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 =  + ;
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 = ; i <= n; i++) scanf("%d", &wi[i]);
        for(int i = ; i <= n; i++) scanf("%d", &vi[i]);
        memset(dp, , sizeof(dp)); //attention
        for(int i = ; i <= n; i++)
            ZeroOnePack(vi[i], wi[i]);
        printf("%d\n", dp[v]);
    }
    return ;
}