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 2009 Multi-University Training Contest 18 - Host by ECNU  

/*
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<iostream>
#include<string.h>
#include<stdio.h>
#define N (1<<18)+5
#define M 20
#define INF 0x3f3f3f3f
using namespace std;
long long dp
[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<m;i++)
{
scanf("%lld%lld",&a,&b);
g[a-1][b-1]=g[b-1][a-1]=1;
}
int tol=(1<<n);
dp[1][0]=1;
for(int i=1;i<tol;i++)
{
for(int j=0;j<n;j++)//枚举你要经过的中间点
{
if(dp[i][j]==0) continue;//上一个状态没有方案数的讨论没意义
//cout<<"ok"<<endl;
for(int k=1;k<n;k++)//枚举你这个状态最终要到达的终点
{
if(!(i&(1<<k))&&g[j][k])
{
//cout<<"ok"<<endl;
dp[i|(1<<k)][k]+=dp[i][j];
}
}
}
}
long long s=0;
for(int i=0;i<n;i++)
{
//cout<<dp[tol-1][i]<<" ";
if(g[0][i])
s+=dp[tol-1][i];
}
//cout<<endl;
printf("%lld\n",s);
}
return 0;
}