hdoj2602 Bone Collector【dp 01背包】

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 ?

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.

One integer per line representing the maximum of the total value (this number will be less than 2 ³¹).

  
   

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

  
   

    14
   
   

    
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
int dp[1010];
struct Node{int value;int wight;
}A[1010];
int Max(int a,int b)
{return a>b?a:b;
}
int main()
{int t,i,j,n,V;scanf("%d",&t);while(t--){scanf("%d%d",&n,&V);for(i=0;i<n;++i){scanf("%d",&A[i].value);}for(i=0;i<n;++i){scanf("%d",&A[i].wight);}memset(dp,0,sizeof(dp));for(i=0;i<n;++i){for(j=V;j>=A[i].wight;--j){dp[j]=Max(dp[j],dp[j-A[i].wight]+A[i].value);}}printf("%d\n",dp[V]);}return 0;
}