroid decomposition.

741D - Arpa’s letter-marked tree and Mehrdad’s Dokhtar-kosh paths : 22796438 A hard problem. You must be very familiar with Dsu on tree to solve it.

For Persian users, there is another problem in Shaazzz contest round #4 (season 2016-2017) problem 3 that is a very hard problem with this technique.

If you have another problem with this tag, give me to complete the list :)).

And after all, special thanks from PrinceOfPersia who taught me this technique.

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Comments Comments (155)
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9 years ago, # |
  Vote: I like it +4 Vote: I do not like it

A2OJ's DSU Section has quite a few tree DSU problems.

Thank you for this post, it explains the theory well and is very easy to read.

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9 years ago, # |
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What does the variable "keep" denote ?

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    9 years ago, # ^ |
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    The way I understand HLD here is basically if a child is the big child, we don't want to recompute answer for it to reduce computation. So we just store the answer for it in the cnt array already so that it's parent doesn't need to re-dfs this subtree. keep denotes whether or not this child is that big child. Please correct me if I'm wrong. :)

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    9 years ago, # ^ |
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    Look at last two lines:

        if(keep == 0)
        add(v, p, -1);

    It means that if keep == false after dfs clear vs subtree information from cnt. And if keep == true, don't clear vs subtree information from cnt. In other word if keep == trueafter calling dfs, for each u from subtree of vertice v, col[u] is in cnt (cnt[ col[u] ]++).

    And NibNalin is right. keep is true if and only if v is biggest child of it's parent.

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      8 years ago, # ^ |
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      Hi Arpa, thanks a ton for this awesome post. I have really learnt lot from it.

      I do have one question though. What is the advantage of having keep=false? If that part is kept as it is, without clearing, doesnt the computation become faster? Can you please help clearing this doubt?

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        8 years ago, # ^ |
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        Hi, Thanks.

        Consider vertex v has two children, q and p. If you call dfs for both of them with keep = true, they will mixed up their information, and queries will be incorrectly answered.

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    9 years ago, # ^ |
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    Observations to understand the complexity:

    1. The dfs function visits each node exactly once.
    2. The problem might seem with the add function. You might think that it is making the algorithm n^2. Note that in the add function, we only go down from a vertex to its children if the edge connecting the vertex to the child is a light edge.

    You can think of it in this way, each vertex v will be visited by a call to the add function for any ancestor of v that is connected to a light edge. Since there are at most log(n) light edges going up from any vertex to the root, each vertex will be visited at most log(n) times.

    So the algorithm is: Say you are at a vertex v, first you find the bigchild, then you run dfs on small childs, passing the value of keep as 0. Why? So they are cleared from cnt. Then you run a dfs on bigchild, and you do not clear it from cnt. Now, cnt stores the results for all vertices in the subtree of bigchild (since we cleared cnt for small childs and didn't do so for the bigchild), so we call the add function to "add" the information of children of current vertex that are connected to it via a light edge. Now we are ready to compute the answer

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      7 years ago, # ^ |
      ← Rev. 3   Vote: I like it +3 Vote: I do not like it

      As you said "add" function goes down using only light edges,Don't these two lines if(bigChild != -1) big[bigChild] = 0; of heavy light decomposition implementation would affect it as if you call "add" after all dfs are done and returned to the root then we only have one heavy edge marked that of root itself others are zero so as "add" goes below it traverses whole tree. Help me here.

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        7 years ago, # ^ |
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        Got it.

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          7 years ago, # ^ |
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          For those who had same doubt as I had: First lets understand why it is wrong to remove it,consider you are at particular node(let it be called A) in recursion,above line is not there,you have 3 children one of them is big child(3rd) while others are normal so you traversed inside 1st and came back to A then you traversed inside 2nd child if you do not have above line then while going inside this children you would have all answer for big children of 1st child which would mess your answer. Now let's understand why complexity is O(n(log(n)):

          Note 1: To calculate complexity you need to measure how many times add function visits the ever y node.

          Note 2: For first add called at a node: A node will visited by add only through its ancestors which are connected by light edges so n nodes log(n) light edges above it this gives us O(n(log(n)))

          Note 3: For second add called at a node: Now somebody may protest that that after above mentioned line(big[bigChild]=0) we are unmarking heavy edge and also calling add after that which may mess up complexity as it travels every node below it which is O(n) but keep==0 condition ensures that for each node there atmost log(n) nodes above in ancestor which have keep=0 function is called.which again gives O(n(log(n)).

