Sokol080808's blog

By Sokol080808, history, 17 months ago, translation, In English

1843A - Sasha and Array Coloring

Idea: Sokol080808, prepared: molney

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1843B - Long Long

Idea: EJIC_B_KEDAX, prepared: molney

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1843C - Sum in Binary Tree

Idea: Sokol080808, prepared: molney

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1843D - Apple Tree

Idea: EJIC_B_KEDAX, prepared: Vladosiya

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1843E - Tracking Segments

Idea: meowcneil, prepared: meowcneil

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1843F1 - Omsk Metro (simple version)

Idea: EJIC_B_KEDAX, prepared: Sokol080808

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1843F2 - Omsk Metro (hard version)

Idea: EJIC_B_KEDAX, prepared: Sokol080808

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17 months ago, # |
  Vote: I like it +2 Vote: I do not like it

Good contest. Ah Why I didn't think of int overflow

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    17 months ago, # ^ |
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    You should try to read statements more carefully. At the end of the output section in B was stated:

    "Pay attention that an answer may not fit in a standard integer type, so do not forget to use 64-bit integer type."

    And also, the problem name was a hint :)

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17 months ago, # |
  Vote: I like it +53 Vote: I do not like it

Once again, thank you for this fun contest! :D

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17 months ago, # |
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Why does my code output 0 in all queries in D problem? Submission Id: 210522059 ~~~~~ void dfs(int apple){

visited[apple] = 1;

for(auto it = adj[apple].begin();it<adj[apple].end();it++){
    if(!visited[*it]){
        dfs(*it);
        leaves[apple]+=leaves[*it];
    }
    // else leaves[apple]+=leaves[*it];
}
if(adj[apple].empty()){
    leaves[apple]++;
    // cout<<"Leaves["<<apple<<"] is : "<<leaves[apple]<<"\n";
    // visited[apple] = 0;
    return;
}
// cout<<"Leaves["<<apple<<"] is : "<<leaves[apple]<<"\n";
// visited[apple] = 0;
return;

