hdoj - 2602 Bone Collector

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

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int a[],b[],dp[];

int main()

{

    int t,n,v;

    scanf("%d",&t);

    for(int k=;k<t;k++)

    {

        memset(dp,,sizeof(dp));

        scanf("%d %d",&n,&v);

        for(int i = ;i<n;i++)

        {

            scanf("%d",&a[i]);

        }

        for(int i = ;i<n;i++)

        {

            scanf("%d",&b[i]);

        }

        for(int i = ;i<n;i++)

        {

            for(int j = v;j>=b[i];j--)

            {

                dp[j] = max(dp[j],dp[j-b[i]]+a[i]);      //关键代码

            }

        }

        printf("%d\n",dp[v]);

    }

    return ;

}

巴特西

hdoj - 2602 Bone Collector

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