将整体期望分成部分期望来做。
F. network
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
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 31int 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\n",ans);}return 0; }
