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C. Journey【dfs + 期望】

2017-08-13 16:28 246 查看
C. Journey

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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.

题意:

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:

由于走过的节点不能再次走,一开始用了vis数组标记(以为有环,结果没有),这样应该是比较麻烦的,由于无环只需要考虑不经过其父亲节点就行,然后深搜子树
4000
长度并记录各个节点的子树个数就可以了。

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#define max_n 100010
using namespace std;
typedef long long LL;
vector<int> v[max_n];

double dfs(int x, int father) {
int p = 0;
double res = 0;
for(int i = 0; i < v[x].size(); i++) {
if(v[x][i] != father) p++;
}
if(p == 0) return 0.0;
for(int i = 0; i < v[x].size(); i++) {
if(v[x][i] == father) continue;
res += dfs(v[x][i], x) + 1.0; //每次走一步,长度加1
}
return res / p; //返回期望路径长度
}

int main() {
int n, p1, p2;
scanf("%d", &n);
for(int i = 0; i < n - 1; i++) {
scanf("%d %d", &p1, &p2);
v[p1].push_back(p2);
v[p2].push_back(p1);
}
double ans = dfs(1, 1);
printf("%.15lf\n", ans);
return 0;
}
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