HDU - 2602 Bone Collector （01背包）

最新推荐文章于 2022-07-27 15:53:18 发布

是奥利奥阿

最新推荐文章于 2022-07-27 15:53:18 发布

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本文介绍了一道经典的01背包问题——骨收集者问题。在一个背包容量限制下，如何选择不同价值和体积的骨头，以获得最大总价值。文章详细解释了使用动态规划解决此问题的方法，并提供了一个通过的C++代码示例。

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 ?
在这里插入图片描述

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
1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output
14

Author
Teddy
问题链接： http://acm.hdu.edu.cn/showproblem.php?pid=2602
问题简述： 有n个骨头，每个骨头有权值跟重量。狗的背包有容量限制，求最多能获得的骨头的权值
问题分析： 01背包模板题，值得注意的是dp用long long（题目要求的最大为2^31），循环j时从0开始（重量可能为0，从1开始会WA）
AC通过的C++语言程序如下：

#include <iostream>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <math.h>
#include <climits>
#include <iomanip>
#include <queue>
#include<vector>
using namespace std;

const int N=10000;

int w[N],v[N];
long long dp[1005][1005];//2的31次方要用long long
int n,m;

int main()
{
    ios::sync_with_stdio(false);
    int T;
    cin>>T;
    while(T--)
    {
        cin>>n>>m;
        for(int i=1;i<=n;i++)
            cin>>v[i];
        for(int i=1;i<=n;i++)
            cin>>w[i];
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<=m;j++)//记得j从0开始，重量可以等于0
            {
                if(j<w[i]) 
                    dp[i][j]=dp[i-1][j];
                else    
                {
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+v[i]);
                }
            }
        }
        cout<<dp[n][m]<<endl;
        memset(dp,0,sizeof(dp));//记得初始化
    }
    return 0;
}