Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 27413 Accepted Submission(s): 11154
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
Sample Output
14
01背包。
AC代码例如以下：
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
int t,n,v;
int i,j;
int a[1005],b[1005],dp[1005];
cin>>t;
while(t--)
{
cin>>n>>v;
for(i=1;i<=n;i++)
cin>>a[i];
for(i=1;i<=n;i++)
cin>>b[i];
memset(dp,0,sizeof dp);
for(i=1;i<=n;i++)
for(j=v;j>=b[i];j--)
dp[j]=max(dp[j-b[i]]+a[i],dp[j]);
cout<<dp[v]<<endl;
}
return 0;
}