POJ 2376 Cleaning Shifts （贪心）

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本文介绍了一个经典的清扫任务调度问题，目标是最少使用多少头牛来确保每个清扫时间段都有牛进行清扫工作。通过输入牛的工作时间区间，利用算法进行最优调度。

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Cleaning Shifts

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17698		Accepted: 4501

Description

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

Sample Output

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

INPUT DETAILS:

There are 3 cows and 10 shifts. Cow #1 can work shifts 1..7, cow #2 can work shifts 3..6, and cow #3 can work shifts 6..10.

OUTPUT DETAILS:

By selecting cows #1 and #3, all shifts are covered. There is no way to cover all the shifts using fewer than 2 cows.

Source

USACO 2004 December Silver

题意：农夫想让奶牛去打扫仓库，将打扫仓库的时间分为从1~T个时间段（注意是时间段，不是时间点），一共有n头牛，每个牛都有自己能工作的时间段（给出开始时间段~结束时间段），要求在T个时间段内每个时间段必须至少有一头牛在打扫仓库，问最少需要安排几头牛，若不能分配牛到每一个时间段，则输出-1。

思路：先排序，第一个点肯定选，依次选择后面的点。选择点的时候，选最优的，即在符合条件的范围内，选择一个结束点最大的。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
	int begin,end;
}t[25100];
bool cmp(node a,node b)
{
	if(a.begin==b.begin)
	return a.end<b.end;
	return a.begin<b.begin;
}
int main()
{
	int n,m,i,j;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i<n;i++)
		scanf("%d%d",&t[i].begin,&t[i].end);
		sort(t,t+n,cmp);
		if(t[0].begin>1)//开始时不能覆盖 
		{
			printf("-1\n");
			continue;
		}
		int ans=1;
		int now=0;
		for(i=0;i<n;)
		{
			int cnt=0;
			for(j=i;j<n;j++)
			{
				if(t[j].begin>t[now].end+1)//中间不能覆盖 
				break;
				if(t[j].begin>=t[now].begin&&t[j].end>=t[now].end+1)
				{
					if(t[j].end>t[cnt].end)//找最优的，即结束时间最大的 
					cnt=j;
				}
			}
			if(cnt==0)
			i++;
			else
			{
				ans++;
				now=cnt;
				i=now;
			}
		}
		if(t[now].end==m)
		printf("%d\n",ans);
		else
		printf("-1\n");
	}
	return 0;
}