HDU 2602-Bone Collector(01背包-一/二维)

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 47666    Accepted Submission(s): 19880


[align=left]Problem Description[/align]
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 ?



 

[align=left]Input[/align]
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.
 

[align=left]Output[/align]
One integer per line representing the maximum of the total value (this number will be less than 231).
 

[align=left]Sample Input[/align]

1
5 10
1 2 3 4 5
5 4 3 2 1

 

[align=left]Sample Output[/align]

14

 

[align=left]Author[/align]
Teddy
 

[align=left]Source[/align]
HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

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一维：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int n[1001],v[1001];
int dp[1001];
int main()
{
int t;
cin>>t;
while(t--)
{
memset(dp,0,sizeof(dp));
int num,vol,i,j;
cin>>num>>vol;
for(i=0; i<num; ++i)
cin>>v[i];
for(i=0; i<num; ++i)
cin>>n[i];
for(i=0; i<num; ++i)
for(j=vol; j>=n[i]; --j)
dp[j]=max(dp[j],dp[j-n[i]]+v[i]);
cout<<dp[vol]<<endl;
}
return 0;
}

二维：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int n[1001],v[1001];
int dp[1001][1001];
int main()
{
int t;
cin>>t;
while(t--)
{
int num,vol,i,j;
cin>>num>>vol;
for(i=0; i<num; ++i)
cin>>v[i];
for(i=0; i<num; ++i)
cin>>n[i];
for(i=0; i<num; ++i)
for(j=0; j<=vol; ++j)
if(j<n[i]) dp[i+1][j]=dp[i][j];
else dp[i+1][j]=max(dp[i][j],dp[i][j-n[i]]+v[i]);
cout<<dp[num][vol]<<endl;
}
return 0;
}