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[],val[];
     int f[];
     int i,N,j;
     cin>>T;
     while(T--)
     {
         memset(f,,sizeof(f));
         cin>>N>>V;
         for(i=;i<=N;i++)
             cin>>val[i];
         for(i=;i<=N;i++)
             cin>>v[i];
         for(i=;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;
     }
 }