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 ?![](https://www.cnblogs.com/data/images/C154-1003-1.jpg)
![](https://www.cnblogs.com/data/images/C154-1003-1.jpg)
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
#include<iostream>
#include<string.h>
using namespace std;
int p[1005];
int val[1005];
int vol[1005];
int mx(int a,int b)
{
return a>b?a:b;
}
int main()
{
int test,i,n,v,j;
cin>>test;
while(test--)
{
cin>>n>>v;
for(i=0;i<n;i++)cin>>val[i];
for(i=0;i<n;i++) cin>>vol[i];
for(i=0;i<=v;i++) p[i]=0;
for(i=0;i<n;i++)
{
for(j=v;j>=vol[i];j--)
{
p[j]=mx(p[j],p[j-vol[i]]+val[i]);
}
}
cout<<p[v]<<endl;
}
}
#include<string.h>
using namespace std;
int p[1005];
int val[1005];
int vol[1005];
int mx(int a,int b)
{
return a>b?a:b;
}
int main()
{
int test,i,n,v,j;
cin>>test;
while(test--)
{
cin>>n>>v;
for(i=0;i<n;i++)cin>>val[i];
for(i=0;i<n;i++) cin>>vol[i];
for(i=0;i<=v;i++) p[i]=0;
for(i=0;i<n;i++)
{
for(j=v;j>=vol[i];j--)
{
p[j]=mx(p[j],p[j-vol[i]]+val[i]);
}
}
cout<<p[v]<<endl;
}
}
为了明天所以选择坚定的执着今天。