杭電 2602 Bone Collector

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Bone Collector Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 27413    Accepted Submission(s): 11154


Problem 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

01背包。

AC代碼例如以下：

#include<iostream>#include<cstring>using namespace std;int main(){    int t,n,v;    int i,j;    int a[1005],b[1005],dp[1005];    cin>>t;    while(t--)    {        cin>>n>>v;        for(i=1;i<=n;i++)            cin>>a[i];        for(i=1;i<=n;i++)            cin>>b[i];        memset(dp,0,sizeof dp);        for(i=1;i<=n;i++)            for(j=v;j>=b[i];j--)            dp[j]=max(dp[j-b[i]]+a[i],dp[j]);            cout<<dp[v]<<endl;    }    return 0;}