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

    题意：n个人，分别有各自的朋友，有的关系好，有的关系不好，如今要求

每一个人的朋友中关系好的与关系不好的数量要同样，问多少种方案？

    思路：由于须要关系好的与关系不好的数量要同样。所以朋友的数量为奇数

的能够直接输出0，（由于奇数没有办法让两者同样），推断完奇数后进行搜索，先找出朋友中关系好的，剩下的自己主动归为关系不好的之中，假设出现两个人之间有一个人好朋友与坏朋友已经满了的，直接return。一个省时间的地方。

比赛的时候让队友做的，当时没有看出来是搜索。啊啊啊啊啊

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

int n,m;
int num[10];
int pa[10],pb[10];
int sum;

struct node
{
    int x;
    int y;
}q[100100];

void DFS(int p)
{
    //printf("p = %d\n",p);
   if(p == m)
   {
       sum++;
       return ;
   }
   int x = q[p].x;
   int y = q[p].y;
   if(pa[x] && pa[y])
   {
       pa[x]--;
       pa[y]--;
       DFS(p+1);
       pa[x]++;
       pa[y]++;
   }
   if(pb[x] && pb[y])
   {
       pb[x]--;
       pb[y]--;
       DFS(p+1);
       pb[x]++;
       pb[y]++;
   }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        sum = 0;
        memset(num,0,sizeof(num));
        scanf("%d%d",&n,&m);
        int x,y;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&x,&y);
            q[i].x = x;
            q[i].y = y;
            num[x]++;
            num[y]++;
        }
        int flag = 0;
        for(int i=1;i<=n;i++)
        {
            pa[i] = num[i]/2;
            pb[i] = num[i]/2;
            if(num[i]%2 == 1)
            {
                flag = 1;
                break;
            }
        }
        if(flag)
        {
            printf("0\n");
            continue;
        }
        DFS(0);
        printf("%d\n",sum);
    }
    return 0;
}