动态规划F

01背包问题
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 31).
Sample Input

15 101 2 3 4 55 4 3 2 1

Sample Output

14

import java.util.ArrayList;import java.util.Scanner;/** * Created by 95112 on 10/24/2017. */public class BoneCollector {    public static void main(String[] args)    {        long[] dp;        Scanner scanner = new Scanner(System.in);        int T = scanner.nextInt();        ArrayList<Long> answers = new ArrayList<>();        while (T-- >0){            int N,V;            N = scanner.nextInt();            V = scanner.nextInt();            dp = new long[V+1];            dp[0] = 0;            int[] boneValue = new int[N];            int[] boneVolume = new int[N];            for (int i = 0 ; i< N;i++)            {                boneValue[i] = scanner.nextInt();            }            for (int i = 0 ;i < N;i++)            {                boneVolume[i] = scanner.nextInt();            }            for (int i = 0 ; i < N; i++){                for (int j=V ; j >= 0 ;j--){                    if (boneVolume[i] <= j)                        dp[j] = max(dp[j],dp[j - boneVolume[i]]+boneValue[i]);                    else                        break;                }            }            answers.add(dp[V]);        }        for (Long answer : answers){            System.out.println(answer);        }    }    public static long max(long a, long b){        if (a>b)            return  a;        else                return b;    }}