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 ?
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.
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
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) {// 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];elsesum[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