线性期望（BUPT2015校赛.F）

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:

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

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[][]=;
    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 ;
}