Milking Cows
来源:互联网 发布:宝宝合成照片软件 编辑:程序博客网 时间:2024/04/28 09:50
Milking Cows
Three farmers rise at 5 am each morning and head for the barn to milk three cows. The first farmer begins milking his cow at time 300 (measured in seconds after 5 am) and ends at time 1000. The second farmer begins at time 700 and ends at time 1200. The third farmer begins at time 1500 and ends at time 2100. The longest continuous time during which at least one farmer was milking a cow was 900 seconds (from 300 to 1200). The longest time no milking was done, between the beginning and the ending of all milking, was 300 seconds (1500 minus 1200).Your job is to write a program that will examine a list of beginning and ending times for N (1 <= N <= 5000) farmers milking N cows and compute (in seconds):
The longest time interval at least one cow was milked.
The longest time interval (after milking starts) during which no cows were being milked.
PROGRAM NAME: milk2
INPUT FORMAT
Line 1: The single integer
Lines 2..N+1: Two non-negative integers less than 1000000, the starting and ending time in seconds after 0500
SAMPLE INPUT (file milk2.in)
3
300 1000
700 1200
1500 2100
OUTPUT FORMAT
A single line with two integers that represent the longest continuous time of milking and the longest idle time.
SAMPLE OUTPUT (file milk2.out)
900 300
/*ID: des_jas1PROG: milk2LANG: C++*/#include <iostream>#include <fstream>#include <string.h>#include <algorithm>//#define fin cin//#define fout coutusing namespace std;const int MAXN=5000+5;int feedl=0,nofeedl=0;struct TT{int begin;int end;}tt[MAXN];bool cmp(TT a,TT b){return a.begin<b.begin;}int main() {ofstream fout ("milk2.out"); ifstream fin ("milk2.in");int N;fin>>N;int i;for(i=1;i<=N;i++)fin>>tt[i].begin>>tt[i].end;sort(tt+1,tt+N+1,cmp);//排个顺,按begin的升序,就只要讨论endint tp;int st,ed;for(i=1;i<=N;i++){st=tt[i].begin;ed=tt[i].end;while(tt[i].end>=tt[i+1].begin && i<N){tt[i+1].end=ed>tt[i+1].end?ed:tt[i+1].end; //如果可以,把每个farmer的工作时间“延长”,为了判断下一个是否ed=tt[i+1].end;//实时更新每连续时间段的最终end时间i++;}tp=ed-st;feedl=feedl>tp?feedl:tp;if(i<N){tp=tt[i+1].begin-tt[i].end; if(tp>nofeedl)nofeedl=tp;}}fout<<feedl<<" "<<nofeedl<<endl;fout.close();fin.close(); return 0;}
- Milking Cows
- Milking Cows
- Milking Cows
- Milking Cows
- Milking Cows
- milking cows
- milking cows
- Milking Cows
- Milking Cows
- Milking Cows
- Milking Cows
- Milking cows
- Milking Cows
- Milking Cows
- Milking Cows
- Milking Cows
- Chap1 Milking Cows
- Milking Cows【转帖】
- 顺序栈的设计与实现
- Webkit中HTML5 Video的实现分析(六) - Safari视频机制分析
- 安装QTP吐血过程
- poj3628
- 链栈的设计与实现
- Milking Cows
- HP官方提供的LR在线基础视频教程
- 滑动取值器
- Java中executeBatch()返回值为-2
- 一元多项式相加
- 硬盘概述
- 用Java操作数据库和excel(一)
- 循环队列的设计与实现
- Cocos2d-x在vs2012下环境部署 (主要解决InstallWizardForVS2012.js问题)