          Giving us finally O(n(log(n))) complexity. Follow this link to understand heavy light decomposition's properties: https://blog.anudeep2011.com/heavy-light-decomposition/

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9 years ago, # |
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In second easy with O(n*lg n)

Why if(keep==false) we delete only vertex from main vector

for(auto u : *vec[v])
            cnt[ col[u] ]--;

but we don't delete vertex from cnt which we changed here:

for(auto u : g[v])
       if(u != p && u != bigChild){
           for(auto u : *vec[u])
               cnt[ col[u] ]++;
       }
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    9 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    There was a mistake in writing. I'm sorry.

    Thanks for reporting this problem.

    code should be:

           if(u != p && u != bigChild){
           for(auto x : *vec[u])
               cnt[ col[x] ]++;
       }

    Instead of:

           if(u != p && u != bigChild){
           for(auto u : *vec[u])
               cnt[ col[u] ]++;
       }

    I have edited that.

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9 years ago, # |
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can someone tell me how 208E is solved with this technique? thanks a lot.

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    9 years ago, # ^ |
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    You need to compute for each pair v p the p-th cousin of v. That is equivalent to finding the number of p-th descendants of the p-th ancestor of v — 1.

    So for each query, replace v p with p_th_ancestor_of_v p. Now you need to store in cnt the number of nodes at a certain depth. In other words, cnt[x] should be equal to number of nodes at depth x in the current subtree.

    Code for Reference: http://codeforces.me/contest/208/submission/17513471

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      9 years ago, # ^ |
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      can't understand why for every vertex v we ans[depth[v]] increase by 1 when we call add function,why must we do it?or why it must be ans[deth[v]] when depth[v] means distance from root to v?

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        9 years ago, # ^ |
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        ans[h] is equal to number of vertices with height h, (with distance h from root).

        Let par, p'th ancestor of v, the answer to query is:

        Consider only subtree of vertice par, print ans[ height[v] ] — 1.

        So with method above we can process all of the queries.

        See my code for better understanding.

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          9 years ago, # ^ |
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          thanks everyone , now i understand.

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9 years ago, # |
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If i haven't read this article, i wouldn't get ac on this problem. It is another problem which can be solved easily by dsu.

here is my code in HLD-style.

Thanks!

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9 years ago, # |
← Rev. 3   Vote: I like it +12 Vote: I do not like it

I can't understand why the second code is correct...

Consider this example:

We wanna calculate the cnt for Vertex 8. These are the steps:

Going to Vertex 1

Going to Vertex 2 by keep=0

Going to Vertex 3 by keep=0,Vec[3]={3}

Going to Vertex 4 by keep=1,Vec[4]={4},Cnt[color[4]]=1

Going back to Vertex 2,Vec[2]={2,4},Cnt[color[4]]=0,Cnt[color[3]]=1

And then when we go to Vertices 5,6,7,8 still Cnt[color[3]]=1.

Also sorry if I did the steps wrong...

UPD Thank you for editing the blog. My problem fixed.

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8 years ago, # |
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Great post. If you explained the idea before showing the code, it would be better to understand. Also commenting the variables meaning in the code would be of great help.

It would be good to mention that most solutions will answer the queries offline, which may be a problem sometime (maybe someone didn't notice this lol).

Also, it would be nice to post hints about the solutions.

Proving explicitly why it is nlogn would be good too (ie. as each node's subtree set gets merged into one set of size equal or greater, and the base set has size 1 and the last set has size n, then we take logn steps to go from 1 to n. Summarizing, each node gets merged logn times, so the total complexity is O(nlogn)).

Here's my solution to 343D, maybe it will be of help to someone: 18958875. A lot easier to code than the one in the tutorial.

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    8 years ago, # ^ |
      Vote: I like it +2 Vote: I do not like it

    Thanks for your suggestions first !

    I proved that it is O(nlogn) : You know that why dsu has O(q log n) time (for q queries); the code uses same method. Merge smaller to greater.

    And about your code (18958875), anyone has a different opinion !

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      8 years ago, # ^ |
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      Thanks for the reply!

