BZOJ 1688: [Usaco2005 Open]Disease Manangement 疾病管理

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/zrts/p/bzoj1688.html

本文介绍了一种通过枚举技术解决农场中选择健康奶牛进行挤奶的问题，确保被选中的奶牛携带的疾病种类不超过限定数量。文章提供了一段C++代码实现，通过2^D的方式枚举所有可能的疾病组合，最终确定可以参与挤奶的最大奶牛数量。

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.

题解：

`2^d`枚举选那些病，统计答案。

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
 
//by zrt
//problem:
using namespace std;
typedef long long LL;
const int inf(0x3f3f3f3f);
const double eps(1e-9);
int n,m,k;
int d[1005];
inline bool judge(int x){
    int c=0;
    while(x){
        c++;
        x&=(x-1);
    }
    return c<=k;
}
int main(){
    #ifdef LOCAL
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
    #endif
    scanf("%d%d%d",&n,&m,&k);
    for(int i=0,c;i<n;i++){
        scanf("%d",&c);
        for(int j=0,x;j<c;j++){
            scanf("%d",&x);
            d[i]|=1<<(x-1);
        }
    }
    int ans=0;
    for(int i=0;i<(1<<m);i++){
         
        if(judge(i)){
            int tot=0;
            for(int j=0;j<n;j++){
                if((d[j]|i)==i) tot++;
            }
            ans=max(ans,tot);
        }
    }
    printf("%d\n",ans);
    return 0;
}