【2018 CCPC网络赛】1009

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=6446

题目给出的数据为一棵树，dfs扫描每条边，假设去掉某条边，则左边 x 个点，右边 n-x 个点，则经过该条边共有 x*(n-x) 种组合，又因为 1~n 全排列有 n! 种，故 a~b，包含 b~a 这条边的贡献为 (n-x)*x*2*(n-1)!*w;

#include<iostream>
#include<cstdio>
#include<vector>
using namespace std;

#define LL long long
const int MOD = 1e9+7;
const int N = 1e5+5;
struct EDGE {
    LL to, length;
};
vector<EDGE> edge[N];
LL n, factorial[N], lchild[N], w[N];

int dfs(int k, int pre) {
    int size = edge[k].size(), ans=1;
    for(int i=0; i<size; i++) {
        if(edge[k][i].to == pre)
            w[k] = edge[k][i].length;
        else
            ans += dfs(edge[k][i].to, k);
    }
    return lchild[k] = ans;
}

int main()
{
    factorial[1] = 1;
    for(int i=2; i<N; i++)
        factorial[i] = factorial[i-1]*i%MOD;
    
    while(~scanf("%d",&n))
    {
        LL  ans=0;
        for(int i=1; i<=n; i++) {
            edge[i].clear();
        }
        for(int x,y,v,i=1; i<n; i++) {
            scanf("%d%d%d", &x, &y, &v);
            edge[x].push_back({y,v});
            edge[y].push_back({x,v});
        }
        
        dfs(1, -1);
        for(int i=2; i<=n; i++) {
            LL tmp=1;
            tmp = lchild[i]*(n-lchild[i])%MOD;
            tmp = 2*factorial[n-1]%MOD*w[i]%MOD*tmp%MOD;
            ans = (ans+tmp)%MOD;
        }
        cout<<ans<<endl;
    }
    return 0;
}