      Yeah, you did prove. People who remember DSU's proof will most likely understand. I stated a more extensive proof would be better thinking about people who don't exactly know the proof. Don't take me wrong, but they may get a little confused reading this proof.

      I mentioned my code exactly because everyone has a different opinion,. Maybe it'll help a later reader, that's all.

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        8 years ago, # ^ |
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        Sorry this might be a stupid question to bring up, but why is the complexity of the heavy-light decomposition style one in O(nlogn)?

        In the case where each node has at most two children: Denote the root node of the tree as u, which is of size s. The child of u connected to the lighter edge is of size at most . So the total number of nodes on which we run the "add" function would be at most . So I don't understand where the log(n) factor comes from.

        The online tutorial for HLD says a new chain is built when we arrive at a child node via a lighter edge, where each chain is stored as a segment tree, and so I can see there is indeed a O(logn) factor involved.

        Regardless can you perhaps elaborate a little bit more on the time complexity of the dsu structure? Thank you!

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          8 years ago, # ^ |
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          Hi !

          The online tutorial for HLD says a new chain is built when we arrive at a child node via a lighter edge, where each chain is stored as a segment tree, and so I can see there is indeed a O(logn) factor involved.

          As you know, if you use segment tree in heavy-light decomposition, each query time complexity will be O(log2(n)). Because in each query you will go O(log(n)) chains and in each chain it will spend O(log(n)) time.

          Now, I will proof that "heavy-light decomposition style implementation" of "dsu on tree" is O(n.log(n)):

          Consider a complete binary tree with n vertices. In dfs function you will run another dfs function in child (T(n / 2) * 2) and you will call add function and it will spend O(n / 2) time. So, T(n) = n / 2 + 2 * T(n / 2) = O(n.log(n))

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      8 years ago, # ^ |
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      You know that why dsu has O(q log n) time (for q queries); the code uses same method. Merge smaller to greater.

      Pardon me , but I don't follow. Which dsu are you talking about? The one with inverse-Ackermann function?

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        8 years ago, # ^ |
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        No. Dsu with size compare. Like this : 

        int find(int x){
    return par[x] == x ? x : find(par[x]);
}
void merge(int v, int u){
    v = find(v), u = find(u);
    if(v == u)  return ;
    if(size[v] < size[u])  swap(v, u);
    par[u] = v;
    size[v] += size[u];
}
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8 years ago, # |
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In easy to code but O(nlog^ 2), I cant't understand why do we store the size of subtrees of vertices in array sz and use it as the criteria for choosing the big child, I think we should store in the array "sz" the number of distinct colors in the subtree of any node v, because that is what we actually iterate on when transferring the map from v to u, why is this wrong?

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    8 years ago, # ^ |
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    Hi !

    It isn't wrong! Both of your method and mine have the same worst case. But your average is better.

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      6 years ago, # ^ |
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      Is child's subtree-size working as a heuristic for the number-of-distinct-colors in that child?

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8 years ago, # |
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Ahh, thanks gabrielsimoes, for anyone struggling to understand: n*log^2n is about answering queries OFFLINE right during the dfs. After the dfs has finished the cnt[v] will no longer be a valid map for vertices that were chosen as bigChild.

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8 years ago, # |
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http://codeforces.me/problemset/problem/291/E 291E - Древесно-строковая задача Arpa This problem can also be done by dsu on trees. Calculate hash value for each suffix value of target string. Then for each suffix of an edge if it is a valid prefix of the target string we would just need the frequency of the hash value of the remaining suffix of the target string in its subtree which can be maintained by this technique. The case when the entire string occurs in an edge can be dealt with separately.

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    8 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    Thanks added to list, but it can be solved very easier : 19827525, just use KMP.

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      7 years ago, # ^ |
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      Just use hashes :) http://codeforces.me/contest/291/submission/29431526

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      5 years ago, # ^ |
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      Hello Arpa,

      The technique which you've used in this question, for the KMP solution, in which you've created the nxt 2D array, is it called something? Is it some recognized idea or did you just came with it on the go?

      Basically, you removed the inner while loop of KMP for matching strings. Instead of jumping multiple times to check whether the current characters match, you created a DP, using which we can do that in a single jump. That saves a lot of time, as we would have to do that for each string in the tree, while matching.

      So, is this some already defined method?

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        5 years ago, # ^ |
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        Hello.

        Do you know automata? I'm building automata from the string.

