POJ 2441 Arrange the Bulls(状态压缩DP)

Arrange the Bulls

Description

Farmer Johnson‘s Bulls love playing basketball very much. But none of them would like to play basketball with the other bulls because they believe that
the others are all very weak. Farmer Johnson has N cows (we number the cows from 1 to N) and M barns (we number the barns from 1 to M), which is his bulls‘ basketball fields. However, his bulls are all very captious, they only like to play in some specific
barns, and don’t want to share a barn with the others.

So it is difficult for Farmer Johnson to arrange his bulls, he wants you to help him. Of course, find one solution is easy, but your task is to find how many solutions there are.

You should know that a solution is a situation that every bull can play basketball in a barn he likes and no two bulls share a barn.

To make the problem a little easy, it is assumed that the number of solutions will not exceed 10000000.

Input

In the first line of input contains two integers N and M (1 <= N <= 20, 1 <= M <= 20). Then come N lines. The i-th line first contains an integer P
(1 <= P <= M) referring to the number of barns cow i likes to play in. Then follow P integers, which give the number of there P barns.

Output

Print a single integer in a line, which is the number of solutions.

Sample Input

3 4
2 1 4
2 1 3
2 2 4

Sample Output

4

题意：有n头牛，m个牛圈，需要给每头牛分配一个合适的牛圈。每头牛都想一个人在一个牛圈里，且都有心中指定的牛圈。 现在给出这些牛心中想住进的牛圈。 每行先给出P，后面有P个数字，表示第i头牛想住进的牛圈的编号。 问需要满足以上条件的分配方法有多少种？

题解：状压dp,在递推式过程中我们把每个牛圈有没有住进牛看成一个状态，一共有(1<<m)-1个状态。 dp[i][j]表示第i头牛在第j种状态下有多少种分配方式。 那么状态转移方程为
dp[i][j] = sigma(dp[i-1][k]) k表示牛圈入住的状态，从0到(1<<m)-1  因为dp[20][1<<20]会超内存，这里要用滚动数组。

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxm 22
int dp[2][1<<20],a[maxm][maxm];
//dp[i][j]表示第i头在第j中状态下的方法数，a[i][j]表示第i头牛能在第j个牛圈里
int main()
{
	int n,m,k,i,j,cnt,num;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(a,0,sizeof(a));
		for(i=1;i<=n;++i)
		{
			scanf("%d",&cnt);
			while(cnt--)
			{
				scanf("%d",&num);
				a[i][num-1]=1;//为了方便运算把牛圈的编号变为从0~m-1
			}
		}
		memset(dp,0,sizeof(dp));
		dp[0][0]=1;
		for(i=1;i<=n;++i)
		{
			for(j=0;j<(1<<m);++j)
			{
				if(dp[1-(i&1)][j])//表示前一头牛在j状态下有值
				{
					for(k=0;k<m;++k)
					{
						if(a[i][k]&&(j!=(j|(1<<k))))//表示第i头牛能放进k位置且k位置没有其他的牛
							dp[i&1][j|(1<<k)]+=dp[1-(i&1)][j];
					}
				}
			}
			memset(dp[1-(i&1)],0,sizeof(dp[1-(i&1)]));
		}
		int ans=0;
		for(i=1;i<(1<<m);++i)//统计所有结果
			ans+=dp[n&1][i];
		printf("%d\n",ans);
	}
	return 0;
}