HDOJ 2602 Bone Collector(01背包)

Bone Collector

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

Total Submission(s): 40238    Accepted Submission(s): 16717

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

模板题，代码如下：

<span style="font-size:14px;">#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){int t,n,v,i,j;int f[1010],cos[1010],val[1010];scanf("%d",&t);while(t--){memset(f,0,sizeof(f));scanf("%d%d",&n,&v);for(i=1;i<=n;++i)   scanf("%d",&val[i]);for(i=1;i<=n;++i)   scanf("%d",&cos[i]);for(i=1;i<=n;++i){for(j=v;j>=cos[i];j--){f[j]=max(f[j],f[j-cos[i]]+val[i]);}}printf("%d\n",f[v]);} return 0;}</span>