Bone Collector

最新推荐文章于 2024-07-19 15:39:37 发布

WQJ0714

最新推荐文章于 2024-07-19 15:39:37 发布

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分类专栏：背包

本文链接：https://blog.youkuaiyun.com/WQJ0714/article/details/43603187

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1 篇文章

订阅专栏

本文介绍了一种经典的背包问题——骨收集者问题。该问题要求根据骨收集者的背包容量及每块骨头的价值与体积，计算出骨收集者能获得的最大总价值。通过使用动态规划方法求解，并给出具体实现代码。

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题目描述

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 ?

输入

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.

输出

One integer per line representing the maximum of the total value (this number will be less than 231).

示例输入

示例输出

#include <bits/stdc++.h>  
using namespace std;  
int main()  
{  
    int dp[1002];  
    int w[1002],v[1002];  
    int i,j,k,l,T,n,m;  
    cin>>T;  
    while(T--)  
    {  
        cin>>n>>m;  
        for(i=1;i<=n;i++)  
        {  
            cin>>w[i];  
        }  
        for(i=1;i<=n;i++)  
           {  
               cin>>v[i];  
           }  
           memset(dp,0,sizeof(dp));  
        for(i=1;i<=n;i++)  
        {  
            for(j=m;j>=v[i];j--)  
            {  
                dp[j]=max(dp[j],dp[j-v[i]]+w[i]);  
            }  
        }  
        cout<<dp[m]<<endl;  
    }  
}