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          5 years ago, # ^ |
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          I do know automata. But damn. That's some beautiful automata you got there. GG.

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    7 years ago, # ^ |
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    Please send me a code of the solution with this techinque)

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8 years ago, # |
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Actually, in China, we call this method as "Heuristic Merge" which always merge the smaller to the bigger. Not hard to understand each vertex will be visited in O(log n) times because when we visited a vertex, then the size of tree which the vertex is in doubled.

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8 years ago, # |
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Hey Arpa,

In your my invented style I'm unable to understand that why in third loop are you not checking for u not being parent of v. Why are you only checking for just u not being the big child.

Thanks in Advance

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8 years ago, # |
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In the easy to code O(nlogn) method vec[v] stores all the vertices in the the subtree rooted at v . How will this fit into memory if we are not deleting the vec of child after merging it with the parent

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    8 years ago, # ^ |
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    Used memory is always less than or equal to time complexity, so when time complexity is , used memory is less than or equal to . In this case, used memory is . Although if you delete useless vec's the memory become .

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      8 years ago, # ^ |
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      Thanks for the reply . I think this problem can also be solved using your approach. (The Grass Type HackerEarth)

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        8 years ago, # ^ |
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        I'll add this problem to the post if I find it related, I'm thankful anyway.

        Edit : Note that this is not my approach, but I'm the first man who publishes a tutorial about this (not sure), Sack has been used in INOI, IOI and ACM several times, so it isn't a new thing, invented be me.

        Edit : Added.

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          8 years ago, # ^ |
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          Can you mention probelms from the IOI that are solved with sack ?

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            8 years ago, # ^ |
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            I'll add one of them tonight.

            Edit : Added.

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              8 years ago, # ^ |
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              Wow, I didn't think of solving it with sack.Thx

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              4 years ago, # ^ |
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              HEY Arpa i've solved all of the problem u mentioned above except digit tree, arpa's letter marked tree and ioi 2011 race.. Can you share a code for ioi 2011 race which uses sack.

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                4 years ago, # ^ |
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                Unfortunately, I don't have enough time.

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8 years ago, # |
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Hi Arpa, I can not understand, why is this approach called dsu on tree? This approach has a nice trick to reduce complexity by saving data about "big child". I can't see any special similarity with general dsu approach. In general dsu problems, we merge 2 subset into 1 set by linked list approach. But, in your tutorial there is no "merge" function. Am I missing something?

Also I see that, in your 600E's solution 14554536 you used merge functon. I can't understand, could you please explain that code?

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    8 years ago, # ^ |
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    In fact we are merging information of small children with big child. Think more.

    In that code, mrg function merges information in u into v.

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8 years ago, # |
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Hi Arpa! Thanks for making this tutorial.

I just want to make sure my understanding is correct: this merging smaller maps into larger ones takes logarithmic time because when a vertex is merged, the new map it is in at least twice its size. Hence, merging can only happen log(n) times for each of the n vertices, leading to a total runtime of O(nlogn)?

Thanks!

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    8 years ago, # ^ |
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    Yes, but note that if you use map, it's .

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      8 years ago, # ^ |
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      If you use unordered_map, does it become O(n·logn), then?

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        8 years ago, # ^ |
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        Unordered_map is theoretically O(n) per query. But you can suppose that it's O(1) per query in code.

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8 years ago, # |
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Why do we need to iterate through the children of v after add(v,p,-1)in the naive approach?

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    8 years ago, # ^ |
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    dfs() solves the problem for all the nodes, not just one. So, after you've gotten the answer for v, it will calculate the answer for its children.

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8 years ago, # |
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101 Hack 47 Summing Tree was solved using this technique by satyaki3794 Submission

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8 years ago, # |
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also 778C - Петрович --- полиглот is solvable with a similar tecnique: is that DSU on tree?

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7 years ago, # |
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Can anyone give me a link to "Shaazzz contest round #4 (season 2016-2017) problem 3" or tell me where can I find it? Thanks.

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    7 years ago, # ^ |
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    It's a Persian contest.

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      7 years ago, # ^ |
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      Can you tell me where can I find it? I searched for it just now but didn't get it.

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7 years ago, # |
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APIO 2016 Fireworks uses this, but is a much harder problem.

