There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they can’t see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link https://en.wikipedia.org/wiki/Expected_value.
The first line contains a single integer n (1 ≤ n ≤ 100000) — number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Print a number — the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if 
.
4 1 2 1 3 2 4
1.500000000000000
5 1 2 1 3 3 4 2 5
2.000000000000000
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.
题意：以1为起点，每次走任意一条路，而且走每条路的概率都随机而且每次走时只会走向未走过的城市。同时，每条路的长度均为1。问：行走路程的期望值是多少？
思路：需要注意的是到达了某一个点时，走每一条可走的路的概率相同而不是最终停在每个终点的概率相同。直接遍历整个图，同时做出相应的计算即可。
AC代码：
#include<iostream>
#include<cstring>
#include<vector>
#define MAX 100005
using namespace std;
bool vis[MAX];
vector<int> G[MAX];
void DFS(int now,double & ans,double val)
{
vis[now] = true;
int size_now = G[now].size();
if(size_now > 1 || (now == 1 && size_now > 0)) ans += val; //需要注意当前点为1的情况。
else return ;
for(int j = 0; j < size_now; j++){
int next = G[now][j];
if(!vis[next]){
if(now == 1)
DFS(next,ans,val/size_now);
else
DFS(next,ans,val/(size_now - 1));
}
}
return ;
}
int main( )
{
int n;
cin >> n;
memset(vis,false,sizeof(vis));
int u,v;
for(int j = 1; j < n; j++)
cin >> u >> v,G[u].push_back(v),G[v].push_back(u);
double ans;
DFS(1,ans,1);
printf("%.10lf\n",ans);
}