动态规划1017

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 …<br>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 ?<br><center><img src=../../../data/images/C154-1003-1.jpg> </center><br>

 

Input

The first line contain a integer T , the number of cases.<br>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
   
   
   
   

代码：

#include<stdio.h>
#include<string.h>
int f[2002];
int max(int x,int y){
 if(x<y)
  x=y;
 return x;
}
int main()
{
 int t,n,V,v[1000],w[1000],i,j;
 scanf("%d",&t);
 while(t--){
  scanf("%d%d",&n,&V);
  for(i=1;i<=n;i++)
   scanf("%d",&w[i]);
  for(i=1;i<=n;i++)
   scanf("%d",&v[i]);
  memset(f,0,sizeof(f));
  for(i=1;i<=n;i++)
   for(j=V;j>=v[i];j--)
    f[j]=max(f[j],f[j-v[i]]+w[i]);
  printf("%d\n",f[V]);
 }
 return 0;
}

这是一个01背包问题，若只考虑第i件物品的策略（放或不放），那么就可以转化为一个只牵扯前i-1件物品的问题。如果不放第i件物品，那么问题就转化为“前i-1件物品放入容量为v的背包中”，价值为f[i-1][v]；如果放第i件物品，那么问题就转化为“前i-1件物品放入剩下的容量为v-c[i]的背包中”，此时能获得的最大价值就是f[i-1][v-c[i]]再加上通过放入第i件物品获得的价值w[i]