HDU 2602-Bone Collector(01背包-一/二维)

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#HDU 2602 #Bone Collector #01背包 #C++

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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，通过给定的案例详细解析了如何利用动态规划算法来求解最大价值。文章提供了两种实现方式，一种是一维DP简化空间复杂度，另一种是传统的二维DP方法。

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 47666 Accepted Submission(s): 19880

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 2 ³¹).

Sample Input

Sample Output

Author

Teddy

Source

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

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一维：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int n[1001],v[1001];
int dp[1001];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        int num,vol,i,j;
        cin>>num>>vol;
        for(i=0; i<num; ++i)
            cin>>v[i];
        for(i=0; i<num; ++i)
            cin>>n[i];
        for(i=0; i<num; ++i)
            for(j=vol; j>=n[i]; --j)
                dp[j]=max(dp[j],dp[j-n[i]]+v[i]);
        cout<<dp[vol]<<endl;
    }
    return 0;
}

二维：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int n[1001],v[1001];
int dp[1001][1001];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int num,vol,i,j;
        cin>>num>>vol;
        for(i=0; i<num; ++i)
            cin>>v[i];
        for(i=0; i<num; ++i)
            cin>>n[i];
        for(i=0; i<num; ++i)
            for(j=0; j<=vol; ++j)
                if(j<n[i]) dp[i+1][j]=dp[i][j];
                else dp[i+1][j]=max(dp[i][j],dp[i][j-n[i]]+v[i]);
        cout<<dp[num][vol]<<endl;
    }
    return 0;
}