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

 

Source

2015 Multi-University Training Contest 2

题目大意：一些朋友关系可离线可在线，要求每个人的离线朋友数等于在线朋友数

题目分析：数据量不大，直接爆搜，记录每个点的度数，奇数的直接不可能，偶数的分成两个数组，c1[i]表示i的在线朋友数，c2[i]表示i的离线朋友数，然后一条边一条边搜就行了，注意一个剪枝，当一条边的两个端点有一个c值为0，则return

#include <cstdio>#include <cstring>bool mp[10][10];int deg[10], c1[10], c2[10];int n, m, cnt, ans;struct EDGE{    int u, v;}e[100000];void DFS(int cur){    if(cur == m + 1)    {        ans ++;        return;    }    int v = e[cur].v;    int u = e[cur].u;    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(){    int T;    scanf("%d", &T);    while(T --)    {        cnt = 0;        ans = 0;        memset(deg, 0, sizeof(deg));        memset(e, 0, sizeof(e));        memset(c1, 0, sizeof(c1));        memset(c2, 0, sizeof(c2));        scanf("%d %d", &n, &m);        for(int i = 1; i <= m; i++)        {            scanf("%d %d", &e[i].u, &e[i].v);            deg[e[i].u] ++;            deg[e[i].v] ++;        }        bool f = false;        for(int i = 1; i <= n; i++)        {            c1[i] = c2[i] = deg[i] / 2;            if(deg[i] & 1)            {                f = true;                break;            }        }        if(f)        {            printf("0\n");            continue;        }        DFS(1);        printf("%d\n", ans);    }}