hdoj 2602（背包）

Problem D

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 9   Accepted Submission(s) : 6

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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

代码：

#include<stdio.h>
#include<string.h>

int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int t,n,m,a[10001],b[10001],dp[10001],i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(i=0;i<n;i++)
        {
            scanf("%d",&b[i]);
        }
        memset(dp,0,sizeof(dp));
        for(i=0;i<n;i++)
        {
            for(j=m;j>=b[i];j--)
            {
                dp[j]=max(dp[j],dp[j-b[i]]+a[i]);
            }
        }
        printf("%d\n",dp[m]);
    }
    return 0;
}