hdu2602 java

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 in = new Scanner(System.in);
		while (in.hasNext()) {
			int q=in.nextInt();
			for (int i1 = 0; i1 < q; i1++) {
				
			
			int n = in.nextInt();
			int m = in.nextInt();
			int v[] = new int[n];
			int w[] = new int[n];
			for (int i = 0; i < w.length; i++) {
				v[i]=in.nextInt();
			}for (int i = 0; i < w.length; i++) {
				w[i]=in.nextInt();
			}
			int dp[][] = new int[n + 1][m + 1];
			for (int i = 0; i < dp.length; i++) {
				dp[i][0] = 0;
			}
			for (int i = 0; i < dp[0].length; i++) {
				dp[0][i] = 0;
			}
			for (int i = 1; i < dp.length; i++) {
				for (int j = 0; j < dp[0].length; j++) {
					if (w[i - 1] <= j) {
						dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - w[i - 1]] + v[i - 1]);
					} else {
						dp[i][j] = dp[i - 1][j];
					}
				}
			}

			System.out.println(dp[n][m]);

		}
		}
	}
}