最新消息:点击查看大S的省钱秘笈

POJ 2376 Cleaning Shifts C++版

POJ题解 Slyar 77浏览 0评论

文章作者:姜南(Slyar) 文章来源:Slyar Home (www.slyar.com) 转载请注明,谢谢合作。

贪心算法

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

3 10
1 7
3 6
6 10

Sample Output

2

Slyar:活动安排问题的进阶版,简化成数学问题就是给你长度为M的区间和N条线段的起点和终点,问你最少需要多少条线段就可以把区间覆盖(线段可以重叠)...

算法不难,按起点排序,起点相同按终点排,反正就是找符合条件的线段里覆盖区间最长的就好...不过陷阱N多,哥没注意一下WA了N-1次...排序后第一个起始点必须是1,有可能出现只有一条线段就完全覆盖的情况等等...还有,丫的数据太大不能用stream,必须用stdio...

转载请注明:Slyar Home » POJ 2376 Cleaning Shifts C++版

发表我的评论
取消评论

表情

Hi,您需要填写昵称和邮箱!

  • 昵称 (必填)
  • 邮箱 (必填)
  • 网址

网友最新评论 (1)

  1. line 64: array[i].start-1 <= start 为什么不是array[i].start <= start?
    anoy5年前 (2012-05-19)回复