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7 years ago, # |
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Arpa, in the Easy to code but O(nlog2n) section code you have written a commented line that is : //now (*cnt)[c] is the number of vertices in subtree of vertex v that has color c. You can answer the queries easily.. But I think it would be //now (*cnt[v])[c] is the number of vertices in subtree of vertex v that has color c. You can answer the queries easily.. Will (*cnt)[c] changed with (*cnt[v])[c] ?

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7 years ago, # |
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You can solve APIO 2012 Dispatching with this technique too.

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7 years ago, # |
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IOI 2011 — Race What is the idea of DSU on tree for this problem? I know of a solution based on Centroid Decomposition.

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    5 years ago, # ^ |
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    keep map<distance from root, depth> for all subtrees. By merging them from smaller to larger, you can get O(nlog^2n) solution.

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7 years ago, # |
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914E - Палиндромы в дереве can solve with sack too, Arpa :)

ps: for this problem centroid decompose works too... :)

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7 years ago, # |
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can anyone please explain how to solve 716E using this technique?

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7 years ago, # |
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In the contest 600, no one can view other's submissions except his own.

That's why no can see your submissions for 600E - Lomsat gelral except you.

So, please give alternating link of your solutions for the first problem in the list, 600E - Lomsat gelral

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7 years ago, # |
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Another problem which can be solved by this technique: Coloring Tree
Its easier than 600E - Lomsat gelral.

One more : 932F - Сбежать через лист

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7 years ago, # |
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No good explanation! Only code! Worst tutorial.

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    7 years ago, # ^ |
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    If you have a doubt why not ask in the comments rather than whining about how bad the tutorial is.

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7 years ago, # |
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Nice tutorial!

and also happy Nowruz:)

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    7 years ago, # ^ |
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    Could you please explain 2no. style ? [Upd : Got it ]

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      6 years ago, # ^ |
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      In 2nd style Arpa can you please explain?..when we do cnt[u]=cnt[heavy child] does this happen in O(1)?

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        6 years ago, # ^ |
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        if yes -> how? else what the benefit of doing this? upd[got it]

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7 years ago, # |
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For those who still confused about the time complexity, I found this explanation by radoslav11 helps a lot.

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7 years ago, # |
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What is col[v]?

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6 years ago, # |
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I loved the way you used the Euler tour sequence to iterate through the subtrees. My 2 cents is a way to use C++ to implement this nicely. The main idea is to use a structure, that keeps pointers instead of st[v] and ft[v], and to give it a begin() and end() member function, so that you can iterate through it with

for(int u : T[i])
    cout << u << " is in the subtree of " << i << "\n";

To actually solve the problem, maybe in a Heavy-Light style, I kept also for each vertices a pointer to the leaf of its heavy-path, so that I could only change the answer in the leaf.

My full implementation of lomsat gelral can be found here

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6 years ago, # |
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Hi , Arpa, I used this technique to solve 600E - Lomsat gelral, its a very neat technique.. but wont this get a MLE? I am getting a MLE with the O(nlog^2n0 method ,.. I saw your solution and I see its the same as mine, but mine gets MLE.. My solution :- 39325497

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    6 years ago, # ^ |
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    Solutions for Educational Codeforces Round 2 is private, nobody can't see your code. Perhaps you do not use pointers, dynamic memory allocation, and memory cleanup. In this technique it is forbidden to copy huge containers.

    I solved this problem with Euler tour on tree and Mo algorithm in time and O(n) memory. Code.

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6 years ago, # |
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Great blog but I am not able to understand the logic behind the O(nlogn) solution it would be a great help if anyone can explain it.

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6 years ago, # |
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I have a question for style 3 i.e HLD style.
I'm not so sure what's happening in there, I have 2 assumptions. Both of them are wrong, so it would great if someone could point out the mistake and explain what's right.

We are in root.

  1. We first go down the light edges of root and when we finish with them, we clear the cnt array, so there is absolutely nothing left in it and then we proceed to the heavy child of the root and then we just update cnt array.
    Now we go back to the light edges of the root and (here's the problem) as the cnt array only contains information for the heavy child of the root, we must go through EVERY vertex in subtrees of light children of the root. If we don't go to the heavy children in the subtrees (as it proposes in tutorial?), then the answer is wrong, as we didn't count them (remember that we cleared the cnt array).

