hdu -- 2602

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 81948    Accepted Submission(s): 33895

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

Author Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

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【分析】

• 01背包问题

  #include<bits/stdc++.h>
  using namespace std;
  int v[1010],w[1010],dp[1010];
  int main()
  {
      int t;
      scanf("%d",&t);
      while(t--)
      {
          memset(dp,0,sizeof(dp));
          int n,m;
          scanf("%d%d",&n,&m);
          for(int i=0;i<n;i++)
  		scanf("%d",&v[i]);
          for(int i=0;i<n;i++)
          	scanf("%d",&w[i]);
          for(int i=0;i<n;i++)
              for(int j=m;j>=w[i];j--)
                  dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
          cout<<dp[m]<<endl;
      }
      return 0;
  }