hdu 3091 Necklace 状态压缩dp *******

Necklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 327680/327680 K (Java/Others)
Total Submission(s): 522    Accepted Submission(s): 168

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

 

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<vector>
 using namespace std;
 
 int n,m;
 vector <int> a[];
 long long dp[(<<)+][];
 void make_ini()
 {
     int i,j;
     int s,r,u;
     memset(dp,,sizeof(dp));
     dp[][]=;
     for(i=;i<(<<n);i++)
     {
         for(j=;j<n;j++)
         {
             if(dp[i][j]==)continue;
             for(s=;s<a[j].size();s++)
             {
                 u=a[j][s];
                 if( (i&(<<u))> ) continue;
                 r=( i | (<<u) );
                 dp[r][u]+=dp[i][j];
             }
         }
     }
     long long Sum=;
     for(i=;i<a[].size();i++)
         Sum+=dp[(<<n)-][a[][i]];
         printf("%I64d\n",Sum);
 }
 int main()
 {
     int i,x,y;
     while(scanf("%d%d",&n,&m)>)
     {
         for(i=;i<n;i++)
         {
             a[i].clear();
         }
         for(i=;i<=m;i++)
         {
             scanf("%d%d",&x,&y);
             x--;
             y--;
             a[x].push_back(y);
             a[y].push_back(x);
         }
         make_ini();
     }
     return ;
 }