暑假集训第四周 阶段一 DP 动态规划 B - Bone Collector

                                         

                                     B - Bone Collector

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 
     
     
     
     

分析：

f(n，m)：花m元买n样东西实现的最大价值。对于任意的f(n，m)，都有下面这两种情况：

情况一，你买了第n件商品，f(n, m)=f(n-1, m-pn)+vn，因为买了第n件商品，所以花费了pn元，也因此得到了vn的价值。f(n, m)就等于第n件商品的价值+用m-pn的钱去买n-1件商品的价值。这样问题就规模就变小了。

情况二，你不买第n件商品，f(n, m)=f(n-1, m)，也就是说f(n, m)等于你用m元钱去买n-1件商品实现的最大价值。

这两种情况，哪个价值大，就取哪一种，所以f(n, m) = max(f(n-1, m-pn)+vn, f(n-1, m))。这便是这第一步的核心。

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#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
struct bone
{
    int a,b;
} v[1010];
int main()
{
    int t,n,m,i,j,f[1010];
    scanf("%d",&t);
    while(t--)
    {
        memset(f,0,sizeof(f));
        scanf("%d %d",&n,&m);
        for(i=0; i<n; i++)
            scanf("%d",&v[i].a);
        for(i=0; i<n; i++)
            scanf("%d",&v[i].b);
        for(i=0; i<n; i++)
            for(j=m;j>=v[i].b;j--)
            if(f[j]<f[j-v[i].b]+v[i].a)
            f[j]=f[j-v[i].b]+v[i].a;
        printf("%d\n",f[m]);
    }
    return 0;
}