题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5305

题面：

Friends

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

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

Source

2015 Multi-University Training Contest 2

解题：

因为28本来就不大，很容易想到搜索，但是直接搜是会超时的，加一个当一个人的现实或者虚拟好友数大于其本身关系数一半时，就返回的剪枝即可。注意开始可以直接判一个人关系数为奇，那么就不比进行搜索了。

代码：

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
int t,n,m,cnt_net[10],cnt_real[10],cnt_all[10],a,b,ans,fm[40],to[40];
void dfs(int No,bool status)
{
  a=fm[No];
  b=to[No];
  if(status)
  {
    cnt_real[a]++;
    cnt_real[b]++;
    if((cnt_real[a]>cnt_all[a]/2)||(cnt_real[b]>cnt_all[b]/2))
        return;
  }
  else
  {
    cnt_net[a]++;
    cnt_net[b]++;
    if((cnt_net[a]>cnt_all[a]/2)||(cnt_net[b]>cnt_all[b]/2))
        return;
  }
  if(No==m)
  {
      bool sign=true;
      for(int i=1;i<=n;i++)
      {
          if(cnt_real[i]!=cnt_net[i])
          {
              sign=false;
              break;
          }
      }
      if(sign)ans++;
      return;
  }
  dfs(No+1,1);
  cnt_real[fm[No+1]]--;
  cnt_real[to[No+1]]--;
  dfs(No+1,0);
  cnt_net[fm[No+1]]--;
  cnt_net[to[No+1]]--;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
      bool flag=true;
      ans=0;
      scanf("%d%d",&n,&m);
      memset(cnt_net,0,sizeof(cnt_net));
      memset(cnt_real,0,sizeof(cnt_real));
      memset(cnt_all,0,sizeof(cnt_all));
      for(int i=1;i<=m;i++)
      {
        scanf("%d%d",&a,&b);
        fm[i]=a;
        to[i]=b;
        cnt_all[a]++;
        cnt_all[b]++;
      }
      for(int i=1;i<=n;i++)
      {
          if(cnt_all[i]%2)
          {
              flag=false;
              break;
          }
      }
      if(!flag)
      {
          printf("0\n");
          continue;
      }
      if(m)
      {
          dfs(1,0);
        cnt_net[fm[1]]--;
        cnt_net[to[1]]--;
    //    cout<<cnt_net[fm[1]]<<" "<<cnt_net[to[1]]<<endl;
        dfs(1,1);
      }
      else ans=1;
      printf("%d\n",ans);
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。