HDU 3091(动态规划-状态压缩)

问题描述:

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

题目题意:给我们n棵珠，给定它们的连接规则，问有多少种项链(注意首尾相连)

题目分析:因为珠子不多，所以我们可以用数字的每一位表示珠子的状态，因为珠子的连接是有规则的，不是所有的珠子都可以相连，假设我们如果知道了前一个珠子是那个，那么后面那些珠子可以相连就确定了，所以还需要一维来表示最后一颗珠子是谁。

dp[i][j]的保存的就是在状态i下，结尾是珠子j的多少种情况。还有一点是首尾的问题，我们可以假设第一个是第一颗珠子，结果不影响，那么尾部那些就可以确定了。

代码如下:

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#define ll long long
using namespace std;

ll dp[(1<<18)][20];
bool G[20][20];

int main()
{
int n,m;
while (scanf("%d%d",&n,&m)!=EOF) {
memset (G,false,sizeof (G));
for (int i=1;i<=m;i++) {
int a,b;
scanf("%d%d",&a,&b);
G[a][b]=G[b][a]=true;//保存图
}
memset (dp,0,sizeof (dp));
dp[1][1]=1;
for (int i=0;i<=((1<<n)-1);i++) {
for (int j=n;j>=1;j--) {
if (!dp[i][j]) continue;
for (int k=1;k<=n;k++) {
if (G[k][j]&&!(i&(1<<(k-1)))) dp[i|(1<<(k-1))][k]+=dp[i][j];
}
}
}
ll ans=0;
for (int i=1;i<=n;i++) {
if (G[1][i])//首尾相连
ans+=dp[((1<<n)-1)][i];
}
printf("%lld\n",ans);
}
return 0;
}