There are  people and  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  people wants to have the same number of online and offline friends (i.e. If one person has  onine friends, he or she must have  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. 

The first line of the input is a single integer , indicating the number of testcases. 

For each testcase, the first line contains two integers  and , indicating the number of people and the number of pairs of friends, respectively. Each of the next  lines contains two numbers  and , which mean  and  are friends. It is guaranteed that  and every friend relationship will appear at most once. 

For each testcase, print one number indicating the answer.

2
3 3
1 2
2 3
3 1
4 4
1 2
2 3
3 4
4 1

0
2

2015 Multi-University Training Contest 2

解题思路：
注意到数据范围很小，因此，暴力搜索+ 剪枝即可。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN = 30;
int x[MAXN],y[MAXN], s[MAXN], s1[MAXN], s2[MAXN], g[MAXN][MAXN];
int n, m;
int ans;
void dfs(int dep)
{
    if(dep > m)
    {
        for(int i=1;i<=n;i++)
        {
            if(s1[i] != s2[i])
                return ;
        }
        ans++;
        return ;
    }
    int u = x[dep], v = y[dep];
    if(s1[u] < s[u] / 2 && s1[v] < s[v] / 2)
    {
        s1[u]++; s1[v]++;
        dfs(dep + 1);
        s1[u]--; s1[v]--;
    }
    if(s2[u] < s[u] / 2 && s2[v] < s[v] / 2)
    {
        s2[u]++; s2[v]++;
        dfs(dep + 1);
        s2[u]--; s2[v]--;
    }
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &m);
        memset(g, 0, sizeof(g));
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d", &x[i], &y[i]);
            g[x[i]][y[i]] = g[y[i]][x[i]] = 1;
        }
        int flag = 1;
        for(int i=1; i<=n; i++)
        {
            s[i] = s1[i] = s2[i] = 0;
            for(int j=1; j<=n; j++)
            {
                if(i != j && g[i][j] == 1)
                    s[i]++;
            }
            if(s[i] & 1) flag = 0;
        }
        if(!flag)
        {
            printf("0\n");
            continue;
        }
        ans = 0;
        dfs(1);
        printf("%d\n", ans);
    }
}