题目大意：

有n个点，有m条边，让你求它的最小生成树，并且在这个生成树上，求任意两点间的距离的期望。

解题思路：

第一问比较裸，就直接求最小生成树就好，第二问需要一些技巧。

现在有一条边ab，其中b的子树个数为num，那么这条边被经过的次数为(n - num) * num

然后乘一下这条边的权值就可以了

代码：

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;const int maxn = 1e5 + 5;typedef long long LL;
typedef struct node{int from, to;int w;bool operator < (node zz){return w < zz.w;}
}Edge;
typedef struct nn{int to;int w;nn(){}nn(LL a, LL b){to = a; w = b;}
}EEE;LL ans;
double res;
int t, n, m;
Edge g[maxn * 10];
vector<EEE> e[maxn];
int vis[maxn], pre[maxn];int findfather(int x){return pre[x] = (pre[x] == x ? x : findfather(pre[x]));
}
void join(int x, int y){pre[findfather(x)] = findfather(y);
}
void kruskal(){int x, y, flag = 0;for(int i = 0; i < m; ++i){x = findfather(g[i].from);y = findfather(g[i].to);if(x == y) continue;join(x, y);++flag;ans += g[i].w;e[g[i].from].push_back(nn(g[i].to, g[i].w));e[g[i].to].push_back(nn(g[i].from, g[i].w));if(flag == n - 1) return;}
}
int dfs(int p){if(vis[p]) return 0;vis[p] = 1;int tt = 1, num;for(int i = 0; i < e[p].size(); ++i){int v = e[p][i].to;num = dfs(v);tt += num;res += (LL)(n - num) * (LL)num * (LL)e[p][i].w;}return tt;
}
int main(){scanf("%d", &t);while(t--){scanf("%d%d", &n, &m);for(int i = 1; i <= n; ++i) {pre[i] = i;vis[i] = 0;e[i].clear();}for(int i = 0; i < m; ++i){scanf("%d%d%d", &g[i].from, &g[i].to, &g[i].w);}sort(g, g + m);ans = 0;kruskal();res = 0;dfs(1);printf("%lld %.2lf\n", ans, res / ((n - 1LL) * n / 2.0));}return 0;
}