O - Necklace （状态dp）

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.

InputIt 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. 

OutputAn 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：

给你几个贝壳，每个贝壳能和特定的贝壳相连，问最多可组成多少种项链？

思路：

状态dp，寻找每个位置

代码;'

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
typedef __int64 ll;
using namespace std;
int n,m;
bool mp[25][25];
ll dp[1<<18][25];

void solve(){
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(!mp[j][k])continue;
dp[i|(1<<k)][k]+=dp[i][j];
}
}
}
ll ans=0;
for(int i=0;i<n;i++)
if(mp[0][i])ans+=dp[bit-1][i];
cout<<ans<<endl;
}
int main(){
ios::sync_with_stdio(false);
int x,y;
int i;
while(cin>>n>>m){
memset(mp,0,sizeof(mp));
for(i=0;i<m;i++){
cin>>x>>y;
--x,--y;
mp[x][y]=mp[y][x]=1;

}
solve();
}
return 0;
}