Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29163    Accepted Submission(s): 11922

 

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 ?

HDOJ 2602 Bone Collector_i++

 

 

 

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

 翻译：有一人爱收集骨头，每个骨头有一个价值和重量，他有一个背包，容量为V，问他装得最大价值量是多少？（经典的01背包问题）

 

重点：01背包、动态规划

难点：动态方程的书写

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
  int T,n,v,i,j,dp[1100],a[1100],b[1100];//a[1100]是价值，b[1100]是重量；
  cin>>T;
  while(T--)
  {
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b)); 
    memset(dp,0,sizeof(dp)); 
    cin>>n>>v;//n是骨头个数，V是背包容量 
    for(i=1;i<=n;i++)
      scanf("%d",&a[i]);
    for(i=1;i<=n;i++)
      scanf("%d",&b[i]);
    for(i=1;i<=n;i++)
      for(j=v;j>=b[i];j--)
      {
        if(dp[j]<dp[j-b[i]]+a[i])
          dp[j]=dp[j-b[i]]+a[i];
      }
    printf("%d\n",dp[v]);
  }
  return 0;
}