zoukankan      html  css  js  c++  java
  • Journey CodeForces

    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 .

    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 100009
    
    
    vector<int> E[MAXN];
    bool vis[MAXN];
    int n;
    double dp(int k,int len)
    {
        vis[k] = true;
        double rate = 1.0, ans = 0;
        int cnt = 0;
        for (int i = 0; i < E[k].size(); i++)
        {
            if (!vis[E[k][i]])
                cnt++;
        }
        if (cnt == 0)
            return len;
        rate /= cnt;
        for (int i = 0; i < E[k].size(); i++)
        {
            if (vis[E[k][i]]) continue;
            ans += dp(E[k][i], len + 1)*rate;
        }
        return ans;
    }
    int main()
    {
        scanf("%d", &n);
        int f, t;
        for (int i = 0; i < n - 1; i++)
        {
            scanf("%d%d", &f, &t);
            E[f].push_back(t);
            E[t].push_back(f);
        }
        printf("%.10lf
    ", dp(1, 0));
    }
  • 相关阅读:
    监控页面所有checkbox改变状态的简单方法
    poorman’sgraphicalboot
    Linux 高精確的時序(sleep, usleep,nanosleep)
    【转】跟我一起写udev规则(译)
    什么情况下可以不创建QCoreApplication
    Linux双网卡bonding举例
    many former solutions has been "deleted"
    文件浏览器
    配置站点集的配额和锁
    HyperV的三种网卡
  • 原文地址:https://www.cnblogs.com/joeylee97/p/7403786.html
Copyright © 2011-2022 走看看