HDU 3006 The Number of set

最新推荐文章于 2020-07-20 20:48:02 发布

原创最新推荐文章于 2020-07-20 20:48:02 发布 · 436 阅读

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本文介绍了一种使用状态压缩动态规划解决组合集合问题的方法。通过给定的若干集合，不破坏原有集合的情况下组合出新的集合，并计算不同新集合的数量。文章提供了完整的C++代码示例，演示了如何利用位运算进行状态压缩，实现高效的DP解决方案。

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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

简单状态压缩dp
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const double pi = acos(-1.0);
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 10;
int n, m, x, y, z;
int dp[maxn];

int main()
{
	while (scanf("%d%d", &n, &m) != EOF)
	{
		for (int i = dp[0] = 1; i < (1 << m); i++) dp[i] = 0;
		while (n--)
		{
			scanf("%d", &x);	z = 0;
			while (x--) { scanf("%d", &y); z |= 1 << y - 1; }
			for (int i = 0; i < (1 << m); i++) dp[i | z] = dp[i] | dp[i | z];
		}
		int ans = 0;
		for (int i = 1; i < (1 << m); i++) ans += dp[i];
		printf("%d\n", ans);
	}
	return 0;
}