Bone Collector hdu2602（经典dp）01背包

Bone Collector

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

Total Submission(s): 59894    Accepted Submission(s): 24996


[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
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int v[100005],w[10005],dp[10005];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
mem(v,0);
mem(w,0);
mem(dp,0);
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%d",&v[i]);
}
for(int i=1;i<=n;i++)
{
scanf("%d",&w[i]);
}
for(int i=1;i<=n;i++)
{
for(int j=m;j>=0;j--)
{
if(j>=w[i])
dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
}
}
printf("%d\n",dp[m]);
}
return 0;
}