01背包——HDOJ2602拾骨者

问题陈述：
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

这是一道01背包模板题，很简单。但是开始提交了几次都是Wronganswer。后来在网上一查，原来dp[][]数组j的下标应该从0而非1开始，果然改完后就通过了。
其中for(j=0;j<=M;j++)中j=0时瑟吉欧这样解释的，如果你一个麻袋东西，然后有一个戒指，它的体积可以忽略也0;所以要从0开始，

源代码：

#include
#include
#include
#include

using namespace std;

int dp[1001][1001];
int value[1001];
int weight[1001];
int N, W;
int main(void)
{
    int n;
    scanf("%d", &n);
    while(n--)
    {
        scanf("%d%d", &N, &W);
        for(int i= 1; i <= N; i++)
        {
            scanf("%d", &value[i]);
        }
        for(int i = 1; i <= N; i++)
            scanf("%d", &weight[i]);
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= N; i++)
            for(int j = 0; j <= W; j++)
            {
                if(j < weight[i])
                    dp[i][j] = dp[i-1][j];
                else
                {
                    dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i]);
                }
            }
        printf("%d\n", dp[N][W]);

    }
    return 0;
}