bzoj1688 疾病管理

最新推荐文章于 2019-07-31 12:42:55 发布

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最新推荐文章于 2019-07-31 12:42:55 发布

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原文链接：http://www.cnblogs.com/Serene-shixinyi/p/7480025.html

本文介绍了一种通过状态压缩的方法解决农场中选择最多数量的健康奶牛进行挤奶的问题，确保所选奶牛携带的疾病种类不超过限定数目。

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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.

Sample Input

6 3 2
0---------第一头牛患0种病
1 1------第二头牛患一种病,为第一种病.
1 2
1 3
2 2 1
2 2 1

Sample Output

5

OUTPUT DETAILS:

If FJ milks cows 1, 2, 3, 5, and 6, then the milk will have only two
diseases (#1 and #2), which is no greater than K (2).

无脑状压

水~

//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
const int maxn=1000+10,maxd=17,maxs=(1<<15)+10;
int n,d,k,mi[maxd],ds[maxn],nowans,ans;
 
int aa;char cc;
int read() {
    aa=0;cc=getchar();
    while(cc<'0'||cc>'9') cc=getchar();
    while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
    return aa;
}
 
bool ok(int x) {
    int tot=0;
    while(x) {
        x-=(x&(-x));
        tot++;
    }
    return tot<=k;
}
 
int main() {
    n=read();d=read();k=read();
    mi[1]=1;int x,y;
    for(int i=2;i<=d;++i) mi[i]=mi[i-1]<<1;
    for(int i=1;i<=n;++i) {
        x=read();
        for(int j=1;j<=x;++j) {
            y=read();
            ds[i]+=mi[y];
        }
    }
    for(int x=0;x<(1<<d);++x) if(ok(x)){
        nowans=0;
        for(int j=1;j<=n;++j) {
            y=(x|ds[j]);
            if(y==x) nowans++;
        }
        ans=max(ans,nowans);
    }
    printf("%d",ans);
    return 0;
}

转载于:https://www.cnblogs.com/Serene-shixinyi/p/7480025.html