【USACO1.2_1】★Milking Cows 挤牛奶

三个农民每天清晨5 点起床,然后去牛棚给3 头牛挤奶.第一个农民在300 时刻(从5 点开始计时,

秒为单位)给他的牛挤奶,一直到1000 时刻.第二个农民在700 时刻开始,在 1200 时刻结束.第三个

农民在1500 时刻开始2100 时刻结束.期间最长的至少有一个农民在挤奶的连续时间为900 秒(从

300 时刻到1200 时刻),而最长的无人挤奶的连续时间(从挤奶开始一直到挤奶结束)为300 秒(从

1200 时刻到1500 时刻).

你的任务是编一个程序,读入一个有N 个农民(1 <= N <= 5000)挤N 头牛的工作时间列表,计算以下

两点(均以秒为单位):

• 最长至少有一人在挤奶的时间段.

• 最长的无人挤奶的时间段.

PROGRAM NAME: milk2

INPUT FORMAT

Line 1: 一个整数N.

Lines 2..N+1: 每行两个小于1000000 的非负整数,表示一个农民的开始时刻与结束时刻.

SAMPLE INPUT (file milk2.in)

3

300 1000

700 1200

1500 2100

OUTPUT FORMAT

一行,两个整数,即题目所要求的两个答案.

SAMPLE OUTPUT (file milk2.out)

900 300

/*
ID: letgoto1
PROG: milk2
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string>
#include<algorithm>

using namespace std;

const int N = 5000;

struct Milk
{
Milk():start(0), end(0){}
long start, end;
};

int cmp(const void *a, const void *b)
{
Milk *m1 = (Milk *)a;
Milk *m2 = (Milk *)b;
return m1->start - m2->start;
}

Milk milk
;

int main()
{
ofstream fout ("milk2.out");
ifstream fin ("milk2.in");

unsigned n = 0; //number of people
fin >> n;

//read the data
int i = 0;
while(!fin.eof())
{
fin >> milk[i].start >> milk[i].end;
i++;
}
qsort(milk, n, sizeof(Milk), cmp);

long start = milk[0].start, end = milk[0].end;
long tOne = end - start, tNone = 0; //tOne is the time of at least one person milking
for(i = 1; i < n; i++)
{
if(milk[i].start <= end)
{end = max(milk[i].end, end);}
else
{
if(milk[i].start - end > tNone)
{tNone = milk[i].start - end;}

if(end - start > tOne)
{tOne = end - start;}

start = milk[i].start;
end = milk[i].end;
}
}

fout << tOne << " " << tNone;

fout <<endl;
return 0;
}