hdu校赛 F

Problem F

Time Limit: 10000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 76    Accepted Submission(s): 25

Problem Description

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 will occur between two nodes.


Figure 1: C knows every one, while D only knows C in this social network.


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

1. She is a female;
2. She knows at least K male friends.

Now we have already taken out several networks from 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), what will be the xpectation of number of social butterflies in the network?

 

Input

The number of test cases T (T <= 10^4) 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×N matrix G followed, where Gij=1 denotes j is a friend of i, otherwise Gij=0. Here, it's always satisfied that Gii=0 and Gij=Gji for all 1≤i,j≤N.

 

Output

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

 

Sample Input

22 10 11 03 10 1 11 0 11 1 0

 

Sample Output

0.5001.125
#include <map>#include <set>#include <stack>#include <queue>#include <cmath>#include <ctime>#include <vector>#include <cstdio>#include <cctype>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define inf -0x3f3f3f3f#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define mem0(a) memset(a,0,sizeof(a))#define mem1(a) memset(a,-1,sizeof(a))#define mem(a, b) memset(a, b, sizeof(a))typedef long long ll;__int64 C[31][31];int P[31];int degree[31];__int64 solve(int n,int m){    if(m==0)        return 1;    __int64 ans=1;    for(int i=1;i<=m;i++)        ans=ans*(n-i+1)/i;    return ans;}int main(){    for(int i=0;i<=30;i++)        for(int j=0;j<=i;j++)            C[i][j]=solve(i,j);    P[0]=1;    P[1]=2;    for(int i=2;i<=30;i++)        P[i]=P[i-1]*2;    int _,n,k,u;    scanf("%d",&_);    while(_--){        scanf("%d%d",&n,&k);        mem0(degree);        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++){                scanf("%d",&u);                if(u==1)                    degree[i]++,degree[j]++;            }        for(int i=1;i<=n;i++)            degree[i]/=2;        __int64 ans=0;        for(int i=1;i<=n;i++){            if(degree[i]>=k){                for(int j=k;j<=degree[i];j++)                    ans=ans+C[degree[i]][j]*P[n-degree[i]-1];            }        }        printf("%.3f\n",(double)ans/(double)P[n]);    }    return 0;}