}

~~~~~

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    17 months ago, # ^ |
    Rev. 2   Vote: I like it +8 Vote: I do not like it

    It is because of the way you are reading the graphs as input, you are making the larger vertex number the child and the smaller one the parent which is not always true

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17 months ago, # |
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I've learned to use binary lifting to store more information thanks to the problem F, really good contest

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    16 months ago, # ^ |
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    can you explain how you have used binary lifting

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      16 months ago, # ^ |
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      For each node, let's store for it some informations, consider the path start from the node to its 2^k-th parents, we store the maximum prefix, suffix, the sum of whole path, and the maximum sum, we can efficiently use these informations to combine the path with another path.

      To get similar with this concept, I suggest you to have a look at the Segment Tree course in Codeforces Edu section, which also have this concept but for an array.

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17 months ago, # |
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nice round,hope for the positive del:)

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17 months ago, # |
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Can someone tell me why my code was giving TLE for 1843D - Apple Tree. All I did was globally define the graph and cnt array similar to the tutorial solution and it worked then.

This is my initial submission: 210536302. This is the final accepted submission: 210538893

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    17 months ago, # ^ |
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    Passing the whole graph as an argument for the helper function each time is costly. C++ basically has to make a new copy of it each time you call that function. After making g global, it no longer has to do this. Alternatively, you could've passed a pointer to the graph as an argument.

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      17 months ago, # ^ |
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      Okay, thanks!

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      2 weeks ago, # ^ |
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      Hi, I know its been a long time after this contest but can you help me understand why my solution is giving WA on test4. My approach and the editorial approach seems similar. Sub:288232403

      thanks

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        13 days ago, # ^ |
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        Hey! Sorry about a late reply, I hope it's still relevant.

        The idea seems correct. I believe your code does not calculate the child[i] correctly. It would be easier to use a DFS here.

        parent[i] seems to be okay, but, since in the last loop you iterate over nodes from n to 1 (for(int i=n;i>=1;i--){...}), it is possible that, when performing child[parent[i]]+=child[i], child[i] is still not fully computed (Meaning that there exist j such that j < i, parent[j] = i).

        I hope this makes sense. Feel free to ask if you have more questions.

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17 months ago, # |
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For problem E, why don’t you need a prefix sum for every possible array. I understand the binary search part but just building the prefix sums would tle right?

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    17 months ago, # ^ |
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    Analyzing the complexity : You binary search on q ( O(log(q)) ) and to check if a certain value is valid, you build the prefix sum array and then iterate through segments ( O(m + n (+q depending on impl) ) ). Combining these, you get O(log(q)(m+n)). If you were to build all prefix sum arrays before hand, that would take you O(q*n) time ( TLE ), because for each query, you would have to build prefix-suns. A way around this is to use a segment tree to achieve O(qlog(n)) for precomputation!

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      17 months ago, # ^ |
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      Thank you for your explanation. I forgot we could build the prefix sums inside of the binary search :(.

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    17 months ago, # ^ |
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    As we are populating the prefix array while we are doing binary search, the maximum number of times we would have to create the prefix array would be O(log Q) and creating each prefix array would take O(N) time, thus total time complexity will be well under limits reaching only O(N*log Q) for prefix array creation and O(M*log Q) for checking every range given to us, making the total time complexity O((N+M)*log Q).

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      17 months ago, # ^ |
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      Thank you for your explanation. I forgot we could build the prefix sums inside of the binary search :(.

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17 months ago, # |
Rev. 3   Vote: I like it -11 Vote: I do not like it
map<int,vector<pair<int/*second node*/,int/*weight*/>>>graph;
pair<int,int> segments[2][N];
int vis[N];
void dfs(int node,int sum,int mx_sum,int mn_sum){
    vis[node]=1;
    for(auto child:graph[node]) {
        if (!vis[child.first]) {
            if (child.second == 1) {
                int val = sum >= 1 ? sum : 0;
                segments[1][child.first].second = mn_sum;
                segments[1][child.first].first = max(mx_sum, val + 1);
                dfs(child.first, val + 1, max(mx_sum, val + 1), mn_sum);
            } else {
                int val = sum <= -1 ? sum : 0;
                segments[1][child.first].second = min(val - 1, mn_sum);
                segments[1][child.first].first = mx_sum;
                dfs(child.first, val - 1, mx_sum, min(mn_sum, val - 1));
            }
        }
    }
}
void solve() {
    int cnt=1;
    int n,size;cin>>n;
    vector<pair<int,int>>queries;
    graph.clear();
    for(int i=0;i<n;++i){
        char t;
        cin>>t;
        if(t=='?') {
            int a, b, val;
            cin>>a>>b>>val;
            queries.emplace_back(b,val);
        }
        else{
            int node,weight;cin>>node>>weight;
            graph[node].emplace_back(cnt+1,weight);
            graph[++cnt].emplace_back(node,weight);
        }
    }
    for(int i=1;i<=cnt;++i)vis[i]=0;
    for(int i=1;i<=cnt;++i)
        segments[1][i]={1,0};
    dfs(1,1,1,0);
    size=queries.size();
    for(int i=0;i<size;++i){
        int b=queries[i].first,weight=queries[i].second;
        cout<<(weight>=segments[1][b].second&&weight<=segments[1][b].first?"YES\n":"NO\n");
    }
}

can someone provide me any hints to detect the error in my code in problem F1?

i fall on test 3 on line 11301st expected yes found no

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    17 months ago, # ^ |
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    same error bro

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      17 months ago, # ^ |
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      + 1 1
      + 2 -1
      + 3 1
      + 2 -1
      + 4 1 
      + 2 -1
      

      and the query is

      ? 1 6 3

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        17 months ago, # ^ |
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        should be no only for this

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          17 months ago, # ^ |
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          actually it's yes because there is a subsegment from 1 to 6 that actually add up to 3

          node 1 ==> 1 node 2 ==> 1 node 3 ==> -1 node 4 ==> 1 node 6 ==> 1

          this adds up to 3

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            17 months ago, # ^ |
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            yes bro spot on. Actually i counted max positive sum on a path as only th length of consecutive ones and set it to 0 again when -1 came in between while this is nt crct as evident feom this tc.such a fool i am

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              17 months ago, # ^ |
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              Ayo bro i did the same XD , i totally forget that not only the consecutive needs to be added Actually this problem is just literally find the maximum and minimum sum of subsegments in array At least we learnt new something today Hope for cyan next div me and you <3

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17 months ago, # |
