【CodeForces - 839C Journey】 DFS

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 ≤ nui ≠ vi) — the cities connected by thei-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:在n个城市之间有n-1条路相连,给出n-1组相连的城市,现在这个人从城市1出发,他不会再走走过的城市,所以他会一直走到一个无法再走下去的城市才停止,现在问这个人这次路途的期望路程是多少。

分析:从点1开始dfs,将点1的期望val设为1,遇到每个节点时将期望val变为val/(G[x].size()-1),距离dis+1,直到找到G[x].size() == 1的点,更新ans += val*dis。

代码如下:
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long

using namespace std;
const int MX = 1e5 + 5;
const int mod = 1e9 + 7;
const int INF = 2e9 + 5;

vector G[MX];
double ans;
int vis[MX];

void dfs(int x, double val, int dis){
    double temp = 1.0;
    vis[x] = 1;
    int sz = G[x].size();
    if(x != 1 && sz > 1){
        temp = temp/(sz-1);
    }
    if(x == 1)  temp = temp/sz;
    for(int i = 0; i < sz; i++){
        int v = G[x][i];
        if(!vis[v]){
            dfs(v, val*temp, dis+1);
        }
    }
    if(G[x].size() == 1){
        ans += val*dis;
    }
}

int main()
{
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n-1; i++){
        int u, v;
        scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    dfs(1, 1.0, 0);
    printf("%.15lf\n", ans);
    return 0;
}


你可能感兴趣的:(CodeForces,搜索)