[Usaco2005 Open]Disease Manangement 疾病管理|状态压缩动态规划
2015-08-30 22:34
441 查看
1688: [Usaco2005 Open]Disease Manangement 疾病管理
Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 427 Solved: 285
[Submit][Status][Discuss]
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 20---------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
5OUTPUT 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).
很简单的题目,二进制表示有没有病,或运算的背包dp。最后判断。
#include<bits/stdc++.h> using namespace std; int n,d,k,ans,a[1005],f[65538]; inline int read() { int a=0,f=1; char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();} while (c>='0'&&c<='9') {a=a*10+c-'0'; c=getchar();} return a*f; } inline bool judge(int x) { int sum=0; for (int i=1;i<=d;i++) if ((1<<(i-1))&x) { sum++; if (sum>k) return 0; } return 1; } int main() { n=read(); d=read(); k=read(); for (int i=1;i<=n;i++) { int x=read(),y; for (int j=1;j<=x;j++) y=read(),a[i]+=(1<<(y-1)); } for (int i=1;i<=n;i++) for (int j=(1<<d)-1;j>=0;j--) f[a[i]|j]=max(f[a[i]|j],f[j]+1); for (int i=0;i<=(1<<d)-1;i++) if (judge(i)) ans=max(ans,f[i]); printf("%d",ans); return 0; }
相关文章推荐
- fopen使用
- centOS 6.3 上配置SVN服务器
- centOS 6.3 上配置SVN服务器
- Linux C语言程序设计(十四)——文件系统与I/O
- nginx standard configure
- Linux之scp报错两例
- 分布式架构的演进
- linux mysql 远程访问权限问题
- Linux下播放rmvb的问题解决
- linux 的shell处理两Excel的比较方法
- poj 2112 Optimal Milking (二分图匹配的多重匹配)
- Hadoop MapReduce核心技术浅析-----RPC框架解析
- 总结一些it学习的网站
- 程序三层架构
- redhat下安装chrome
- linux系统常用命令
- shell脚本那点事儿6
- php linux部署相关
- 初识NSQ分布式实时消息架构
- linux学习笔记2---gstreamer1.0 安装与简介