Bone Collector(ZeroOnebag)

时间:2022-02-16 07:30:42

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 ?
Bone Collector(ZeroOnebag)
 
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;
}
}