Codeforces Round #428 (Div. 2) C.Journey（dfs求期望）

C. Journey

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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.

Output

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 .

Examples

input

41 21 32 4

output

1.500000000000000

input

51 21 33 42 5

output

2.000000000000000

Note

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.

题解：

题意：

给你一颗树，设每两个节点间距离为1，从1开始，让你求这棵树的路程的期望

思路：

一开始我想用bfs求出每个叶子节点的长度和个数，用长度的和除以个数的做法在第五组就过不了。。。这里应该是涉及一堆数学推导这样求的结果不等同于期望吧，然后看了题解才知道这里的期望要一个个节点递归相加来求解。。具体原理还是不太懂，哎只会图论和数据结构的弱渣

代码：

#include<algorithm>#include<iostream>#include<cstring>#include<stdio.h>#include<math.h>#include<string>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<deque>#define M (t[k].l+t[k].r)/2#define lson k*2#define rson k*2+1#define ll long long#define INF 100861111;using namespace std;vector<int>p[100005];//放邻接表int vis[100005];//遍历数组double dfs(int index)//index表示当前节点下标{    vis[index]=1;    int i;    double res=0,ans=0;    for(i=0;i<p[index].size();i++)    {        if(!vis[p[index][i]])//没有遍历        {            res+=dfs(p[index][i]);            ans++;        }    }    if(ans==0)//当前为叶子节点，由于已经不能走了，路径条数为0，所以是0        return 0;    res=1+res/ans;//不是叶子节点的话用自身的期望1（这里可以稍微画图推一下）加上叶子节点的期望    return res;}int main(){    int i,j,k,n,x,y;    double s=0,ans=0;    memset(vis,0,sizeof(vis));    scanf("%d",&n);    for(i=0;i<n-1;i++)    {        scanf("%d%d",&x,&y);        p[x].push_back(y);//注意是双向        p[y].push_back(x);    }    printf("%.15lf\n",dfs(1));    return 0;}