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[][];
 int dp[];
 inline int read()
 {
     int x=,f=;char c=getchar();
     while(c<''||c>''){if(c=='-')f=-;c=getchar();}
     while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
     return x*f;
 }

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