activity select problem(greedy algorithms)

Posted Wujunde

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了activity select problem(greedy algorithms)相关的知识,希望对你有一定的参考价值。

many activities will use the same place, every activity ai has its‘  start time si and finish time fi.let the number of activities to be as many as possible.

1. dynamic programming

use ak be a knife to cut the set activities into two parts and recursive to find the max subset

c[i,j](star after ai finish and finish before aj star) = max {1+c[i,k] + c[k,j]} or 0(haven‘t ak);

2.greedy programming

let ai ranked by their finish time. earlier finish time ranked front than the later.

then choose the activities by its finish time, keep they are not contradictory.

 1 public class activity_select {
 2      int[] s = {1,3,0,5,3,5,6,8,8,2,12};
 3     int[] f = {4,5,6,7,9,9,10,11,12,14,16};
 4     private static class activity{
 5         private int sta ;
 6         private int fin ;
 7         public activity(){
 8             sta = 0;
 9             fin = 0;
10         }
11     }
12 
13     public   activity[] select(){
14     activity[]  act = new activity[s.length];
15     for(int i = 0;i<s.length;i++){   //initial
16         act[i] = new activity();
17         act[i].sta = s[i];
18         act[i].fin = f[i];
19     }    
20     for(int i = 0;i<s.length;i++){   //insert sort from early fin to later fin
21         for(int j = i;j < s.length;j++){
22             if(act[i].fin > act[j].fin){
23                 int testa = act[j].sta;
24                 int tefin = act[j].fin;
25                 act[j].sta = act[i].sta;
26                 act[j].fin = act[i].fin;
27                 act[i].fin = tefin;
28                 act[i].sta = testa;
29             }
30         }
31     }    
32     activity[] res = new activity[s.length];
33     res[0] = act[0];
34     int j = 0;
35     for(int i = 0;i < s.length -1;i++){
36         if(act[i+1].sta > res[j].fin){
37             res[++j] = act[i + 1];
38         }
39     }
40     activity[] res1 = new activity[j+1];
41     for(int i = 0;i <=j;i++){
42         res1[i] = res[i];
43     }
44     return res1;
45     }
46     
47     
48 
49     public static void main(String[] args){
50         activity_select  ac = new activity_select();
51         activity[] a = ac.select();
52         int n = a.length;
53         for(int i = 0;i < n;i++){
54             System.out.println(a[i].sta + " " +a[i].fin);
55         }
56     }
57     
58     
59 }

 

以上是关于activity select problem(greedy algorithms)的主要内容,如果未能解决你的问题,请参考以下文章

G: A Towering Problem

HDU4291—A Short problem

G - Mike and gcd problem

实验12:Problem G: 强悍的矩阵运算来了

Problem G: 圆周率

Activiti_Activiti核心数据表的作用