HDU 2602 Bone Collector（01背包）

最新推荐文章于 2021-01-18 14:47:36 发布

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最新推荐文章于 2021-01-18 14:47:36 发布

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分类专栏：背包文章标签： HDU 背包

本文链接：https://blog.youkuaiyun.com/b_r_e_a_d/article/details/75573182

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本文介绍了一道典型的01背包问题，并提供了完整的AC代码实现。该问题是关于如何选择不同价值和体积的骨头以最大化骨袋的价值，同时不超过袋子的容量限制。

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Bone Collector

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

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

这道题没什么好说的，就最基本的01背包问题。

AC代码：

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    int w[1111];
    int v[1111];
    int res[1111];
    int T,N,V;
    cin>>T;
    while(T--)
    {
        cin>>N>>V;
        memset(res,0,sizeof(res));
        for(int i=0;i<N;i++)
        {
            cin>>v[i];
        }
        for(int i=0;i<N;i++)
        {
            cin>>w[i];
        }
        for(int i=0;i<N;i++)
        {
            for(int j=V;j>=w[i];j--)
            {
                res[j]=max(res[j],res[j-w[i]]+v[i]);
            }
        }
        cout<<res[V]<<endl;
    }
    return 0;
}