Bone Collector（杭电2602）（01背包）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31604    Accepted Submission(s): 13005

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 2 ³¹).

Sample Input

1 5 10 1 2 3 4 5 5 4 3 2 1

Sample Output

14

/*简单背包问题*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
  int test,n,v,i,j;
  int a[1100],s[1100],dp[1100];
  scanf("%d",&test);
  while(test--){
    memset(dp,0,sizeof(dp));
  scanf("%d %d",&n,&v);
  for(i=1;i<=n;i++)
  {
    scanf("%d",&a[i]);
  }
  for(i=1;i<=n;i++)
  {
    scanf("%d",&s[i]);
  }
  for(i=1;i<=n;i++)
  {
    for(j=v;j>=s[i];j--)
    {
      dp[j]=max(dp[j],dp[j-s[i]]+a[i]);
    }
  }
  printf("%d\n",dp[v]);}
  return 0;
}