第四周1001题解

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 2[sup]31[/sup]).

Sample Input

1

5 10

1 2 3 4 5

5 4 3 2 1

Sample Output

14

Author

Teddy

import java.util.Scanner;

public class Main1001 {
//  public static int value[];
//  public static int volume[];
//  public static int res[][];
//
//public static int getDp(int N,int V){//递归dp
//  if(N==0)
//      return 0;
//      res
[1] = value
+getDp(N-1,V-volume
);
//      res
[0] =getDp(N-1,V);
//      return Math.max(res
[1], res
[0]);
//}

public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int T = sc.nextInt();
for (int x = 0; x < T; x++) {
int N = sc.nextInt();// 宝石数量
int V = sc.nextInt();// 背包容量
int []value = new int
;// 宝石的价值
int []volume = new int
;// 宝石的体积
int [][]res = new int[1001][1001];
for(int i = 0;i<N;i++){
value[i] = sc.nextInt();
}
for(int i = 0;i<N;i++){
volume[i] = sc.nextInt();
}
for(int i=0; i<=N; ++i){
for(int j=0; j<=V; ++j){
res[i][j] = i==0 ? 0 : res[i-1][j];//假设不取
if(i>0 && j>=volume[i-1]){
if(res[i][j]<res[i-1][j-volume[i-1]]+value[i-1])
res[i][j] = res[i-1][j-volume[i-1]]+value[i-1];
}
}
}

//          System.out.println(getDp(N-1, V));
System.out.println(res
[V]);
}
}
}