hdu 2602 Bone Collector 背包入门题

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602

题目分析：0-1背包  注意dp数组的清空， 二维转化为一维后的公式变化

/*Bone Collector



Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 34192    Accepted Submission(s): 14066





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

 



Author

Teddy

 



Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

*/

//ZeroOnePack

#include <cstdio>

#include <cstring>

const int maxn = 1000 + 10;

int dp[maxn], n, v, wi[maxn], vi[maxn];

int Max(int a, int b)

{

    return a > b ? a : b;

}

void ZeroOnePack(int C, int W)

{

    for(int i = v; i >= C; i--)

        dp[i] = Max(dp[i], dp[i-C]+W);

}



int main()

{

    int T;

    scanf("%d", &T);

    while(T--){

        scanf("%d%d", &n, &v);

        for(int i = 1; i <= n; i++) scanf("%d", &wi[i]);

        for(int i = 1; i <= n; i++) scanf("%d", &vi[i]);

        memset(dp, 0, sizeof(dp)); //attention

        for(int i = 1; i <= n; i++)

            ZeroOnePack(vi[i], wi[i]);

        printf("%d\n", dp[v]);

    } 

    return 0;

}