  2. We first go down the light edges of the root, but this time, for every heavy child, we keep the information.
    But then as we proceed to the heavy child of the root, the array cnt won't be empty and the answer for heavy child will be incorrect.

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    6 years ago, # ^ |
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    Consider you entered vertex v, let it's heavy child h. First, go to all child of v except h and calculate the answer for them and each time clear the cnt array. Then go to h and calculate the answer and don't clear the cnt array. Then traverse all of the subtree of v except subtree of h, and add them to cnt.

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6 years ago, # |
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Another problem can be solved with this technique. Arpa please add this one in the list.

http://codeforces.me/contest/1009/problem/F (Dominant Indices)

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6 years ago, # |
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Can someone explain why this solution 40510312 is getting TLE on problem 600E - Lomsat gelral. I'm using style 4.

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    6 years ago, # ^ |
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    Can't view your submission

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      6 years ago, # ^ |
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      Firstly, thank you for reaching out to help :). I got my mistake.

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6 years ago, # |
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I'm continuously getting TLE in test case 30 in 570 : D : Tree Request (The second one given in the blog practice problem) while implementing through DSU similar to what is given here. Submission Id : 40780911 , can somebody make a look over this and provide hint to optimize it.

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    6 years ago, # ^ |
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    My solution passed the time limit after replacing endl with "\n".

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6 years ago, # |
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This blog was a bit code-heavy for me when I first read this. Hence I have tried to simplify the concept a bit more in a more textual fashion in my own tutorial spin-off. I have tried to provide the intuition behind small-to-large merging including small to large on trees, also knows as DSU on trees.

However I. haven't provided any code as the code given by the OP is more than enough.
DSU-on-Tree-Intuition

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6 years ago, # |
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sorry but you absolutely wrong about the complexity, i know the trick merge smaller to greater but it works when you don't for(i = 0; i < greater.size(); i++) again, but here in O(nlogn ^ 2) you use cnt[v] = cnt[bigchild], it equal to "for(int i = 0; i < cnt[bigchild].size(); i++) {} ", so it's not O(nlogn ^2)

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6 years ago, # |
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Can someone explain, why will the complexity of the first code will be O(nlog^2n)?

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    6 years ago, # ^ |
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    Every vertex will appear in cnt's, every time we want to insert a vertex it costs , so the whole time complexity is .

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      6 years ago, # ^ |
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      1. That means, if I use unordered_map my complexity will reduce to O(nlogn)?
      2. How can you say that every vertex will appear in log(n) cnt's?
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        6 years ago, # ^ |
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        1. Yes.
        2. Read the proof of that for each vertex there is at most light edges in the path between this vertex to root: Link.
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6 years ago, # |
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This particular problem can be solved in linear time:

int cnt[maxn];

void dfs(int v, int p){
    // Subtract cnt[c] for query here, e.g. answer[query] -= cnt[c];
    
    cnt[col[v]]++;
    for(auto u : g[v])
        if(u != p)
            dfs(u, v);

    // Add cnt[c] for query here, e.g. answer[query] += cnt[c];
}
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6 years ago, # |
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Can somebody help me understand why my solution 51874074 for problem 246E - Братья по крови возвращаются gets WA on test 50? I tried to implement the Heavy Light Decomposition style dsu

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6 years ago, # |
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How to answer online queries using this technique??

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6 years ago, # |
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why did he declare every time using pointer, can't we simply declare array of map for first approach as map<int,int> cnt[maxx] instead of map<int,int>*cnt[maxx] ?

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    6 years ago, # ^ |
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    Due to this line, cnt[v] = cnt[bigChild]. Because we are using pointer, we only pass the reference. In other case, whole map will be copied(time complexity: linear in size of map).

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      5 years ago, # ^ |
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      Then, will the information for the child be lost when we modify the parent's map which actually references the child's map... If so then how to answer queries in case when I need to access the nodes' map later?

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        4 years ago, # ^ |
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        Yes, it is an offline algorithm. You can use it if you can read all the queries and solve them together.

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6 years ago, # |
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Why is my code getting TLE for problem 600E — Lomsat gelral? I'm using the technique no. 1 from this tutorial.

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    5 years ago, # ^ |
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    your code is working in $$$O(N^2)$$$ .You can't go through the whole $$$(*cnt[u])$$$ to answer queries.

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      5 years ago, # ^ |
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      after 2 months!!!

      you think he / she, will check this blog again looking for his answer?

