HDU 5305(Friends-暴搜)

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探讨了在一个社交网络中，如何让每个人在线下和线上拥有相同数量的朋友，并计算满足条件的组合方式总数的问题。

Friends

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1717 Accepted Submission(s): 854

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

Sample Output

0
2

Author

XJZX

Source

2015 Multi-University Training Contest 2

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暴搜

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100+10)
#define MAXM (100+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,m;
int e[MAXM][2];
int degree[MAXN],totdeg[MAXN];
ll ans;
bool check(int x,int y) 
{
	return  ( ( totdeg[x]||( !degree[x] ))   &&  ( totdeg[y]||( !degree[y] ))  );

}
void dfs(int p)
{
	if (p==m)
	{
		For(i,n)
			if (i!=e[p][0]&&i!=e[p][1]&°ree[i]) return ;
		if (degree[e[p][0]]==degree[e[p][1]]&&abs(degree[e[p][0]])==1) {
			ans++;
		} 
		return ;	
	}
//	if (p==m+1) 
//	{
//		ans++;
//		return;
//	}
	int x=e[p][0],y=e[p][1];
	totdeg[x]--;totdeg[y]--;
	degree[x]++;degree[y]++;
	if (check(x,y)) dfs(p+1);
	degree[x]-=2;degree[y]-=2;
	if (check(x,y)) dfs(p+1);
	degree[x]++;degree[y]++;
	totdeg[x]++;totdeg[y]++;
}
int main()
{
//	freopen("F.in","r",stdin);
	
	int T;cin>>T;
	while(T--) {
		ans=0; MEM(degree) MEM(totdeg)  
		cin>>n>>m;
		For(i,m) scanf("%d%d",&e[i][0],&e[i][1]),totdeg[e[i][0]]++,totdeg[e[i][1]]++;
		
		bool flag=0;
		For(i,n) if (totdeg[i] & 1) {
			flag=1;puts("0");break;
		}
		if (flag) continue;
		
		if (m) dfs(1); else ans=1;
		printf("%lld\n",ans);
	}
	
	return 0;
}