HDU 5305 Friends（2015 Multi-University Training Contest 2）

Friends

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1946    Accepted Submission(s): 975

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
   
   
   
   

 

Author

XJZX

 

Source

2015 Multi-University Training Contest 2

题目大意：

有 n个人，m对朋友关系，朋友之间可以选择成为 在线朋友 或者 离线朋友，每个人都想有相同数目的 在线朋友 和 离线朋友。（比如一个人有 x 个在线朋友，那么他必须有 x 个离线朋友）但是不同的人 x 可以不同。求有多少种方案可以满足他们的要求。

样例解释

2 两个测试用例
3 3 三个人，三对朋友关系
1 2 1 2 是朋友，他们可以选择成为离线或者在线朋友
2 3 2 3 是朋友，他们可以选择成为离线或者在线朋友
3 1 3 4 是朋友，他们可以选择成为离线或者在线朋友
4 4 同理
1 2
2 3
3 4
4 1

解题思路

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxm=100;
int x[maxm];
int y[maxm];
int n,m;
int ans;
int vis1[maxm];
int vis2[maxm];
int sum[maxm];
void dfs(int pos)
{
    if(pos==m+1)
    {
        int ok=1;
        for(int i=1; i<=n; i++)
        {
            if(vis1[i]!=vis2[i])
            {
                ok=0;
                break;
            }
        }
        if(ok)
        {
            ans++;
            return;
        }
    }
    if(vis1[x[pos]]<(sum[x[pos]]>>1)&&vis1[y[pos]]<(sum[y[pos]]>>1))
    {
        vis1[x[pos]]++;
        vis1[y[pos]]++;
        dfs(pos+1);
        vis1[x[pos]]--;
        vis1[y[pos]]--;
    }
    if(vis2[x[pos]]<(sum[x[pos]]>>1)&&vis2[y[pos]]<(sum[y[pos]]>>1))
    {
        vis2[x[pos]]++;
        vis2[y[pos]]++;
        dfs(pos+1);
        vis2[x[pos]]--;
        vis2[y[pos]]--;
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(sum,0,sizeof(sum));
        memset(vis1,0,sizeof(vis1));
        memset(vis2,0,sizeof(vis2));
        scanf("%d%d",&n,&m);
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d",&x[i],&y[i]);
            sum[x[i]]++;
            sum[y[i]]++;
        }
        int ok=1;
        for(int i=1; i<=n; i++)
        {
            if(sum[i]%2)
            {
                ok=0;
                printf("0\n");
                break;
            }
        }
        if(ok)
        {
            ans=0;
            dfs(1);
            printf("%d\n",ans);
        }
    }
    return 0;
}