zoukankan      html  css  js  c++  java
  • [BZOJ1688][Usaco2005 Open]Disease Manangement 疾病管理

    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).


    又是用ysq学长的号交的...设f[i]表示疾病的集合为i的最多的奶牛数量。

    然后...没有然后了...

    bitset真好用)美滋滋


    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <bitset>
    using namespace std;
    #define gc getchar()
    inline int read(){
        int res=0;char ch=gc;
        while(!isdigit(ch))ch=gc;
        while(isdigit(ch)){res=(res<<3)+(res<<1)+(ch^48);ch=gc;}
        return res;
    }
    int n, m, k;
    int eve[1005];
    int f[1<<16];
    int ans;
    
    int main()
    {
        n = read(), m = read(), k = read();
        for (int i = 1 ; i <= n ; i ++)
        {
            int x = read();
            while(x--)
            {
                int y = read();
                eve[i] |= (1<<(y-1));
            }
        }
        for (int i = (1<<m)-1 ; i >= 0 ; i --)
        {
            for (int j = 1 ; j <= n ; j ++)
            {
                f[i|eve[j]] = max(f[i|eve[j]], f[i] + 1);
            }
        }
        for (int i = 0 ; i <= (1<<m)-1 ; i ++)
        {
            bitset <16> bit = i;
            if (bit.count() > k) continue; 
            ans = max(ans, f[i]);
        }
        cout << ans << endl;
        return 0;
    }
  • 相关阅读:
    Mybatis框架学习_6_mapper.xml 文件中的输入参数详解 (paraterType)
    Mybatis框架学习_5_自定义类型转换器
    Mybatis框架学习_4_属性文件、全局参数、别名
    Mybatis框架学习_3_基于约定或动态代理实现增删改查
    Mybatis框架学习_2_增删改查的简单实现
    Mybatis框架学习_1_简介以及入门示例
    Linux 系统下启动命名的书写过程
    spring-boot-Web学习2-模板引擎 Thymeleaf
    spring-boot-Web学习1-简介
    MacBook无法开机问题
  • 原文地址:https://www.cnblogs.com/BriMon/p/9356988.html
Copyright © 2011-2022 走看看