HDU 5723/2016多校1A Abandoned country

题意：给你n个点m条边，求最小生成树，和任意两点距离的期望

题解：最小生成树的裸题，生成树有了以后期望再dfs跑一下就可以了。

注意：这题爆ll的地方很多

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<list>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<deque>
#define mem(x,y) memset(x,y,sizeof(x))
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,double> pii;
#define bug puts("===========");
#define zjc puts("");
const double pi=(acos(-1.0));
const double eps=1e-8;
const ll INF=1e18+10;
const ll inf=1e9+10;
const int mod=1e9+7;
const int maxn=100000+10;
const int maxm=1e6+5;

struct node{
int x, y, v;
bool operator<(const node &b)const{
return v<b.v;
}
} e[maxm];
int f[maxn],num[maxn];
int find(int x){ return f[x]= f[x]==x?x:find(f[x]); }
ll ans2;
int n,m;
vector<pii>vec[maxn];
ll dfs(int x,int fa){
ll sum=1;
for(int i=0;i<vec[x].size();i++){
int next=vec[x][i].first;
if(next==fa) continue;
ll z=dfs(next,x);
ans2 += z*(n-z)*vec[x][i].second;
sum += z;
}
return sum;
}
ll Kruscal(){
sort(e, e + m);
int cnt = n;
ll ans = 0;
for(int i = 0; i <= n; i++) {
vec[i].clear();
f[i] = i;
num[i] = 1;
}
for (int i = 0; i < m; i++){
int x = find(e[i].x);
int y = find(e[i].y);
if (x != y){
f[x] = y;
vec[e[i].x].pb((pii){e[i].y,e[i].v});
vec[e[i].y].pb((pii){e[i].x,e[i].v});
cnt--;
ans = ans + e[i].v;
if (cnt == 1) break;
}
}
dfs(1,-1);
return ans;
}

int main()
{
int T_T;
scanf("%d",&T_T);
while(T_T--){
ans2=0;
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++) scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].v);
printf("%I64d ",Kruscal());
double z=ans2;
z/=((ll)n*(n-1))/2;
printf("%.2f\n",z);
}
return 0;
}