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  • 线性期望(BUPT2015校赛.F)

    将整体期望分成部分期望来做。

    时间限制 3000 ms 内存限制 65536 KB

    题目描述

    A social network is a social structure made up of a set of social actors (such as individuals or organizations) and a set of the relationships between these actors. In simple cases, we may represent people as nodes in a graph, and if two people are friends, then an edge occurs between two nodes.

    There are many interesting properties in a social network. Recently, we are researching on the  SocialButterfly. A social butterfly should satisfy the following conditions:

                                     A simple social network,where C knows everyone but D knows just C.

    Now we have already had several networks in our database, but since the data only contain nodes and edges, we don't know whether a node represents a male or a female. We are interested, that if there are equal probabilities for a node to be male and female (each with 1/2 probability).A node is a social butterfly if and only if this node is a female and connects with at least K males.What will be the expectation of number of social butterflies in the network?

    输入格式

    The number of test cases T(T104) will occur in the first line of input.

    For each test case:

    The first line contains the number of nodes N(1N30)and the parameter K (0 <= K < N))

    Then an N×Nmatrix G followed, where Gij=1 denotes j as a friend of i, otherwise Gij=0. Here, it's always satisfied that Gii=0and Gij=Gji for all 1i,jN.

    输出格式

    For each test case, output the expectation of number of social butterflies in 3 decimals.

    ##Hint

    In the first sample, there are totally 4 cases: {Female, Female}, {Female,
    Male},{Male, Female} and {Male, Male}, whose number of social butterflies
    are respectively 0, 1, 1, 0. Hence, the expectation should be

    E=14×0+14×1+14×1+14×0=12

    输入样例

    2
    2 1
    0 1
    1 0
    3 1
    0 1 1
    1 0 1
    1 1 0

    输出样例

    0.500
    1.125

    //
    //  main.cpp
    //  160323.F
    //
    //  Created by 陈加寿 on 16/3/25.
    //  Copyright © 2016年 chenhuan001. All rights reserved.
    //
    
    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    #include <algorithm>
    #include <math.h>
    using namespace std;
    #define N 31
    
    int mat[N][N];
    double C[N][N];
    
    
    int main() {
        C[0][0]=1;
        for(int i=1;i<=30;i++)
        {
            C[i][0]=1;
            for(int j=1;j<=i;j++)
            {
                C[i][j] = C[i-1][j-1]+C[i-1][j];
            }
        }
        
        int T;
        cin>>T;
        while(T--)
        {
            int n,k;
            scanf("%d%d",&n,&k);
            for(int i=0;i<n;i++)
                for(int j=0;j<n;j++)
                    scanf("%d",&mat[i][j]);
            double ans=0;
            for(int i=0;i<n;i++)
            {
                int cnt=0;
                for(int j=0;j<n;j++)
                {
                    cnt+=mat[i][j];
                }
                double tmp=0;
                for(int j=k;j<=cnt;j++)
                    tmp += C[cnt][j];
                tmp = tmp/pow(2.0,cnt);
                tmp *= 0.5;
                ans += tmp;
            }
            printf("%.3lf
    ",ans);
        }
        return 0;
    }
     
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  • 原文地址:https://www.cnblogs.com/chenhuan001/p/5319838.html
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