HDU3091 Necklace 【状压】

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题意：有许多珠子，能和别的相邻，求哈密顿回路的方案数

思路：n不大，考虑状压DP，dp[s][i],表示到s状态时，i在最后一个。从任意一个点开始，递推到s包含所有点，最后答案累加时，判断结尾和你选定的起点是否相连

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<stdlib.h>#include<math.h>#include<vector>#include<list>#include<map>#include<stack>#include<queue>#include<algorithm>#include<numeric>#include<functional>using namespace std;typedef long long ll;const int maxn = 1<<20;ll dp[maxn][19];int dis[20][20];int main(void){int n,m,i,j;while(scanf("%d%d",&n,&m)!=EOF){memset(dis,0,sizeof dis);while(m--){int a,b;scanf("%d%d",&a,&b);dis[a][b] = dis[b][a] = 1;}memset(dp,0,sizeof dp);dp[1][1] = 1;int en = (1<<n)-1;for(int s = 1; s < en; s++){for(j = 1; j <= n; j++){if(!dp[s][j]) continue;for(i = 1; i <= n; i++) // new point{if(s&(1<<(i-1))) continue;if(!dis[i][j]) continue;dp[ s|(1<<(i-1)) ][i] += dp[s][j];}}}ll ans = 0;for(i = 1; i <= n; i++){if( dis[i][1] )ans += dp[en][i];}printf("%lld\n",ans);}return 0;}

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