HDU2602 DP + 裸 + 01背包

1）

#include <iostream>
#include <string.h>

using namespace std;
const int maxn=1010;
int value[maxn];
int volume[maxn];
long long int dp[maxn];
int main()
{
    int kase;
    cin>>kase;
    while(kase--){
        memset(dp,0,sizeof(dp));
        int n,v;
        cin>>n>>v;
        for(int i=1;i<=n;i++){
            cin>>value[i];
        }
        for(int i=1;i<=n;i++){
            cin>>volume[i];
        }
    
        for(int i=1;i<=n;i++){
            for(int j=v;j>=volume[i];j--){
                dp[j]=max(dp[j],dp[j-volume[i]]+value[i]);
            }
        }

        cout<<dp[v]<<endl;
    }
}

2）

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 2 ³¹).

 

Sample Input

     
     
     
     

      
      
      
      1
5 10
1 2 3 4 5
5 4 3 2 1 
     
     
     
     

 

Sample Output

   
   
   
   

    
    
    
    14