2015 Multi-University Training Contest 2 1006(DFS)

Friends

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

There are n people and m pairs of friends. For every pair of friends, they can choose to become online friends (communicating using online applications) or offline friends (mostly using face-to-face communication). However, everyone in these n people wants to have the same number of online and offline friends (i.e. If one person has x onine friends, he or she must have x offline friends too, but different people can have different number of online or offline friends). Please determine how many ways there are to satisfy their requirements.

 

 

Input

The first line of the input is a single integer T (T=100), indicating the number of testcases.

For each testcase, the first line contains two integers n (1≤n≤8) and m (0≤m≤n(n−1)2), indicating the number of people and the number of pairs of friends, respectively. Each of the next m lines contains two numbers x and y, which mean x and y are friends. It is guaranteed that x≠y and every friend relationship will appear at most once.

 

 

Output

For each testcase, print one number indicating the answer.

 

 

Sample Input

2

3 3

1 2

2 3

3 1

4 4

1 2

2 3

3 4

4 1

 

 

Sample Output

0

2

 

题意：有N个人有M个关系，每两个人的关系有线上关系和线下关系两种，要使每个人与其他人的线上好友和线下好友数量相等，求有多少种安排方法？

分析：考虑到人数很少，所以可以考虑暴力枚举每一条边，但总边数也有(8 * 7) / 2 = 28条边，最后要求线上好友和线下好友人数相等，所以每个人的好友不能为奇数，再去枚举每一条边。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
using namespace std;
const int MAXN=10+5;
int kase,n,m,ans;
int e[50][2],l[MAXN],r[MAXN],tot[MAXN];

void DFS(int cur)
{
    if(cur==m)
    {
        for(int i=1;i<=n;i++)
        {
            if(l[i]!=r[i])
                return ;
        }
        ans++;
        return ;
    }
    int tmpl=e[cur][0],tmpr=e[cur][1];
    if(l[tmpl]+1 <= tot[tmpl]/2 && l[tmpr]+1 <= tot[tmpr]/2)//线上好友
    {
        l[tmpl]++;
        l[tmpr]++;
        DFS(cur+1);
        l[tmpl]--;
        l[tmpr]--;
    }
    if(r[tmpl]+1 <= tot[tmpl]/2 && r[tmpr]+1 <= tot[tmpr]/2)//线下好友
    {
        r[tmpl]++;
        r[tmpr]++;
        DFS(cur+1);
        r[tmpl]--;
        r[tmpr]--;
    }

}
int main()
{
    scanf("%d",&kase);
    while(kase--)
    {
        int i;
        memset(tot,0,sizeof(tot));
        memset(e,0,sizeof(e));
        memset(l,0,sizeof(l));
        memset(r,0,sizeof(r));

        scanf("%d %d",&n,&m);
        for(i=0;i<m;i++)
        {
            scanf("%d %d",&e[i][0],&e[i][1]);//一条边上的两个点
            tot[e[i][0]]++;
            tot[e[i][1]]++;//点的度
        }
        for(i=1;i<=n;i++)
            if(tot[i]%2)
                break;
        if(i<=n) {printf("0\n"); continue;}
        ans=0;
        DFS(0);
        printf("%d\n",ans);
    }
    return 0;
}