POJ 2376 Cleaning Shifts

最新推荐文章于 2022-02-03 22:01:25 发布

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本文探讨了如何通过合理分配牛的工作时间来确保农场全天候清洁。目标是最少使用牛的数量，同时保证每个工作时段都有牛在岗。算法首先将牛的工作时间排序，然后逐个检查，确保工作时间无缝对接，实现资源最大化利用。

题目
Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on cleaning things up and has divided the day into T shifts (1 <= T <= 1,000,000), the first being shift 1 and the last being shift T.

Each cow is only available at some interval of times during the day for work on cleaning. Any cow that is selected for cleaning duty will work for the entirety of her interval.

Your job is to help Farmer John assign some cows to shifts so that (i) every shift has at least one cow assigned to it, and (ii) as few cows as possible are involved in cleaning. If it is not possible to assign a cow to each shift, print -1.
Input
Line 1: Two space-separated integers: N and T
Lines 2…N+1: Each line contains the start and end times of the interval during which a cow can work. A cow starts work at the start time and finishes after the end time.
Output
Line 1: The minimum number of cows Farmer John needs to hire or -1 if it is not possible to assign a cow to each shift.
Sample Input
3 10
1 7
3 6
6 10
Sample Output
2

题意：判断能否从头到尾都有牛在工作，如果能，最少要有几头牛

#include <stdio.h>
#include <algorithm>
using namespace std;
int n,t;
struct node  //结构体保存开始时间和结束时间
{
	int s,e;
};
node arr[1000010];
bool cmp(node a,node b)
{
	if(a.s!=b.s) return a.s<b.s;  //按开始时间从小到大排序
	else return a.e<b.e;  //如果开始时间相同就按结束时间排序
}
int main()
{
	scanf("%d%d",&n,&t);
	for(int i=0;i<n;i++)
	    scanf("%d%d",&arr[i].s,&arr[i].e);
	sort(arr,arr+n,cmp);
	if(arr[0].s!=1)  //最小的开始时间不等于1说明牛的工作时间不能覆盖整个时间段
	{
		printf("-1\n");
		return 0;
	}
	int start=1,end=-1,c=1;
	for(int i=0;i<n;i++)
	{
		if(arr[i].s<=start)  //除start==1时，start实质上是上一头牛工作结束时间+1，所以这条语句实质上是判断两头牛的工作时间是否可以衔接上（实际上下面的else if(arr[i].s<=end+1)  中已经判断过）
		{
			if(arr[i].e>end)  //当前牛的工作结束时间大于已存在结束时间时才更新结束时间，防止同一时段有多头牛工作，最大限度利用每头牛的工作时间
			    end=arr[i].e;
			if(end>=t)  //当前结束时间大于规定结束时间时说明已覆盖全部工作时区，结束程序
			{
				printf("%d\n",c);
				return 0;
			}
		}
		else if(arr[i].s<=end+1)  //arr[i].s<=end+1保证两头牛的工作时间可以衔接上
		{
			start=end+1; 
			c++;
			i--;  //执行这条语句后才能依据这头牛的工作结束时间更新整个区间结束时间，否则会未判断这一头牛的结束时间就直接跳到下一头牛，无法更新结束时间
		}  //这个else if 其实在为上面的if做前提工作
		else 
		{
			printf("-1\n");
		    return 0;
		}
	}
	if(end>=t)
	{
		printf("%d\n",c);
		return 0;
	}
	printf("-1\n");
	return 0;
}