Abandoned country

A Abandoned country
最小生成树 + 树上任意点对距离之和

树上任意点对距离之和：
计算时考虑每一个边，它的贡献值为它的左端点以左的点的个数 * 右端点以右的点的个数 * 该边的权值。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long ll;
typedef pair<int,ll> PIL;
const int N =1e5+10;
const int NN = 1e6+10;
int father[N];
int n,m,cntt;
int vis[N];
ll tot;
double di;
struct Edge
{int a, b;ll v;bool operator< (const Edge &W) const{return v < W.v;}
}edge[NN];vector< pair<int,ll> >tedges[N];int find(int x)
{if (father[x] != x) father[x] = find(father[x]);return father[x];
}ll Kruskal()
{ll res =0;for (int i = 1; i <= n; i ++ ) father[i] = i;sort(edge, edge + m);for (int i = 0; i < m; i ++ ){int a = edge[i].a, b = edge[i].b;ll v =edge[i].v;if (find(a) != find(b)){tedges[a].push_back({b,v});tedges[b].push_back({a,v});res += v;father[find(a)] = find(b);}}return res;
}
ll dfs(int cur)
{vis[cur]=1;ll now = 0 ,ans = 1;for(int i=0;i<tedges[cur].size();i++){if(!vis[tedges[cur][i].first]){now = dfs(tedges[cur][i].first);ans += now;tot += tedges[cur][i].second * (n-now) * now;   //tot保存边的贡献和//下方的点数可由dfs求得，其余的点为(n-下方的点)}}return ans;
}
int main()
{int T;scanf("%d",&T);while(T--){tot=0;cntt = 0;memset(vis,0,sizeof(vis));scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)tedges[i].clear();//tedges[0] = {1,0};di = (double)n*(n-1)/2;for(int i=0;i<m;i++){scanf("%d%d%lld",&edge[i].a,&edge[i].b,&edge[i].v);}ll ans = Kruskal();dfs(1);printf("%lld %.2f\n",ans,(double)tot/di);}return 0;
}

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