杭电oj（Java版）2602 Bone Collector 01背包问题

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 60730    Accepted Submission(s): 25315

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

import java.util.Scanner;public class Main{   public static void main(String[] args){   Scanner sc=new Scanner(System.in);   int nums=sc.nextInt();   for(int i=0;i<nums;i++){   int count=sc.nextInt();   int weight=sc.nextInt();   int[] weigh=new int[count];   int[] value=new int[count];   for(int j=0;j<count;j++){   value[j]=sc.nextInt();   }   for(int j=0;j<count;j++){   weigh[j]=sc.nextInt();   }   backBag(weigh,value,weight);   }   } public static void backBag(int[] weigh, int[] value, int weight) { int[] dp=new int[weight+1]; for(int i=0;i<weigh.length;i++){ for(int j=weight;j>=weigh[i];j--){ dp[j]=Math.max(dp[j], dp[j-weigh[i]]+value[i]); } }System.out.println(dp[weight]);}}