Given a tree, calculate the average distance between two vertices in the tree. For example, the average distance between two vertices in the following tree is (d 01 + d 02 + d 03 + d 04 + d 12 +d 13 +d 14 +d 23 +d 24 +d 34)/10 = (6+3+7+9+9+13+15+10+12+2)/10 = 8.6. 

 

Input

On the first line an integer t (1 <= t <= 100): the number of test cases. Then for each test case: 

One line with an integer n (2 <= n <= 10 000): the number of nodes in the tree. The nodes are numbered from 0 to n - 1. 

n - 1 lines, each with three integers a (0 <= a < n), b (0 <= b < n) and d (1 <= d <= 1 000). There is an edge between the nodes with numbers a and b of length d. The resulting graph will be a tree. 
 

Output

For each testcase: 

One line with the average distance between two vertices. This value should have either an absolute or a relative error of at most 10 -6 
 

Sample Input

1
5
0 1 6
0 2 3
0 3 7
3 4 2

Sample Output

8.6

题意：就是求上面那个式子的值，看式子和图就能懂。

题解：用树形dp跑，把每条边用的次数算出来，然后乘这条边的值就好了，用过的次数=该边的尾节点后边所连点的个数（包括该点）*该边的尾节点前面所连点的个数（不包括该点），然后用次数乘价值，最后把每条边所求加和，就可以了。

注意：不能用cin，cout 输入输出流，要用标准输入，输出，要不然会T（超时）。  上代码：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int MAX = 1e5;
struct hh{
	int u,v;
	double w;
	int nt;
}a[MAX];//用来保存树
int head[MAX];
double num[MAX];
int cnt,n;
double ar;
void add(int u,int v,double w){//建树，记住就好了
	a[cnt].u=u;
	a[cnt].v=v;
	a[cnt].w=w;
	a[cnt].nt=head[u];
	head[u]=cnt++;
}
void dfs(int foot,int fz){
	num[foot]=1;
	for (int i = head[foot]; ~i; i = a[i].nt){
		int vv=a[i].v;
		int ww=a[i].w;
		if(vv!=fz){
			dfs(vv,foot);
			num[foot]+=num[vv];//统计用的次数
			ar+=num[vv]*(n-num[vv])*ww;//后边*前边*价值，然后相加
		}
	}
} 
int main(){
	int t;
	scanf("%d", &t);
	while(t--){
		memset(head,-1,sizeof(head));//存为-1，建图常用
		memset(num,0,sizeof(num));//初始化
		cnt=0;
		scanf("%d",&n);
		for (int i = 0; i < n-1;i++){
			int u,v;
			double w;
			scanf("%d%d%lf",&u,&v,&w);
			add(u,v,w);//建立双向 ，看题目里的图就知道
			add(v,u,w);
		}
		ar=0;
		dfs(0,-1);
		printf("%.7f\n",ar/(1.0*n*(n-1)/2.0));//输出注意精度
	}	
	return 0;
}