2015暑假多校联合---Friends（dfs枚举）

原题链接

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

 

Recommend

wange2014

 

题意：输入n，m，n表示有n个人，m表示m对朋友关系，现在要使每个人的朋友划分为在线朋友和离线朋友，且在线朋友和离线朋友数量相等（一对朋友之间只能是在线朋友或者离线朋友），求方案数；

 

思路：用dfs深搜枚举每一条边（即每一对朋友关系），若能深搜进行完最后一条边，即当前边cnt==m+1  则ans++；

 

代码如下：

#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
using namespace std;
int n,m,cnt,ans;
int c1[],c2[],d[];
struct Node
{
    int u,v;
}node[];

void dfs(int i)
{
    if(i-==m)
    {
        ans++;
        return ;
    }
    if(c1[node[i].u]&&c1[node[i].v])
    {
        c1[node[i].u]--;
        c1[node[i].v]--;
        dfs(i+);
        c1[node[i].u]++;
        c1[node[i].v]++;
    }
    if(c2[node[i].u]&&c2[node[i].v])
    {
        c2[node[i].u]--;
        c2[node[i].v]--;
        dfs(i+);
        c2[node[i].u]++;
        c2[node[i].v]++;
    }
}

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        cnt=;
        ans=;
        scanf("%d%d",&n,&m);
        memset(node,,sizeof(node));
        memset(c1,,sizeof(c1));
        memset(c2,,sizeof(c2));
        memset(d,,sizeof(d));
        for(int i=;i<=m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            node[++cnt].u=u;
            node[cnt].v=v;
            d[u]++;
            d[v]++;
        }
        int flag=;
        for(int i=;i<=n;i++)
        {
            c1[i]=c2[i]=d[i]/;
            if(d[i]&)
            {
                flag=;
                break;
            }
        }
        if(flag)
        {
            puts("");
            continue;
        }
        dfs();
        printf("%d\n",ans);
    }
    return ;
}