HDU 2602 Bone Collector (01背包)

最新推荐文章于 2021-05-13 11:08:06 发布

SCaryon

最新推荐文章于 2021-05-13 11:08:06 发布

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文章标签： dp HDU

本文链接：https://blog.youkuaiyun.com/SCaryon/article/details/61922167

本文通过一个名为“骨收集者”的趣味性问题介绍了01背包问题的经典案例。问题描述了一个骨收集者如何根据骨的价值和体积来最大化其背包中骨头总价值的过程。文章提供了具体的输入输出示例及详细的算法实现过程。

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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
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14

大致思路：

标准的01背包，创建一个二维数组dp[i][j]，i代表第i件物品，j代表剩余的背包容量。
但不知道为什么用滚动数组做就是WA。先记录一下。

代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
using namespace std;
int dp[1000][1000];
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int N,V,value[1001],volume[1001];
        cin>>N>>V;
        for(int i=1;i<=N;++i)
            cin>>value[i];
        for(int i=1;i<=N;++i)
            cin>>volume[i];
        for(int i=1;i<=V;++i){
            if(volume[1]<=i)
                dp[1][i]=value[1];
            else
                dp[1][i]=0;
        }
        for(int i=2;i<=N;++i){
            for(int j=V;j>=0;--j){
                if(j>=volume[i])//状态转移方程
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-volume[i]]+value[i]);
                else
                    dp[i][j]=dp[i-1][j];
            }
        }
        cout<<dp[N][V]<<endl;
    }
    return 0;
}