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5 years ago, # |
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Love this.

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5 years ago, # |
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Thank you for the article.Now I understand it.

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5 years ago, # |
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Small nitpick: In solution 2, the comment

//now (*cnt[v])[c] is the number of vertices in subtree of vertex v that has color c.

should be:

//now cnt[c] is the number of vertices in subtree of vertex v that has color c.
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5 years ago, # |
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For everyone who needs an explanation, here is a blog.

And obviously, thank you Arpa for this wonderful content!

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5 years ago, # |
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Thanks a lot really.

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5 years ago, # |
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For heavy-light decomposition style for the first add(v,p,1) except the big child sub-tree we are visiting the whole sub-tree of all the light children. Also for second add we are visiting again all the sub-tree if keep==0. How does the complexity becomes $$${O(nlogn)}$$$. Having a hard time to understand, please can anyone tell me how is the complexity calculated.

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    5 years ago, # ^ |
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    For each vertex like $$$v$$$, let $$$l(v)$$$ number of light edges in the path from the root to $$$v$$$. Add function called on $$$v$$$, $$$l(v)$$$ times. $$$l(v) \in \mathcal{O}(\log n)$$$, so overall complexity is $$$\mathcal{O}(n \log n)$$$.

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      5 years ago, # ^ |
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      For second add function we have made big[bigChild]=0 for all bigChilds in subtree. Now when we call add(v,p,-x) the whole subtree is visited. How does this make complexity $$${O(nlogn)}$$$

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        5 years ago, # ^ |
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        We're deleting all the children only when the current node is not a bigChild, so that it doesn't mess up the global data structure. When the current node is a bigChild, then keep=1, and no data regarding the current subtree is deleted.

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5 years ago, # |
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Can we know the shortest distance from a vertex using dsu ?

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5 years ago, # |
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I tried solving 208E - Братья по крови using the first approach as explained in this blog but getting MLE on test case 59. I'm not able to figure out why it's giving MLE. Please help!

Here is the link to my submission : 80790329

Thanks!

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4 years ago, # |
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https://www.hackerearth.com/problem/algorithm/monkeys-on-a-tree-0fe98ed9/

This problem can also be done by dsu on trees.

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4 years ago, # |
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cnt[v] = new map<int, int> ();

Why are we doing this? My code gave segmentation fault without this.

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    4 years ago, # ^ |
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    cnt[v] is not a map. It is "pointing" to a map. This is because we don't want to re-assign all the values of the bigChild to v, but rather just change the "pointer", which is a lot faster. Of course by doing that, we are affecting the subtree of the bigChild as well, but we have already replied to queries related to the bigChild's subtree. That is, because here we have a pointer, we need to assign to cnt[v] a "new map", whenever it is pointing to a NULL, else you will get seg-fault.

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4 years ago, # |
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Very late, but this problem (Educational Round 91 E) is a very recent problem that can be solved with this technique.

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4 years ago, # |
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Nice

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4 years ago, # |
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Hello AmirAz, I tried to submit a similar solution to yours for the problem Water Tree but it was giving tle on case 20. Can you tell me what to did you exactly took care of to reduce time complexity?

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    4 years ago, # ^ |
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    Arpa Can you tell me why his solution got accepted but mine failed (tle) on test 20?

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4 years ago, # |
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Great practice problems! Thanks for the tutorial Arpa.

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3 years ago, # |
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Sorry for necroposting.

I felt this could be helpful for people trying to learn this great technique.

https://codeforces.me/problemset/problem/1363/E

My submission: https://codeforces.me/contest/1363/submission/125449811

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3 years ago, # |
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Very useful blog! Thank for Arpa. I also looking forward to seeing an update on the problem list as there are many new forms of problems to be solved by this technique nowadays. As far as I know, some problems can even be associated with data structures, such as this.

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Can anyone please help, why we need bigChild here, why don't we have direct DFS and where child returning map of cnt.

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    47 hours ago, # ^ |
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    That's the main crux of the small-to-large merging.

    Let's say you have n children maps with sizes x1, x2 , ... , xn. Then instead of creating a new set for parent and then merging all n sets one by one, We can start with the bigChild and merge all other children's maps into it.

    Same thing is happening here, we are not clearing the bigChild's info so it can be use as base and then we are adding all other children's info.

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