首页 技术 正文
技术 2022年11月8日
0 收藏 543 点赞 2,097 浏览 2359 个字

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 ≤ nui ≠ 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 Journey CodeForces – 839C.

Example

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.

树形DP

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<memory>
#include<bitset>
#include<string>
#include<functional>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define INF 0x3f3f3f3f
#define MAXN 100009vector<int> E[MAXN];
bool vis[MAXN];
int n;
double dp(int k,int len)
{
vis[k] = true;
double rate = 1.0, ans = ;
int cnt = ;
for (int i = ; i < E[k].size(); i++)
{
if (!vis[E[k][i]])
cnt++;
}
if (cnt == )
return len;
rate /= cnt;
for (int i = ; i < E[k].size(); i++)
{
if (vis[E[k][i]]) continue;
ans += dp(E[k][i], len + )*rate;
}
return ans;
}
int main()
{
scanf("%d", &n);
int f, t;
for (int i = ; i < n - ; i++)
{
scanf("%d%d", &f, &t);
E[f].push_back(t);
E[t].push_back(f);
}
printf("%.10lf\n", dp(, ));
}
相关推荐
python开发_常用的python模块及安装方法
adodb:我们领导推荐的数据库连接组件bsddb3:BerkeleyDB的连接组件Cheetah-1.0:我比较喜欢这个版本的cheeta…
日期:2022-11-24 点赞:878 阅读:9,492
Educational Codeforces Round 11 C. Hard Process 二分
C. Hard Process题目连接:http://www.codeforces.com/contest/660/problem/CDes…
日期:2022-11-24 点赞:807 阅读:5,907
下载Ubuntn 17.04 内核源代码
zengkefu@server1:/usr/src$ uname -aLinux server1 4.10.0-19-generic #21…
日期:2022-11-24 点赞:569 阅读:6,740
可用Active Desktop Calendar V7.86 注册码序列号
可用Active Desktop Calendar V7.86 注册码序列号Name: www.greendown.cn Code: &nb…
日期:2022-11-24 点赞:733 阅读:6,494
Android调用系统相机、自定义相机、处理大图片
Android调用系统相机和自定义相机实例本博文主要是介绍了android上使用相机进行拍照并显示的两种方式,并且由于涉及到要把拍到的照片显…
日期:2022-11-24 点赞:512 阅读:8,132
Struts的使用
一、Struts2的获取  Struts的官方网站为:http://struts.apache.org/  下载完Struts2的jar包,…
日期:2022-11-24 点赞:671 阅读:5,295