HDU 2602 Bone Collector(01背包）

Bone Collector

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

Total Submission(s): 51035 Accepted Submission(s): 21446

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 ?

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

01背包：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<math.h>
using namespace std;
int dp[1100];
struct node
{
int value,volume;
} a[1100];
int main()
{
int t;
cin>>t;
while(t--)
{
int n,v,i,j;
memset(dp,0,sizeof(dp));
cin>>n>>v;
for(i=1; i<=n; i++)
{
cin>>a[i].value;
}
for(i=1;i<=n;i++)
{
cin>>a[i].volume;
}
for(i=1; i<=n; i++)
{
for(j=v; j>=a[i].volume; j--)
{
dp[j]=max(dp[j],dp[j-a[i].volume]+a[i].value);
}
}
cout<<dp[v]<<endl;
}
return 0;
}