程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> hdu3006——The Number of set

hdu3006——The Number of set

編輯：C++入門知識

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

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

同學讓我看的這道題，剛看還以為是dp，後來仔細看了結果發現自己想不出如何簡潔地表示狀態，然後換了思路

我們可以發現，m <=14,這個條件給我很大的啟發，為什麼我不把一個集合看成一個整數A呢？集合裡出現某個數我就在A的對應位上置1，然後，合並集合的操作就是或運算了，這時候只要加上標記然後dfs就可以得到答案了，數據比較小

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

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;
}