7.23多校——5305DFS——Friends

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

 

Source

2015 Multi-University Training Contest 2

 

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/*

cnt1[i]表示i顶点编号为1的度数

cnt2[i]表示i顶点编号为2的度数

显然得到如果存在一个点的度数为奇数，那么输出0（要平分）

如果顶点度数都是偶数,进行dfs，dfs边的数目，终止条件为cur == m （从0开始）

判断之前添的边能否满足

*/

#include <cstdio>

#include <algorithm>

#include <cstring>

using namespace std;



int n,m;

int cnt1[10], cnt2[10];

pair <int, int > a[100];



int ret;



bool check(int x)

{

    for(int i = 1; i <= x; i++){

        if(cnt1[i] != cnt2[i])

                return false;

    }

     return true;

}



void dfs(int cur)

{

    if(cur == m ){

        if(check(n))

            ret++;

        return ;

    }

    int u = a[cur].first;

    int v = a[cur].second;

    cnt1[u]++;

    cnt1[v]++;

    if(check(u-1)){

        dfs(cur + 1);

    }

    cnt1[u]--;

    cnt1[v]--;

    cnt2[u]++;

    cnt2[v]++;

    if(check(u-1)){

        dfs(cur + 1);

    }

    cnt2[u]--;

    cnt2[v]--;

}



int main()

{

    int T;

    int x, y;

    scanf("%d", &T);

    while(T--){

        scanf("%d%d", &n, &m);

        memset(cnt1, 0, sizeof(cnt1));

        memset(cnt2, 0, sizeof(cnt2));

        for(int i = 0; i < m ; i++){

            scanf("%d%d", &x, &y);

            if( x > y)

                swap(x, y);

            cnt1[x]++;

            cnt1[y]++;

            a[i] = make_pair(x, y);

        }

        int flag = 0;

        sort(a , a + m );

        for(int i = 1; i <= n ;i++)

            if(cnt1[i] % 2 == 1){

                flag = 1;

                break;

            }

        memset(cnt1, 0, sizeof(cnt1));

        ret = 0;

        dfs(0);

        if(flag == 0) printf("%d\n", ret);

        else printf("0\n");

    }

    return 0;

}