Bone Collector（ZeroOnebag）

Bone Collector

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

1

5 10

1 2 3 4 5

5 4 3 2 1

 

Sample Output

14

　模板题，找到状态转移方程，使用一维01背包，对于背包的体积要逆序遍历，这样是的右边的max(f[j],f[j-v[i]]+val[i])能够表示为上一层的状态，而对于当j<v[i时，默认f[j]不变，是上一层的值

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
int main()
{
    int T,V;
    int v[1001],val[1001];
    int f[1001];
    int i,N,j;
    cin>>T;
    while(T--)
    {
        memset(f,0,sizeof(f));
        cin>>N>>V;
        for(i=1;i<=N;i++)
            cin>>val[i];
        for(i=1;i<=N;i++)
            cin>>v[i];
        for(i=1;i<=N;i++)
        {
            for(j=V;j>=v[i];j--)
                f[j]=max(f[j],f[j-v[i]]+val[i]);
        }
        cout<<f[V]<<endl;
    }
}