hdu 3006 The Number of set 状态压缩

最新推荐文章于 2019-10-06 08:21:08 发布

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本文探讨了一个关于集合组合的问题，给定多个不包含超过特定范围的正整数集合，求这些集合通过组合能形成多少种不同的新集合。采用位运算进行集合的表示与合并操作，实现了高效计算。

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用一个二进制数表示一个集合如 {1,4,7} 可以用 1001001，每个位置上有0或1代表这个数存在不存在；

用"|"可以实现两个集合的合并。

The Number of set

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1264 Accepted Submission(s): 770

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

Sample Output

15
2

#include <stdio.h>
#include <string.h>
int main()
{
	int n, m, k, w, t, i, ans;
	int set[1<<14];
	while(~scanf("%d%d", &n, &m))
	{
		memset(set,0,sizeof(set));
		while(n--)
		{
			scanf("%d", &k);
			w = 0;
			while(k--)
			{
				scanf("%d", &t);
				w += 1<< (t-1);
			}
			set[w] = 1;
			for(i = 1;i < 1<<14;i++)
			{
				if(set[i])
					set[w|i] = 1;
			}
			ans = 0;
		}
		for(i = 1;i < 1<<14;i++)
		{
			if(set[i])
				ans++;
		}
		printf("%d\n", ans);
	}
}