Introduction

It's been around 18 months since this problem first crossed my way and yesterday I've finally understood the author's intended solution (thanks to thiagocarvp's patient explanation!). I think this solution is somehow explained here, but I also think it can be explained in a better way. This is the goal of this post.

Problem statement

First, you're given a connected graph with n vertices and m weighted edges. And then a sequence of q new edges is added to the graph. For each of these q new edges, output the weight of a minimum spanning tree considering only this and the previous edges. For example, take V = {1, 2}, E = {({1, 2}, 5)} and the sequence (({1, 2}, 7), ({1, 2}, 3)), i.e., n = 2, m = 1 and q = 2. The answers are 5 and 3, respectively.

Naive approach

Let's try to answer the queries online. First, build a MST for the initial graph. All we can do with new edges is try to improve the current MST, i.e., the MST can only become lighter, never heavier. It's not hard to see that the required procedure is the following greedy algorithm.

There are only two possibilities for a new edge ({u, v}, w):

1. An edge between u and v is already present in the MST. In this case, just update its weight taking the minimum between the new and the old weight.
2. There's no edge between u and v in the current MST. In this case, the new edge will create a cycle. Then just remove the heaviest edge from this cycle.

The first situation can be easily handled in using maps. The second, however, takes more effort. A simple DFS could find and remove the heaviest edge of the cycle, but it would cost O(n) operations, resulting in a total running time of at least operations in the worst case. Alternatively, it's possible to augment a link cut tree to do all this work in per new edge, resulting in a much better running time.

So the naive approach is either too slow (DFS), or too much code (link cut tree).

Author's intended solution

The naive approach might be hard to apply, but it certainly helps us to make an important observation:

Two consecutive MSTs will differ in at most one edge.

In other words, the changes in the solution are very small from one query to the next. And we are going to take advantage of this property, like many popular offline algorithms do. In particular, we'll do something like the square root decomposition of Mo's algorithm. Usually, this property is achieved by sorting the queries in a special way, like Mo's algorithm itself requires. In our case, we have just noticed that this is not necessary. Hence, we'll process the queries in a very straightforward way (and I keep asking myself what took me so long to understand this beautiful algorithm!).

The observation is used as follows. We'll split the queries in consecutive blocks of consecutive queries. If we compute the edges that simultaneously belong to all the MSTs of one block, we'll be able to reduce the size of the graph for which we should compute minimum spanning trees. In other words, we're going to run Kruskal's algorithm q times, once per new edge, but it will run for much smaller graphs. Let's see the details.

First, imagine the MST T_i computed right after adding the edge e of the i-th query. Now, if e belongs to T_i, consider . What does it look like? Sure, T_i' is a forest with two trees (components). And if we condense these two components, we'll get a much smaller graph with precisely two vertices and no edges. Right now, this much smaller graph do not seems to be useful, but let's see what happens if we consider this situation for not only one, but for a block of new edges.

Now, imagine the MST M_i computed right after adding all the edges of the i-th block B_i. The graph is a minimum spanning forest with at most components, because the removal of an edge increases the number of components in exactly one and we are considering the removal of at most edges. Therefore, a condensation would produce a set S_i of at most vertices. Let's write X to denote the total sum of the weights of the condensed edges (the internal edges of the components).

Compute a MST for the set S_i considering only the edges added before the i-th block. This MST will have at most edges. If use the edges of this MST to initialize and maintain a multiset M of edges, we can insert a new edge in M and run Kruskal's algorithm times, once per query. Over all blocks, we'll run Kruskal's algorithm q times for graphs with at most vertices and edges. For the j-th query, we should output X + Y_j, where Y_j is the total sum of the weights of the edges chosen by Kruskal's algorithm.

In a step-by-step description, the algorithm is as follows:

1. Store the m initial edges in a multiset edges.
2. Compute large, an array with the edges of a MST for the initial m edges (Kruskal's for m edges).
3. For each block [l, r]:
4. 1. Create an empty array initial and swap the contents with large.
5. 1. Insert edges e[i] in the multiset edges, l ≤ i ≤ r.
6. 1. Recompute large for the new state of edges (Kruskal's for O(m + q) edges).
7. 1. Use large to find the forest, condense its components and to find the value of X.
8. 1. Create a multiset M of edges and use initial and the condensed components to fill it with at most edges.
9. 1. For each edge e[i], l ≤ i ≤ r:
10. 1. 1. Insert e[i] in M.
11. 1. 1. Compute Kruskal's minimum weight Y for the multiset M and output X + Y (Kruskal's for edges).

We run Kruskal's algorithm times for a graph with O(m + q) edges and q times for a graph with edges, so the total running time is around , if we have a fast DSU implementation.

