題目地址：點擊打開鏈接
題目大意：有n個城市要修路，最後希望它們互相直接或間接聯通，給了m條備選的路，問最小費用是多少還有就是路長的期望是多少。
解題思路：這題其實和省賽中的挺像，本質就是求最小生成樹，對於後面的期望，每一條邊的期望等於邊兩端的點的數量的積乘上邊的長度，sigma求和在除以從n個點中取2個點建邊的情況即可。
隊友的圖論水平是越來越高了，直接排序的最小生成樹+並查集（果然我還是太弱了），隨便令一個作爲頂點，在將點數延邊傳遞。
代碼：
#include<iostream>
#include<vector>
#include<cmath>
#include<algorithm>
#include<ctime>
#include "cstdio"
#include "string"
#include "string.h"
using namespace std;
const int maxn = 1e5 + 50;
struct Edge {
int from, to;
long long cost;
bool operator<(const Edge&rhs)const {
return cost < rhs.cost;
}
}edges[maxn*10],cs[maxn*10];
vector<int>G[maxn];
int n, m, cnt, fa[maxn], node[maxn];
void init()
{
for (int i = 1;i <= n;i++)
{
G[i].clear();
fa[i] = i;
}
}
int find(int x)
{
return x == fa[x] ? x : fa[x] = find(fa[x]);
}
int dfs(int now, int parent)
{
int siz = G[now].size();
int ans = 1;
for (int i = 0;i < siz;i++)
{
int next = G[now][i];
if (next != parent)
ans += dfs(next, now);
}
node[now] = ans;
return ans;
}
int main()
{
int T;
scanf("%d",&T);
while (T--)
{
scanf("%d %d", &n, &m);
init();
for (int i = 0;i < m;i++)
scanf("%d %d %d", &edges[i].from, &edges[i].to, &edges[i].cost);
sort(edges, edges + m);
long long sum = 0;
cnt = 0;
memset(node, 0, sizeof(node));
for (int i = 0;i < m;i++)
{
int x = edges[i].from;
int y = edges[i].to;
int dx = find(x);
int dy = find(y);
if (dx != dy)
{
fa[dx] = dy;
sum += edges[i].cost;
G[x].push_back(y);
G[y].push_back(x);
cs[cnt++] = edges[i];
}
}
dfs(1, -1);
double exp = 0;
for (int i = 0;i < cnt;i++)
{
int x = cs[i].from;
int y = cs[i].to;
int dMin = min(node[x], node[y]);
int dMax = n - dMin;
exp += (long long)dMin*dMax*cs[i].cost;
}
exp = exp / n / (n - 1) * 2;
printf("%lld %.2lf\n", sum, exp);
}
return 0;
}