一、视频讲解

蓝桥杯真题讲解：网络稳定性（Kruskal重构树+LCA）

二、正解代码

//kruskal重构树 + lca
#include<bits/stdc++.h>
#define endl "\n"
#define deb(x) cout << #x << " = " << x << '\n';
#define INF 0x3f3f3f3f
using namespace std;
const int N = 2e5 + 10;
const int M = 3e5 + 10;
int n, m, q;struct edge{int x, y, w;
}e[M];
vector<int>g[N];
int p[N];
int val[N];//重构树节点的权值。
int find(int x){if(x != p[x])p[x] = find(p[x]);return p[x];
}
void kruskal(){for(int i = 1; i <= n * 2; i ++){p[i] = i;}sort(e, e + m, [&](edge x, edge y){return x. w > y.w;});int id = n;for(int i = 0; i < m; i ++){int x = e[i].x, y = e[i].y, w = e[i].w;int px = find(x), py = find(y);if(py == px)continue;id ++;p[px] = id, p[py] = id;g[id].push_back(py);g[id].push_back(px);val[id] = w;}
}int dep[N], fa[N], siz[N];
int top[N], hs[N];
void dfs1(int u, int f){dep[u] = dep[f] + 1;siz[u] = 1;fa[u] = f;for(auto s: g[u]){if(s == f)continue;dfs1(s, u);siz[u] += siz[s];if(siz[s] > siz[hs[u]])hs[u] = s;}
}void dfs2(int u, int t) {top[u] = t;if(!hs[u])return;dfs2(hs[u], t);for(auto s: g[u]){if(s == fa[u] || s == hs[u])continue;dfs2(s, s);}
}int lca(int u, int v){while(top[u] != top[v]){if(dep[top[u]] < dep[top[v]])swap(u, v);u = fa[top[u]];}return dep[u] < dep[v] ? u : v;
}void solve()
{cin >> n >> m >> q;for(int i = 0; i < m; i ++){cin >> e[i].x >> e[i].y >> e[i].w;}kruskal();for(int i = 1; i <= 2 * n; i ++){if(p[i] == i){//该连通块中lca的预处理dfs1(i, 0);dfs2(i, i);}}while(q --){int a, b; cin >> a >> b;if(find(a) != find(b)){cout << -1 << endl;continue;}cout << val[lca(a, b)] << endl;}}signed main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t = 1;//cin >> t;while(t--)solve();
}