看上去一个很复杂的问题，解法却出奇的简单，题目描述如下（懒得翻译了……）：

We are given as input a set of n jobs, where job j has a processing time pj and a deadline dj. Recall the definition of completion times 

Cj
 from the video lectures. Given a schedule (i.e., an ordering of the jobs), we define the 
lateness
 

lj
 of job 

j
 as the amount of time 

Cj−dj
 after its deadline that the job completes, or as 0 if 

Cj≤dj
. Our goal is to minimize the maximum lateness, 

maxjlj

. Which of the following greedy rules produces an ordering that minimizes the maximum lateness? You can assume that all processing times and deadlines are distinct

.

只要按照结束时间对任务进行升序排列，得到的便是所求的解，非常简洁的贪心。正确性的证明，可以通过反证法，即假设最优的排列顺序不是升序，那么必定存在一个逆序对，交换这对任务，可以得到更优的解，这样就和假设矛盾了。