HDU 2602 Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7774 Accepted Submission(s): 2884

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

1

5 10

1 2 3 4 5

5 4 3 2 1

Sample Output

14

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

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#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define max 1010
struct bone 
{
     int ve , value ;       
}num[max] ;
int T , N , V , Mv[max] ; 
int MAX ( int a , int b )
{
    return  a > b ? a : b ;
}

int main ()
{ 
    scanf ( "%d" , &T ) ;
    while ( T -- ) {
          scanf ( "%d%d" , &N , &V ) ; 
          for ( int i = 1 ; i <= N ; i ++ )
              scanf ( "%d" , &num[i].value ) ;
          for ( int i = 1 ; i <= N ; i ++ )
              scanf ( "%d" , &num[i].ve ) ;
          for ( int i = 0 ; i <= V ; i ++ )
              Mv[i] = 0 ;
          for ( int i = 1 ; i <= N ; i ++ ) 
              for ( int j = V ; j >= num[i].ve ; j -- )
                  Mv[j] = MAX ( Mv[j] , Mv[j - num[i].ve] + num[i].value ) ;  // 状态方程 
          printf ( "%d\n" , Mv[V] ) ;
              }
    return 0 ;
}