题目链接：点击打开链接
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.
题目意思：给你一个共有n个点无向图，其中有n-1条路，根节点为1，每条边 长为1，从根节点1出发，每到一个结点，等概率的往其子树走，到叶子则终止，同一种状态下走过的节点不再走，求走过的路径长度的期望。
思路：深搜
AC代码：
#pragma GCC diagnostic error "-std=c++11"
#include<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<string>
#include<map>
#include<climits>
#include<vector>
#include<queue>
#include<set>
#include<cstring>
using namespace std;
#define rep(i,a,b) for(int i=a;i<b;i++)
#define CLR(a,b) memset(a,b,sizeof(a))
#define maxn 1000010
const int MOD = 1000000007;
typedef long long ll;
double sum=0.0;
vector<int>v[100010];
//now为当前节点
//fa为当前节点的父节点
//len为当前路径的长度
//pr为当前路径的概率
void dfs(int now,int fa,int len,double pr)
{
if(now!=1&&v[now].size()==1)//当走到路径末端时
{
sum+=(pr*len);
return;
}
double pr1=pr;
if(now==1)
pr1*=1.0/(double)v[now].size();
else pr1*=1.0/((double)v[now].size()-1);
for(auto i:v[now])
if(i!=fa)
dfs(i,now,len+1,pr1);
}
int main()
{
//ios::sync_with_stdio(false);
//cin.tie(0);
double ans;
int n;
cin>>n;
n--;
while(n--)
{
int a,b;
cin>>a>>b;
v[a].push_back(b);
v[b].push_back(a);
}
dfs(1,-1,0,1.0);
printf("%.15lf\n",sum);
return 0;
}