状态压缩DP

K - Necklace

Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:327680KB     64bit IO Format:%I64d & %I64u

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Description

 

Input

 

Output

 

Sample Input

 

Sample Output

 

Hint

 

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

题意：

给出N个珠子，要把他们连成串，每个珠子都要用上，有多少种连法。重点是dp状态转移方程。

代码：

 /*
 给出的珠子要全部用上，因此只要最后的状态dp[i][j]中的j能够与第一个珠子相连就行，其中i表示取珠子
 的状态，每多加一颗珠子他的dp数就是加上没加之前的dp数(重点！).
 */
 #include<iostream>
 #include<string>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<vector>
 #include<iomanip>
 #include<queue>
 #include<stack>
 using namespace std;
 int n,m,a,b;
 bool d[][];
 long long dp[<<][]; //用int过不了
 int main()
 {
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         memset(dp,,sizeof(dp));
         memset(d,,sizeof(d));
         for(int i=;i<m;i++)
         {
             scanf("%d%d",&a,&b);
             a--;         //这里减一是为了后面写循环简便，也可以不减一
             b--;
             d[b][a]=;
             d[a][b]=;
         }
         dp[][]=;     //初始化第一个当放一个钥匙时总数为1
         for(int i=;i<(<<n);i++)
         {
             for(int j=;j<n;j++)
             {
                 if(dp[i][j]==) continue; //也可以用!(i&(1<<j))来判断，但后者会更费时
                 for(int k=;k<n;k++)      //k也可以从0开始
                 {
                     if(!d[k][j]) continue;  //判断k与j是否能连接
                     if(i&(<<k)) continue;  //判断第K个钥匙是否已算过
                     dp[i|(<<k)][k]+=dp[i][j];
                 }
             }
         }
         long long ans=;
         for(int i=;i<n;i++)
         {
             if(d[][i]) ans+=dp[(<<n)-][i];
         }
         cout<<ans<<endl;
     }
     return ;
 }