hdu3091之状态压缩dp

Necklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 327680/327680 K (Java/Others)
Total Submission(s): 542    Accepted Submission(s): 173

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 31 21 32 3

 

Sample Output

2

 

本题和TSP旅行商差不多，只不过本题求构成项链个数（环）则只需要从某一点出发即可，还有就是1-2-3-1和1-3-2-1是两种

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <queue>#include <algorithm>#include <map>#include <cmath>#include <iomanip>#define INF 99999999typedef __int64 LL;using namespace std;const int MAX=(1<<18)+10;int n,m;LL dp[MAX][20];bool edge[20][20];void DP(){int bit=1<<n;memset(dp,0,sizeof dp); dp[1][0]=1;for(int i=1;i<bit;++i){for(int j=0;j<n;++j){if(dp[i][j] == 0)continue;for(int k=1;k<n;++k){if(i&(1<<k))continue;if(!edge[j][k])continue;dp[i|(1<<k)][k]+=dp[i][j];}}}LL sum=0;for(int i=0;i<n;++i)if(edge[0][i])sum+=dp[bit-1][i];printf("%I64d\n",sum);}int main(){int u,v;while(~scanf("%d%d",&n,&m)){memset(edge,false,sizeof edge);for(int i=0;i<m;++i){scanf("%d%d",&u,&v);--u,--v;edge[u][v]=edge[v][u]=true;}DP();}return 0;}