Bone Collector hdu2602（经典dp）01背包

最新推荐文章于 2020-04-02 16:55:44 发布

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本文介绍了一个经典的背包问题变种——骨收集者问题。该问题是关于如何在限定容量下选取骨头以获得最大价值。输入包括骨头数量、背包容量及每根骨头的价值与体积，输出为能获得的最大总价值。

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Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 59894 Accepted Submission(s): 24996

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

Sample Output

14

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int v[100005],w[10005],dp[10005];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        mem(v,0);
        mem(w,0);
        mem(dp,0);
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&v[i]);
        }
         for(int i=1;i<=n;i++)
        {
            scanf("%d",&w[i]);
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=m;j>=0;j--)
            {
                if(j>=w[i])
                    dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
            }
        }
        printf("%d\n",dp[m]);
    }
    return 0;
}