杭電oj(Java版)—— 2602 Bone Collector—— 01揹包問題
阿新 • • 發佈:2019-02-09
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 59998 Accepted Submission(s): 25024
Problem Description 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 ?
Input 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.
Output One integer per line representing the maximum of the total value (this number will be less than 231
Sample Input 1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output 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]; //價值 int W = scanner.nextInt(); //包能容納的總重量 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]); } } }