hdu3006——The Number of set

The Number of set

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1160    Accepted Submission(s): 709

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
   
   
   
   

 

Source

2009 Multi-University Training Contest 11 - Host by HRBEU

 

Recommend

gaojie   |   We have carefully selected several similar problems for you:   3007  3003  3004  3002  3005

同学让我看的这道题，刚看还以为是dp，后来仔细看了结果发现自己想不出如何简洁地表示状态，然后换了思路

我们可以发现，m <=14,这个条件给我很大的启发，为什么我不把一个集合看成一个整数A呢？集合里出现某个数我就在A的对应位上置1，然后，合并集合的操作就是或运算了，这时候只要加上标记然后dfs就可以得到答案了，数据比较小

#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int jihe[2222];
bool vis[33333];
int ans, n;

void dfs(int x, int v)
{
	for (int i = 1; i <= n; ++i)
	{
		if (vis[v | jihe[i]] || i == x)
		{
			continue;
		}
		vis[v | jihe[i]] = 1;
		ans++;
		dfs(i, v | jihe[i]);
	}
}

int main()
{
	int m, k, pos;
	while (~scanf("%d%d", &n, &m))
	{
		memset (vis, 0, sizeof(vis));
		memset (jihe, 0, sizeof(jihe));
		ans = 0;
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d", &k);
			for (int j = 0; j < k; ++j)
			{
				scanf("%d", &pos);
				jihe[i] |= (1 << pos);
			}
			if (vis[jihe[i]])
			{
				continue;
			}
			vis[jihe[i]] = 1;
			ans++;
		}
		for (int i = 1; i <= n; ++i)
		{
			dfs(i, jihe[i]);
		}
		printf("%d\n", ans);
	}
	return 0;
}