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

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

4
1 2
1 3
2 4

output

1.500000000000000

input

5
1 2
1 3
3 4
2 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.

題意： 给你一棵树（注意读题，无向图且无环），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：本来想的直接将无向树转化成有向树，然后找到初度为零的点就是叶子，然后记录从1到叶子的分叉，都计算一遍累加起来就是结果，可是莫名其妙的wa在样例四，很无奈，看了题解，很萌，最痛苦的是不知道自己哪里错了。

附上正确解的博客：

点击打开链接

还有就是可能错误的原因：

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.

另附上正确代码：

#include<cstdio>#include<cstring>#include<vector>#include<iostream> using namespace std;vector<int> op[100010];double dfs(int x,int f){double k = 0;double tmp = 0;int flag = 1;for(int i=0; i<op[x].size(); i++){int pp = op[x][i];if(pp==f)continue;flag = 0;k++;tmp += dfs(pp,x);}if(flag) return 0.0;    else return 1.0 + tmp/k;}int main(){int n;    int a,b;    cin >> n;    for(int i=1; i<n; i++) {cin >> a >> b;op[a].push_back(b);op[b].push_back(a);}printf("%.15lf\n",dfs(1,0));  return 0;} 

水波.