[背包问题][第三阶段-初见dp][HDU-2504]Bone Collector

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) {// TODO Auto-generated method stubScanner in = new Scanner(System.in);int T = in.nextInt();while(T>0){int N = in.nextInt();int V = in.nextInt();int[] value = new int[N+1];int[] volume = new int[N+1];int[][] sum = new int[N+10][V+10];int i,j;for(i=1;i<N+1;i++)value[i]=in.nextInt();for(i=1;i<N+1;i++)volume[i]=in.nextInt();   for(j=0;j<=V;j++)   for(i=1;i<=N;i++){   if(j<volume[i]){   sum[i][j]=sum[i-1][j];   continue;   }else if(sum[i-1][j-volume[i]]+value[i]>sum[i-1][j])   sum[i][j]=sum[i-1][j-volume[i]]+value[i];   else   sum[i][j]=sum[i-1][j];   }System.out.println(sum[N][V]);T--;}}}

一定要从0开始检索

因为有下面这组变态的数据

1
2     0
20   1
0     1

答案是： 20