Milking Cows

最新推荐文章于 2019-03-01 20:55:57 发布

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Problem:

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)

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

My Answer:

/*
ID:Iby07
PROG:milk2
LANG:C++
*/

# include<iostream>
# include<fstream>
using namespace std;

struct Time
{	int start;
	int end;
};

int divide(Time a[],int low,int high)						//测试数据不一定都跟已给出的这一组似的，按开始工作时间顺序，所以写快排函数
{	int k=a[low].start,l=a[low].end;
	do
	{	while(low<high && a[high].start>=k) --high;
		if(low<high) 
			{a[low].start=a[high].start; a[low].end=a[high].end; ++low;}
		while(low<high && a[low].start<=k) ++low;
		if(low<high)
			{a[high].start=a[low].start; a[high].end=a[low].end; --high;}
	}while(low!=high);
	a[low].start=k;
	a[low].end=l;
	return low;
}

void quicksort(Time a[],int low,int high)
{	int mid;
	if(low>=high) return;
	mid=divide(a,low,high);
	quicksort(a,low,mid-1);
	quicksort(a,mid+1,high);
}

int main ()
{	ifstream fin("milk2.in");
	ofstream fout("milk2.out");
	int n,max1=0,max2=0,low,high,tmp,interval;
	fin>>n;
	Time *time=new Time [n];

	for(int i=0;i<n;++i)
		fin>>time[i].start>>time[i].end;

	quicksort(time,0,n-1);

	low=time[0].start;
	high=time[0].end;
	max1=high-low;
	for(int j=1;j<n;++j)
	{	if(time[j].start<=high && time[j].end>high)			//上一个工人尚在工作，后一个工人已经加入
		{	high=time[j].end; 
			tmp=high-low;
			max1=(tmp>max1)?tmp:max1;
		}
		if(time[j].start>high)								//直到上个工人离开，下一个工人还没来
		{	interval=time[j].start-high;
			max2=(interval>max2)?interval:max2;
			low=time[j].start;								
			high=time[j].end;
			tmp=high-low;
			max1=(tmp>max1)?tmp:max1;						
		}
	}

	fout<<max1<<' '<<max2<<endl;

	return 0;
}