hdoj -2603 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 2³¹).      

              

Sample Input

 15 101 2 3 4 55 4 3 2 1 

              

Sample Output

 14 

         一维数组

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>using namespace std;int dp[1000];int p[1000];int w[1000];int max(int x,int y){    if(x>y)        return x;    else        return y;}int main(){    int n,v,i,j,t;    while(scanf("%d",&t)!=EOF)    {        while(t--)        {            memset(dp,0,sizeof(dp));            scanf("%d %d",&n,&v);            for(i=1;i<=n;i++)            {                scanf("%d",&w[i]);            }            for(i=1;i<=n;i++)            {                scanf("%d",&p[i]);            }            for(i=1;i<=n;i++)            {                for(j=v;j>=p[i];j--)                {                    dp[j]=max(dp[j],dp[j-p[i]]+w[i]);                }            }            printf("%d\n",dp[v]);        }    }    return 0;}