将整体期望分成部分期望来做。

F. network

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:

线性期望（BUPT2015校赛.F） 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(T≤104) will occur in the first line of input.

For each test case:

The first line contains the number of nodes N(1≤N≤30)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 1≤i,j≤N.

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

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[][]=;

    for(int i=;i<=;i++)

    {

        C[i][]=;

        for(int j=;j<=i;j++)

        {

            C[i][j] = C[i-][j-]+C[i-][j];

        }

    }

    int T;

    cin>>T;

    while(T--)

    {

        int n,k;

        scanf("%d%d",&n,&k);

        for(int i=;i<n;i++)

            for(int j=;j<n;j++)

                scanf("%d",&mat[i][j]);

        double ans=;

        for(int i=;i<n;i++)

        {

            int cnt=;

            for(int j=;j<n;j++)

            {

                cnt+=mat[i][j];

            }

            double tmp=;

            for(int j=k;j<=cnt;j++)

                tmp += C[cnt][j];

            tmp = tmp/pow(2.0,cnt);

            tmp *= 0.5;

            ans += tmp;

        }

        printf("%.3lf\n",ans);

    }

    return ;

}

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线性期望（BUPT2015校赛.F）