hdu 2602（01背包裸题）

Bone Collector
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
Submit Status

Appoint description: 
System Crawler  (2014-04-03)

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 2 31).
 

Sample Input

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

 

Sample Output

14

 

转移方程

dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i]]+w[i])

代码如下

#include <stdio.h>
#include <math.h>
#include<algorithm>
#include<cstring>
using std::max;
int dp[1005],w[1005],v[1005];
int main()
{
//freopen("input.txt","r",stdin);
int N,T,V;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&N,&V);
for(int i=1;i<=N;i++)
scanf("%d",w+i);
for(int i=1;i<=N;i++)
scanf("%d",v+i);
memset(dp,0,sizeof(dp));
for(int i=1;i<=N;i++)
for(int j=V;j>=v[i];j--)
dp[j]=max(dp[j],dp[j-v[i]]+w[i]);
printf("%d\n",dp[V]);
}
return 0;
}