hdu 2602 Bone Collector (01背包)

Bone Collector
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 84339    Accepted Submission(s): 34917

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

C/C++:

#include <bits/stdc++.h>

using namespace std;

const int MAX = 1010;

int main()
{
    int t;
    scanf("%d", &t);
    while (t --)
    {
        int n, v, dp[MAX] = {0}, val[MAX], vol[MAX];
        scanf("%d%d", &n, &v);
        for (int i = 0; i < n; ++ i) scanf("%d", &val[i]);
        for (int i = 0; i < n; ++ i) scanf("%d", &vol[i]);

        for (int i = 0; i < n; ++ i)
            for (int j = v; j >= vol[i]; -- j)
                dp[j] = max(dp[j], dp[j - vol[i]] + val[i]);
        printf("%d
", dp[v]);
    }
    return 0;
}