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I don't know how the tree given in the example in question D is drawn, can someone explain it to me, thank you very much.

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    17 months ago, # ^ |
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    This is the first testcase:

    5 1 2 3 4 5 3 3 2 4 3 4 5 1 4 4 1 3

    The first number is n, so there are n — 1 edges with follow. Let us start with the first edge. "1 2" means an edge is drawn between vertices 1 and 2 (depicted by a line connected 1 and 2). We do the same for edges "3 4", "5 3", and "3 2".

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17 months ago, # |
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My solution for F2 https://youtu.be/FZJxWhOoxo0

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17 months ago, # |
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E can be solved with wavelet tree:

create an array, initially all INF. then for the ith update, set array[index] = i

now for each segment, the (segment len/2 + 1)th-smallest number is either INF, or the id of the query which makes it beautiful

210572758

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    17 months ago, # ^ |
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    Any solution possible with — segment trees concept only ?

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      17 months ago, # ^ |
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      It boils down to finding k-th smallest element in a range, there's a lot of ways of doing it: wavelet tree, persistent segment tree, parallel binary search...

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    17 months ago, # ^ |
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    I solved E during the contest using wavelet tree too, I'm looking for the (len/2 + 1 + (number of zeros))th-smallest in each segment

    my submission: 210413964

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    10 months ago, # ^ |
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    I guess wavlet tree is better than the Editorial solution

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17 months ago, # |
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AnyOne tried to use — segment trees ? for problem E

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17 months ago, # |
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In F1 and F2 the realize that you only need to find maximum and minimum sum can create make everything much easier.

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17 months ago, # |
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I am struggling hard to figure out what's wrong with my code for problem D.210612596 Unlike other solutions, I have only connected nodes u and v from u to v such that u < v. I am guessing that's the bug but I am not sure exactly how.

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    17 months ago, # ^ |
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    What's your logic to add edge u->v only if u < v ??

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      17 months ago, # ^ |
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      From what I thought after reading the testcases is that when an edge v1, v2 is given and v1 < v2, then that means v1 is the parent node of v2, otherwise v2 is the parent node. So I made a directed adjacency list such that adj[v] contains only the nodes which are it's children. Using this, finding the leafnodes is simple as adj[v] will be empty if v is leafnode.

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        17 months ago, # ^ |
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        "From what I thought after reading the testcases is that when an edge v1, v2 is given and v1 < v2, then that means v1 is the parent node of v2, otherwise v2 is the parent node."

        That's the problem: You assumed something that wasn't guranteed in the statement. The assumption that smaller vertices are parents of larger ones is incorrect. You need to treat the given tree as a graph of undirected edges and find the parent nodes yourself (if you need them).

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17 months ago, # |
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You can also solve F2 with HLD for the same reason, if you could not figure out (like me) how to do it with binlifts. Code: https://codeforces.me/contest/1843/submission/210677217

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    17 months ago, # ^ |
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    I'm trying the HLD approach but I'm keeping it WA at test 3, here is my submission: 210725677. Could you help me to point out where am I missing, thanks.

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    14 months ago, # ^ |
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    Hi I am curious how did you avoid stack overflow when running dfs? I tried HLD approach but still stuck at testcase 3 with Runtime error due to stack overflow

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17 months ago, # |
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Can someone help me debug my solution for F2? https://codeforces.me/contest/1843/submission/211222013

WA on the third test (180th token)

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16 months ago, # |
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211316049 Can anyone tell me why this code results in Exit code is-1073741571.thanks.

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16 months ago, # |
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An even shorter solution for C: 2*x - popcount(x)

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    16 months ago, # ^ |
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    Is this formula derived purely from observation or is there some intuition to it?

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      16 months ago, # ^ |
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      Intuition is what I used to find this solution, but it can be easily proven using basic number theory.

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        16 months ago, # ^ |
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        Can you give some hints about the intuition behind it?

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          16 months ago, # ^ |
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          Consider the contribution of each bit to the answer $$$\sum \lfloor \frac{n}{2^k} \rfloor$$$. If the $$$j$$$-th bit is set in $$$n$$$, the $$$(j-1)$$$-th bit will be set in $$$\lfloor \frac{n}{2} \rfloor$$$, the $$$(j-2)$$$-th bit will be set in $$$\lfloor \frac{n}{4} \rfloor$$$ and so on until $$$(j-j)=0$$$-th bit is set. We also know that $$$\sum_{0 \le k \le j} 2^k = 2^{j+1}-1$$$.

          Putting everything together, you get $$$2 \cdot n - (\text{no. of set bits in } n)$$$.

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16 months ago, # |
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Can anyone pls tell me why in sol for F2 the parent of LCA is also merged??

Thanks in Advanced.

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16 months ago, # |
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Can anyone please tell why this solution gives TLE on test case 5

https://codeforces.me/contest/1843/submission/213606130

Thanks

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16 months ago, # |
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why my code works for first 4 cases but then gives different result, even though the logic is correct. code

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    16 months ago, # ^ |
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    Instead of int try using long long, because answer may be greater than the size of int (2^31 = 2147483648) 214366788

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16 months ago, # |
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I solved E using K-th order statistics on segment. Let's sort queries by X. For each segment (let it be [l, r]) we use binary search to find all queries, laying on it. Easy to understand these queries form a segment (let this segment be [l1, r1]). Suppose K = (r-l+1) / 2 + 1 (it's the number of 1 for segment to become beautiful). Then we are to find on segment [l1, r1] the number of query, which would be on position K, if we sorted these queries by their numbers.

P.S. Yeah, you can tell me I'm sick)

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10 months ago, # |
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when will Div3 change by not givving horrible implementation in last question They just dump it inot div3

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6 months ago, # |
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i am stupid, i used persistence segment tree to solve E and solution of tutorial just