3006 HDU The Number of set

http://www.cnblogs.com/lonelycatcher/archive/2011/05/27/2060158.html

Given you n sets.All positive integers in sets are not less than 1 and not greater than m.If use these sets to combinate the new set,how many different new set you can get.The given sets can not be broken.

 

Input

There are several cases.For each case,the first line contains two positive integer n and m(1<=n<=100,1<=m<=14).Then the following n lines describe the n sets.These lines each contains k+1 positive integer,the first which is k,then k integers are given. The input is end by EOF.

 

Output

For each case,the output contain only one integer,the number of the different sets you get.

 

Sample Input

4 4 1 1 1 2 1 3 1 4 2 4 3 1 2 3 4 1 2 3 4

 

Sample Output

15 2

//第一次接触这种题目网上说是状态压缩DP。

'|'的运算，

集合 中的元素当成是 2进制中的名次， 比如1 2 3 代表 111。 1 3代表101。这样就可以求出集合的个数了

不过感觉有点小疑问，题目说的是新组成的集合，但是算的时候却包含了本来的，这个就、、、

#include <iostream>
#include <string.h>
#include <cstdio>
using namespace std;
int hash[1<<15];
int main()
{  // freopen("in.txt","r",stdin);
    int n,m,k,i,x,y;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(hash,0,sizeof(hash));
        m=1<<m;
        while(n--)
           {
               scanf("%d",&k);
               y=0;//开始时y=0写到上面去了，郁闷，
               while(k--)
               {
                scanf("%d",&x);
                y=y|(1<<(x-1));
               }
             hash[y]=1;
             for(i=0;i<=m;i++)
              if(hash[i])
                hash[i|y]=1;  //算是得到新集合吧
           }n=0;

        for(i=0;i<=m;i++)
          if(hash[i])
          n++;
        printf("%d\n",n);
    }
    return 0;
}