hdu3006之位压缩

The Number of set

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1002    Accepted Submission(s): 626

Problem Description

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

 

分析：由于m较小，所以可以用相应的位记录出现的数，这样每一个集合就可以用一个数表示了，能够成的集合也就是这些书进行|运算后得到的所有不同数个数了

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#define INF 99999999
using namespace std;

const int MAX=1<<15;
bool mark[MAX];
int s[MAX];

int main(){
	int n,m,k,a;
	while(cin>>n>>m){
		m=1<<(m+1);
		memset(s,0,sizeof s);
		memset(mark,false,sizeof mark);
		mark[0]=true;
		for(int i=0;i<n;++i){
			cin>>k;
			while(k--){
				cin>>a;
				s[i]=s[i]|(1<<a);
			}
			for(int j=0;j<m;++j){
				if(mark[j])mark[s[i]|j]=true;
			}
		}
		int num=0;
		for(int j=1;j<m;++j)num+=mark[j];
		cout<<num<<endl;
	}
	return 0;
}