杭电oj(Java版)—— 2602 Bone Collector—— 01背包问题
2017-04-26 22:09
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Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 59998 Accepted Submission(s): 25024
[align=left]Problem Description[/align]
Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the
maximum of the total value the bone collector can get ?
![](http://acm.hdu.edu.cn/data/images/C154-1003-1.jpg)
[align=left]Input[/align]
The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third
line contain N integers representing the volume of each bone.
[align=left]Output[/align]
One integer per line representing the maximum of the total value (this number will be less than 231).
[align=left]Sample Input[/align]
1
5 10
1 2 3 4 5
5 4 3 2 1
[align=left]Sample Output[/align]
14
import java.util.Scanner; public class Main{ public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int n = scanner.nextInt(); for (int i=0;i<n;i++) { int a = scanner.nextInt(); int v[] = new int[a]; //重量 int w[] = new int[a]; //价值 int W = scanner.nextInt(); //包能容纳的总重量 long dp[] = new long[1005]; for (int j=0;j<a;j++) { w[j] = scanner.nextInt(); } for (int j=0;j<a;j++) { v[j] = scanner.nextInt(); } for (int j=0;j<a;j++) { for (int k=W;k>=0;k--) { if (v[j]<=k) { dp[k] = Math.max(dp[k], dp[k-v[j]]+w[j]); } } } System.out.println(dp[W]); } } }
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