HDU2602(背包)

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 52296    Accepted Submission(s): 22040

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

 

裸的01背包

//2016.9.6
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int N = 1005;
int dp[N], c[N], w[N];

int main()
{
    int T, n, v;
    cin>>T;
    while(T--)
    {
        scanf("%d%d", &n, &v);
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i < n; i++)
              scanf("%d", &w[i]);
        for(int i = 0; i < n; i++)
              scanf("%d", &c[i]);
        for(int i = 0; i < n; i++)
              for(int j = v; j >= c[i]; j--)
                  dp[j] = max(dp[j], dp[j-c[i]]+w[i]);
        printf("%d\n", dp[v]);
    }

    return 0;
}