状压dp——Disease Manangement 疾病管理

题目描述
Alas! A set of D (1 <= D <= 15) diseases (numbered 1..D) is running through the farm. Farmer John would like to milk as many of his N (1 <= N <= 1,000) cows as possible. If the milked cows carry more than K (1 <= K <= D) different diseases among them, then the milk will be too contaminated and will have to be discarded in its entirety. Please help determine the largest number of cows FJ can milk without having to discard the milk.
输入格式
Line 1: Three space-separated integers: N, D, and K

Lines 2..N+1: Line i+1 describes the diseases of cow i with a list of 1 or more space-separated integers. The first integer, d_i, is the count of cow i"s diseases; the next d_i integers enumerate the actual diseases. Of course, the list is empty if d_i is 0.

有N头牛,它们可能患有D种病,现在从这些牛中选出若干头来,但选出来的牛患病的集合中不过超过K种病.
输出格式
Line 1: M, the maximum number of cows which can be milked.

#include<bits/stdc++.h>
using namespace std;

const int maxn=1e3+10;
int n,d,k;
int dp[1<<16],v[1<<16],s[1<<16];
int tot,ans;

int main()
{
	int x,y;
	scanf("%d%d%d",&n,&d,&k);
	for(int i=1;i<(1<<d);i++)//处理出患病<=k的所有情况 
	{
		s[i]=s[i-(i&-i)]+1;
		if(s[i]<=k) v[++tot]=i;
	}
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&x);
		int z=0;
		for(int j=1;j<=x;j++)
		{
			scanf("%d",&y);
			z+=(1<<(y-1));
		}
		if(x>k) continue;//如果患病>k直接跳过 
		for(int j=1;j<=tot;j++)//循环所有情况如果牛符合某种情况就加1 
		{
			if((v[j]&z)==z)
			{
				++dp[j];
				ans=max(ans,dp[j]);//最后遍历一下所有情况，取牛的个数最多的一种输出 
			}
		}
	}
	printf("%d",ans);
	
	return 0;
}
/*
样例输入
6 3 2
0
1 1
1 2
1 3
2 2 1
2 2 1
样例输出
5
*/