hdu oj 2602 Bone Collector

hdu oj 2602 Bone Collector

题目：

Bone Collector

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

Total Submission(s): 39870    Accepted Submission(s): 16533

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 2 ³¹).

 

Sample Input

    
    
    
    

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

 

Sample Output

    
    
    
    

     
     
     
     14
    
    
    
    

 

解析：

• 01背包问题，01背包教程点此。

代码：

• 方法一：

• #include <iostream>
  #include <string.h>
  #include <stdio.h>
  
  int dp[1001][1001],c[1001],w[1001];
  int main()
  {
      int t,k,v,i,j;
      scanf("%d",&t);
      while(t--)
      {
          scanf("%d %d",&k,&v);
          memset(dp,0,sizeof(dp));
          for(i=1;i<=k;i++)
              scanf("%d",&w[i]);
          for(i=1;i<=k;i++)
              scanf("%d",&c[i]);
          for(i=1;i<=k;i++)
          {
              for(j=0;j<=v;j++)
              {
                  if(c[i]<=j)
                      dp[i][j]=dp[i-1][j]>(dp[i-1][j-c[i]]+w[i])?dp[i-1][j]:(dp[i-1][j-c[i]]+w[i]);
                  else dp[i][j]=dp[i-1][j];
              }
          }
          printf("%d\n",dp[k][v]);
      }
      return 0;
  }

• 方法二：

• #include <iostream>
  #include <string.h>
  #include <stdio.h>
  using namespace std;
  int dp[1001],c[1001],w[1001];
  int main()
  {
      int t,k,v,i,j;
      scanf("%d",&t);
      while(t--)
      {
          scanf("%d %d",&k,&v);
          memset(dp,0,sizeof(dp));
          for(i=1;i<=k;i++)
              scanf("%d",&w[i]);
          for(i=1;i<=k;i++)
              scanf("%d",&c[i]);
          for(i=1;i<=k;i++)
          {
              for(j=v;j>=c[i];j--)
              {
                  dp[j]=dp[j]>(dp[j-c[i]]+w[i])?dp[j]:(dp[j-c[i]]+w[i]);
              }
          }
          printf("%d\n",dp[v]);
      }
      return 0;