0-1背包问题(经典)HDU2602 Bone Collector
2024-10-21 20:32:58
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 36479 Accepted Submission(s): 15052
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 ?
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.
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
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
Author
Teddy
Source
AC code:
#include <iostream>
using namespace std;
int main()
{
int t,n,v,i,j;
int weight[1005],value[1005],record[1005];
cin>>t;
while(t--)
{
memset(record,0,sizeof(record));
cin>>n>>v;
for(i=0; i<n; i++)
cin>>value[i];
for(i=0; i<n; i++)
cin>>weight[i];
for(i=0; i<n; i++)
{
for(j=v;j>=weight[i]; j--)
{
if(record[j-weight[i]]+value[i]>record[j])
record[j] = record[j-weight[i]]+value[i];
}
}
cout<<record[v]<<endl;
} return 0;
}
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