Here is my implementation:

#include <bits/stdc++.h>
using namespace std;

// =============================================================================
// BEGIN template
// =============================================================================
#ifdef ONLINE_JUDGE
  #define fastio        std::ios::sync_with_stdio(false)
  #define dbg(X)
  #define _             _
#else
  #include <unistd.h>
  #define fastio
  #define dbg(X)        cerr << ">>> (" << #X << ") = (" << X << ")\n"
  #define _             << " _ " <<
#endif
#define   ff            first
#define   ss            second
#define   pb            push_back
#define   eb            emplace_back
#define   em            emplace
#define   pq            priority_queue
#define   pqmin(X)      priority_queue<X,vector<X>,greater<X>>
#define   all(X)        (X).begin(),(X).end()
#define   sci1(X)       scanf("%d",&(X))
#define   sci2(X,Y)     scanf("%d%d",&(X),&(Y))
#define   sci3(X,Y,Z)   scanf("%d%d%d",&(X),&(Y),&(Z))
#define   scl(X)        scanf("%lld",&(X))
#define   scs(X)        scanf("%s",X)
#define   flush         fflush(stdout)
#define   lg(X)         (63 - __builtin_clzll(X))
#define   LG(X)         (lg(X)+((1<<lg(X)) < (X)))
#define    rp(i,L,R)    for (ll i = L, __R = R; i <= __R; i++)
#define   rpd(i,R,L)    for (ll i = R, __L = L; __L <= i; i--)

// dps
#define DP1(type,X)\
  static type dp[X];\
  static bool mark[X];\
  auto& ans = dp[i];\
  if (mark[i]) return ans;\
  mark[i] = true;
#define DP2(type,X,Y)\
  static type dp[X][Y];\
  static bool mark[X][Y];\
  auto& ans = dp[i][j];\
  if (mark[i][j]) return ans;\
  mark[i][j] = true;
#define DP3(type,X,Y,Z)\
  static type dp[X][Y][Z];\
  static bool mark[X][Y][Z];\
  auto& ans = dp[i][j][k];\
  if (mark[i][j][k]) return ans;\
  mark[i][j][k] = true;

// overloads
#define GET_SCI(_1,_2,_3,NAME,...) NAME
#define sci(...) GET_SCI(__VA_ARGS__,sci3,sci2,sci1)(__VA_ARGS__)
#define GET_DP(_1,_2,_3,_4,NAME,...) NAME
#define DP(...) GET_DP(__VA_ARGS__,DP3,DP2,DP1)(__VA_ARGS__)

// types
typedef   long long     ll;
typedef   pair<int,int> ii;
typedef   vector<int>   vi;
typedef   vector<bool>  vb;
typedef   set<int>      si;

// constants
const ll  oo =          0x3f3f3f3f3f3f3f3fll;
const int LGN =         25;
const int MOD =         1e9+7;
const int N =           3e4+5;
// =============================================================================
// END template
// =============================================================================

struct edge {
  int u,v,w,id;
  edge() : id(0) {}
  bool operator<(const edge& o) const { return w < o.w; }
  void read() { sci(u,v,w); }
};

struct dsu {
  int mark[N], p[N], pass;
  dsu() : pass(1) {}
  void reset() { pass++; }
  int Find(int x) {
    if (mark[x] != pass) {
      mark[x] = pass;
      p[x] = x;
    }
    return p[x] == x ? x : p[x] = Find(p[x]);
  }
  void Union(int x, int y) { p[Find(x)] = Find(y); }
};

int kruskal(const multiset<edge>& edges, vector<edge>* mst = nullptr) {
  static dsu uf;
  uf.reset();
  int ans = 0;
  for (auto& e : edges) if (uf.Find(e.u) != uf.Find(e.v)) {
    uf.Union(e.u,e.v);
    if (mst) mst->pb(e);
    ans += e.w;
  }
  return ans;
}

int main() {
  fastio;
  int t;
  sci(t);
  while (t--) {
    // input
    int n,m,q;
    sci(n,m,q);
    multiset<edge> edges;
    rp(i,1,m) {
      edge e;
      e.read();
      edges.insert(e);
    }
    static edge query[N];
    rp(i,1,q) {
      query[i].read();
      query[i].id = i;
    }
    // initial large mst
    vector<edge> largemst;
    kruskal(edges,&largemst);
    // answer each block
    for (int l = 1, b = sqrt(q)+1; l <= q; l += b) {
      int r = min(l+b-1,q);
      // current large mst is the initial mst for the queries of this block
      vector<edge> initial;
      largemst.swap(initial);
      // compute next large mst
      rp(i,l,r) edges.insert(query[i]);
      kruskal(edges,&largemst);
      // compute forest
      static dsu uf;
      uf.reset();
      int forest = 0;
      for (auto& e : largemst) if (e.id < l) {
        uf.Union(e.u,e.v);
        forest += e.w;
      }
      // compute initial compressed mst
      multiset<edge> eds;
      for (auto& e : initial) if (uf.Find(e.u) != uf.Find(e.v)) {
        auto tmp = e;
        tmp.u = uf.Find(e.u), tmp.v = uf.Find(e.v);
        eds.insert(tmp);
      }
      // answer each query
      rp(i,l,r) {
        auto tmp = query[i];
        tmp.u = uf.Find(tmp.u), tmp.v = uf.Find(tmp.v);
        eds.insert(tmp);
        printf("%d\n",forest+kruskal(eds));
      }
    }
  }
  return 0;
}

Problems

I think that this problem can also be solved with some adaptation of this algorithm. If anyone knows any other problems suitable for this technique, just comment and I'll add them here!