항전 OJ2602------Bone Collector-------01 가방

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 2
31).
 
Sample Input

   
   
   
   

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

 
Sample Output

   
   
   
   

    
    
    
    14
   
   
   
   

가장 간단한 01 가방은 최적화된 1차원 그룹 해법을 사용합니다.
중요한 단계는

for i=1..N

 for v=V..0

        f[v]=max{f[v],f[v-c[i]]+w[i]};

dev에서는 컴파일할 수 있지만 OJ에서는 안 되기 때문에 그룹 정의를 동적 그룹으로 바꿉니다.
코드:

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		int N,V;
		cin>>N>>V;
		vector<int>w(N+1);
		vector<int>c(N+1);
		vector<int>f(V+1,0);
		for(int i=1;i<=N;i++)
			cin>>w[i];    //  
		for(int i=1;i<=N;i++)
			cin>>c[i];    // 		
		for(int i=1;i<=N;i++)
			for(int v=V;v-c[i]>=0;v--)
			{
				f[v]=max(f[v],f[v-c[i]]+w[i]);
			} 
		cout<<f[V]<<endl;		
	}	
	return 0;
}