Find non-overlap jobs with max cost

　　

Given a set of n jobs with [start time, end time, cost] find a subset so that no 2 jobs overlap and the cost is maximum.
Job: （start_time, end_time] --- cost

如果只是求maxCost, 一维就可以做。

但是如果要知道有选了哪些job，则需要存成二维。

 

 1 package leetcode;
 2 
 3 import java.util.ArrayList;
 4 import java.util.Arrays;
 5 import java.util.Comparator;
 6 
 7 class Job{
 8     Integer start_time;
 9     Integer end_time;
10     Integer cost;
11     public Job(Integer s, Integer e, Integer c){
12         start_time = s;
13         end_time = e;
14         cost = c;
15     }
16     
17     public String toString(){
18         StringBuilder sb = new StringBuilder();
19         sb.append("Job: start [" + start_time + "], end ["+ end_time + "], cost [" + cost + "];");
20         return sb.toString();
21     }
22 }
23 
24 public class FindNonOverlapJobs {
25     public static ArrayList<Job> findJobsWithMaxCost(Job[] jobList){
26         ArrayList<Job> result = new ArrayList<Job> ();
27         if(jobList == null || jobList.length == 0) return result;
28         Arrays.sort(jobList, new Comparator<Job>(){
29             public int compare(Job j1, Job j2){
30                 return j1.end_time > j2.end_time ? 1 : (j1.end_time == j2.end_time ? 0 : -1);
31             }
32         });
33         int len = jobList.length;
34         int[][] dp = new int[len + 1][jobList[len - 1].end_time + 1];
35         for(int i = 1; i <= len; i ++){
36             Job tmp = jobList[i - 1];
37             int start = tmp.start_time;
38             int end = tmp.end_time;
39             for(int j = 0; j < dp[0].length; j ++){
40                 if(j < end){
41                     dp[i][j] = dp[i - 1][j];
42                 }else if(j == end){
43                     dp[i][j] = Math.max(dp[i - 1][start] + tmp.cost, dp[i - 1][j]);
44                 }else{
45                     dp[i][j] = dp[i][j - 1];
46                 }
47             }
48         }
49 
50         int i = dp[0].length - 1;
51         while(i > 0){
52             if(dp[len][i] == dp[len][i - 1]) i --;
53             else{
54                 int j = len;
55                 while(j > 0 && dp[j][i] == dp[j - 1][i]) j --;
56                 result.add(jobList[j - 1]);
57                 i --;
58             }
59         }
60         return result;
61     }
62     
63 public static void main(String[] args){
64     Job[] test = new Job[5];
65     test[0] = new Job(1,3,4);
66     test[1] = new Job(3,5,2);
67     test[2] = new Job(2,3,3);
68     test[3] = new Job(1,2,2);
69     test[4] = new Job(2,6,3);
70     ArrayList<Job> result = findJobsWithMaxCost(test);
71     for(int i = 0; i < result.size(); i ++){
72         System.out.println(result.get(i).toString());
73     }
74 }
75 }

Output:

Job: start [3], end [5], cost [2];

Job: start [2], end [3], cost [3];

Job: start [1], end [2], cost [2];

 

posted on 2015-04-03 15:15  Step-BY-Step  阅读(301)  评论(0)  编辑  收藏  举报

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