hdu-2602-Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40361    Accepted Submission(s): 16764

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      

Author

Teddy

Source

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

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题目分析：

就是简单的背包，但是要读清题，第一行是 石头数（也就是价值）  第二行是 需要的背包体积。但是一定要搞清楚 石头数可以是 0  背包体积也可以是 0.而且石头数（就是价值）可以不是0 ，但消耗的体积可以0

5  0

1 2 3 4 5

0 0 0 1 0

结果是： 11  

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int f[1010][1010];
int cost[1010],val[1010];
int main()
{
  int t;
  scanf("%d",&t);
  while(t--)
  {
    int n,v;
    
    scanf("%d%d",&n,&v);
    memset(f,0,sizeof(f));
    
    cost[0]=val[0]=0;
    for(int i=1;i<=n;i++)
        scanf("%d",&val[i]);
    for(int i=1;i<=n;i++)
        scanf("%d",&cost[i]);
    for(int i=1;i<=n;i++)
      for(int j=0;j<=v;j++)  // 记住要从0开始  石头是可以不消耗体积的
      {
        f[i][j]=f[i-1][j];
        if(j>=cost[i])
          f[i][j]=max(f[i-1][j],f[i-1][j-cost[i]]+val[i]);
      }
      printf("%d\n",f[n][v]);
  }
  return 0;
}

一维数组优化代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int f[1010];
int cost[1010],val[1010];
int main()
{
  int t;
  scanf("%d",&t);
  while(t--)
  {
    int n,v;
    
    scanf("%d%d",&n,&v);
    memset(f,0,sizeof(f));
    
    cost[0]=val[0]=0;
    for(int i=1;i<=n;i++)
        scanf("%d",&val[i]);
    for(int i=1;i<=n;i++)
        scanf("%d",&cost[i]);
    for(int i=1;i<=n;i++)
      for(int j=v;j>=0;j--)//  记者要到 0 
      {
        if(j>=cost[i])
          f[j]=max(f[j],f[j-cost[i]]+val[i]);
      }
      printf("%d\n",f[v]);
  }
  return 0;
}