[Coding Made Simple] Weighted Job Scheduling

Given certain jobs with start and end time and amount you make on finishing the job, find the maximum value you can make by scheduling jobs in non-overlapping way.

 

Dynamic programming solution.

First sort all the jobs in ascending order based on their end time. This makes checking if two jobs overlap or not easier. 

State: maxGain[i]: Given the first i jobs, the max profit we can get by selecting jobs[i]. 

Function: maxGain[i] = max {maxGain[i], maxGain[j] + jobs[i].gain}, for j in [0, i)

Initialization: maxGain[i] = jobs[i].gain, as we know by selecting jobs[i], we at least have a profit of jobs[i].gain.

Answer:  the max value of maxGain[i]. 

 

 1 import java.util.Arrays;
 2 
 3 class Job implements Comparable<Job>{
 4     int start;
 5     int end;
 6     int gain;
 7     Job(int start, int end, int gain) {
 8         this.start = start;
 9         this.end = end;
10         this.gain = gain;
11     }
12     public int compareTo(Job o) {
13         return this.end - o.end;
14     }
15 }
16 public class WeightedJobScheduling {
17     public int getMaxGain(Job[] jobs) {
18         Arrays.sort(jobs);
19         int[] maxGain = new int[jobs.length];
20         for(int i = 0; i < maxGain.length; i++) {
21             maxGain[i] = jobs[i].gain;
22         }
23         for(int i = 1; i < maxGain.length; i++) {
24             for(int j = 0; j < i; j++) {
25                 if(jobs[i].start >= jobs[j].end) {
26                     maxGain[i] = Math.max(maxGain[i], maxGain[j] + jobs[i].gain);
27                 }
28             }
29         }
30         
31         int max = Integer.MIN_VALUE;
32         for(int i = 0; i < maxGain.length; i++) {
33             if(maxGain[i] > max) {
34                 max = maxGain[i];
35             }
36         }
37         return max;
38     }
39 }

 

转载于:https://www.cnblogs.com/lz87/p/7288799.html