codeforces839C Journey（DFS）

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#dfs+概率问题

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本文介绍了一个基于概率的旅行期望值计算问题，通过DFS算法解决从指定起点出发到达各个城市的平均路径长度问题。考虑了不同路径概率对结果的影响。

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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 $\text{[math]}$ .

   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.

题意：大概就是，给出一棵树，从节点1出发，问到达根节点的所有长度的平均值（注意概率问题）

比赛的时候一直wa，就是因为没有考虑概率的问题，总以为走到每个根节点的概率都是相等的，结果wa到怀疑人生，比赛结束之后，才想到概率这个事。。。想清楚就很简单啦，就是一个DFS

#include<stdio.h>
#include<iostream>
#include<set>
#include<queue>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<string>
#include<vector>
using namespace std;
const int inf = 0x3f3f3f3f ;
typedef long long ll ;
const int maxn = 1e5 + 5 ;

int n,fa[maxn];
vector<int>G[maxn];
double sum ;
void dfs(int x , int d , double p){
    if (G[x].size()==1&&x!=1) {///到达根节点时进行计算
        sum += (d-1)*p;
        return;
    }
    for (int i=0;i<G[x].size();i++){
        int son=G[x][i];
        if (son==fa[x]) continue;
        fa[son]=x;
        if(x == 1)///走到所有子节点的概率都是相等的
            dfs(son , d + 1 , p/G[x].size()) ;///注意根节点和其他节点分开判断
        else
            dfs(son , d + 1 , p/(G[x].size()- 1) ) ;
    }
    return;
}
int main() {
    //freopen ("in.txt", "r", stdin);
    while (~scanf ("%d",&n)){
        for (int i=0;i<=n;i++)
            G[i].clear();
        sum=0;
        memset(fa,0,sizeof(fa));
        int u,v;
        n--;
        while (n--){
            scanf ("%d%d",&u,&v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        dfs(1 , 1 , 1.0);///刚开始概率为1
        printf ("%.15lf\n" , sum);
    }
    return 0;
}