Necklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 327680/327680 K (Java/Others)

Total Submission(s): 709    Accepted Submission(s): 245

Problem Description

One day , Partychen gets several beads , he wants to make these beads a necklace . But not every beads can link to each other, every bead should link to some particular bead(s). Now , Partychen wants to know how many kinds of necklace he can make.

 

 

Input

It consists of multi-case .

Every case start with two integers N,M ( 1<=N<=18,M<=N*N )

The followed M lines contains two integers a,b ( 1<=a,b<=N ) which means the ath bead and the bth bead are able to be linked.

 

 

Output

An integer , which means the number of kinds that the necklace could be.

 

 

Sample Input

3 3 1 2 1 3 2 3

 

 

Sample Output

2

 

 

Source

 

/*	已经给了时间看题解了，正式练习一道题三天之内不准再看题解！*/#include
      
       #include
       
        #include
        
         #define N (1<<18)+5#define M 20#define INF 0x3f3f3f3fusing namespace std;long long dp[N][M],g[M][M],n,m;//dp[i][j]表示在i个状态下,你到达j这个点的方案数int main(){	//freopen("in.txt", "r", stdin);	while(scanf("%lld%lld",&n,&m)!=EOF)	{		memset(dp,0,sizeof dp);		memset(g,0,sizeof g);		int a,b;		for(int i=0;i