HDU2602——Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9023    Accepted Submission(s): 3467

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

 

 

 

 

 

基本的01背包问题，状态方程f[v]=max{f[v],f[v-c[i]]}

 

代码：

#include <stdio.h>
#include <string.h>
#define N 1001

struct _bone
{
    int value;
    int volume;
}bone[N];

int f[N];

int max(int a, int b)
{
    return a > b ? a : b;
}
int main()
{
    int t, n, v;
    int i, j, k;
    scanf("%d", &t);
    for(k=0;k<t;k++)
    {
        memset(f, 0, sizeof(f));
        scanf("%d%d", &n, &v);
        for(i=0;i<n;i++)
        {
            scanf("%d", &bone[i].value);
        }
        for(i=0;i<n;i++)
        {
            scanf("%d", &bone[i].volume);
        }
        for(i=0;i<n;i++)
        {
            for(j=v;j>=0;j--)
            {
                if(j >= bone[i].volume)
                {
                    f[j] = max(f[j], f[j-bone[i].volume]+bone[i].value);
                }
            }
        }
        printf("%d\n", f[v]);
    }
    return 0;
}