杭电 2603 Bone Collector（简单01背包）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 35296    Accepted Submission(s): 14531

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 2³¹).

 

Sample Input

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

 

Sample Output

14

 
很裸的一道01背包，注意内层for循环（for(int j = v ;j>=weight[i];j--)）

AC代码：

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std ;

int dp[10000] ;
int value[1005],weight[1005] ;
int t,n,v;

int main()
{
    while(scanf("%d",&t)!=EOF)
    {
        while(t--)
        {
            scanf("%d%d",&n,&v) ;
            for(int i = 0 ;i<n ;i++)
                scanf("%d",&value[i]) ;
            for(int j = 0 ;j<n ;j++)
                scanf("%d",&weight[j]) ;
            memset(dp,0,sizeof(dp)) ;
            for(int i = 0; i<n ;i++)
            {
                for(int j = v ;j>=weight[i] ;j--)
                {
                    dp[j] = max(dp[j],dp[j-weight[i]]+value[i]) ;
                }
            }
            printf("%d
",dp[v]) ;
        }
    }
    return 0 ;
}