HDU 5305 Friends (DFS)

Friends

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

 Total Submission(s): 163    Accepted Submission(s): 61

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 thesen people wants to have the same number of online and offline friends (i.e. If one person hasx 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 integerT (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 nextm 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

23 31 22 33 14 41 22 33 44 1

 

Sample Output

02

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>#include <queue>#include <stack>using namespace std;struct edge{    int u, v;};edge e[30];int deg[30], c1[30], c2[30];int t, n, m, ans;void dfs(int cur){    if(cur==m)    {        ans++;        return;    }    int u=e[cur].u, v=e[cur].v;    if(c1[u] && c1[v])    {        c1[u]--, c1[v]--;        dfs(cur+1);        c1[u]++, c1[v]++;    }    if(c2[u] && c2[v])    {        c2[u]--, c2[v]--;        dfs(cur+1);        c2[u]++, c2[v]++;    }    return;}int main(){    scanf("%d", &t);    while(t--)    {        memset(deg, 0, sizeof(deg));        scanf("%d%d", &n, &m);        for(int i=0; i<m; i++)        {            scanf("%d%d", &e[i].u, &e[i].v);            deg[e[i].u]++;            deg[e[i].v]++;        }        bool flag=true;        for(int i=1; i<=n; i++)        {            if(deg[i]%2)            {                flag=false;                break;            }            c1[i]=deg[i]/2;            c2[i]=deg[i]/2;        }        ans=0;        if(flag)        {            dfs(0);            printf("%d\n", ans);        }        else        {            printf("0\n");        }    }    return 0;}