HDU 2602 Bone Collector（01背包）

標籤：acm   dp   hdu   01背包   

Bone CollectorProblem DescriptionMany 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 ?

 
InputThe 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. 
OutputOne integer per line representing the maximum of the total value (this number will be less than 231). 
Sample Input

15 101 2 3 4 55 4 3 2 1

 
Sample Output

14

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#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 1005;int n, v, d[N], val[N], vol[N], cas;int main(){    scanf ("%d", &cas);    while (cas--)    {        memset (d, 0, sizeof (d));        scanf ("%d%d", &n, &v);        for (int i = 1; i <= n; ++i)            scanf ("%d", &val[i]);        for (int i = 1; i <= n; ++i)        {            scanf ("%d", &vol[i]);            for (int j = v; j >= vol[i]; --j)                d[j] = max (d[j], d[j - vol[i]] + val[i]);        }        printf ("%d\n", d[v]);    }    return 0;}