Arrange the Bulls [POJ2441] [状压DP]

题意

n头牛，m个房间，每头牛有自己喜欢的房间，问每头牛都住进自己喜欢的房间有多少种分配方法？

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

Analysis

首先，这种数据我们很容易想到是状压DP

我们可以比较轻松的写出状态转移方程

if(status&(1<<room)==0)

　　dp[i][status|(1<<room)]+=dp[i-1][status];

但是我们发现这样是会MLE的，我们就可以采用滚动数组，或是直接上一维

一维要注意status从大到小循环，因为是每次小的更新大的，status用完之后要清零。

Code

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define RG register int
#define rep(i,a,b)    for(RG i=a;i<=b;++i)
#define per(i,a,b)    for(RG i=a;i>=b;--i)
#define ll long long
#define inf (1<<29)
using namespace std;
int n,m,ans;
int a[25][25];
int dp[1048600];
inline int read()
{
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}


int main()
{
    n=read(),m=read();
    rep(i,1,n)
    {
        a[i][0]=read();
        rep(j,1,a[i][0])
            a[i][j]=read();
    }
    dp[0]=1;
    rep(i,1,n)
    {
        per(j,(1<<m)-1,0)
        {
            if(!dp[j]) continue;
            rep(k,1,a[i][0])
            {
                if(!(j&(1<<(a[i][k]-1))))
                {
                    dp[j|(1<<(a[i][k]-1))]+=dp[j];
                }
            }
            dp[j]=0;
        }
    }
    int ans=0;
    per(j,(1<<m)-1,0) ans+=dp[j];
    cout<<ans;
    return 0;
}