Bone Collector(01背包）

最新推荐文章于 2020-04-02 16:55:44 发布

fanlinqiang

最新推荐文章于 2020-04-02 16:55:44 发布

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分类专栏：动态规划文章标签： integer each output less input

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本文探讨了如何计算骨收集者在收集过程中所能获得的最大总价值，涉及不同骨头的价值和体积，通过动态规划方法解决实际问题。

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Bone Collector

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

Total Submission(s) : 46 Accepted Submission(s) : 28

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

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int main(int ac,char av[])
{
    int cases,n,bag_w,i,j,bone_w[1002],val[1002],dp[1002];
    cin>>cases;
    while(cases--)
    {
        cin>>n>>bag_w;
        for(i=1;i<=n;i++)
        cin>>val[i];
        for(i=1;i<=n;i++)
        cin>>bone_w[i];
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
         for(j=bag_w;j>=bone_w[i];j--)
         {
             dp[j]=max(dp[j],dp[j-bone_w[i]]+val[i]);
         }
        cout<<dp[bag_w]<<endl;
    }
    return 0;
}