# BZOJ 1688: [Usaco2005 Open]Disease Manangement 疾病管理 状压DP + 二进制 + 骚操作

## Description

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.

## Input

* 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.

## Output

* Line 1: M, the maximum number of cows which can be milked.

$lowbit(i)$ 的定义为 $i$ 的最后一个 $1$ 所对应的那个二进制串(我好像也不太能用一句话解释得清).

#include <bits/stdc++.h>
#define setIO(s) freopen(s".in","r",stdin)
#define maxn 100000
using namespace std;
int F[1 << 16], s[1 << 16], v[1 << 16], f[1 << 16];
int n, D, k, tot = 0, ans = 0;
int main()
{
// setIO("input");
scanf("%d%d%d",&n,&D,&k);
for(int i = 1; i < (1 << D); ++i)
{
s[i] = s[i - (i & -i)] + 1;
if(s[i] <= k) v[++tot] = i;
}
for(int i = 1; i <= n; ++i)
{
int a, b;
scanf("%d",&a);
int t = 0;
for(int j = 1; j <= a; ++j)
{
scanf("%d",&b);
t += (1 << (b - 1));
}
for(int j = 1; j <= tot; ++j)
{
if((v[j] & t) == t)
++f[j], ans = max(ans, f[j]);
}
}
printf("%d\n",ans);
return 0;
}


posted @ 2019-06-06 20:01  EM-LGH  阅读(114)  评论(0编辑  收藏  举报