#805. 【模板】最小瓶颈生成树（数据加强版）

考虑最小瓶颈生成树的性质

因此我们可以直接在最小生成树上求 \(LCA\) 以及 路径上的边的最大值

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

template <typename T> void chkmax(T &x, T y) { x = x >= y ? x : y; }
template <typename T> void chkmin(T &x, T y) { x = x <= y ? x : y; }

const int N = 100001, M = 1000001;

struct Edg {
  int u, v, w;
  bool operator <(const Edg &T) const {
    return w < T.w;
  }
}edge[M];

int n, m, Q;
int h[N], e[N * 2], ne[N * 2], w[N * 2], idx;
int p[N];
int q[N], dep[N], fa[N][20], mx[N][20];

int find(int x) {
  if(p[x] != x) p[x] = find(p[x]);
  return p[x];
}

void add(int a, int b, int c) {
  e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}

void bfs(int root) {
  memset(dep, 0x3f, sizeof dep);
  dep[0] = 0; dep[root] = 1;
  int hh = 0, tt = 0; q[0] = root;
  while(hh <= tt) {
    int t = q[hh ++];
    for(int i = h[t]; ~i; i = ne[i]) {
      int j = e[i];
      if(dep[j] <= dep[t] + 1) continue;
      dep[j] = dep[t] + 1;
      fa[j][0] = t;
      mx[j][0] = w[i];
      q[++ tt] = j;
      for(int k = 1; k < 20; k ++ ) {
        fa[j][k] = fa[fa[j][k - 1]][k - 1];
        mx[j][k] = max(mx[j][k - 1], mx[fa[j][k - 1]][k - 1]);
      }
    }
  }
}

int lca(int a, int b) {
  if(dep[a] < dep[b]) swap(a, b);
  for(int k = 19; k >= 0; k -- ) {
    if(dep[fa[a][k]] >= dep[b]) {
      a = fa[a][k];
    }
  }
  if(a == b) return a;
  for(int k = 19; k >= 0; k -- ) {
    if(fa[a][k] != fa[b][k]) {
      a = fa[a][k];
      b = fa[b][k];
    }
  }
  return fa[a][0];
}

int query (int u, int v) {
  int res = -1e9;
  for(int i = 19; i >= 0; i -- ) {
    if(dep[fa[u][i]] >= dep[v]) {
      res = max(res, mx[u][i]);
      u = fa[u][i];
    }
  }
  return res;
}

void solve() {

  scanf("%d%d", &n, &m);

  for(int i = 0; i <= n; i ++ ) h[i] = -1, p[i] = i;
  
  for(int i = 0; i < m; i ++ ) {
    int u, v, w; scanf("%d%d%d", &u, &v, &w);
    edge[i] = {u, v, w};
  }

  sort(edge, edge + m);

  for(int i = 0; i < m; i ++ ) {
    int u = edge[i].u, v = edge[i].v, w = edge[i].w; 
    int fu = find(u), fv = find(v);
    if(fu == fv) continue;
    add(u, v, w); add(v, u, w);
    p[fu] = fv; 
  }

  bfs(1);

  scanf("%d", &Q);

  while(Q -- ) {
    int u, v; scanf("%d%d", &u, &v);
    int p = lca(u, v);
    printf("%d\n", max(query(u, p), query(v, p)));
  }

}

int main() {
  solve();
  return 0;
}

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来源： https://www.cnblogs.com/c972937/p/16196599.html