HDU 2602 骨头收集者

最新推荐文章于 2018-12-11 22:08:09 发布

HDU-Dade

最新推荐文章于 2018-12-11 22:08:09 发布

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分类专栏： c语言入门文章标签： HDU ACM 2602 01背包

本文链接：https://blog.youkuaiyun.com/qq_43130927/article/details/84498763

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本文详细解析了骨收集者问题，这是一个经典的0-1背包问题变种，涉及到最大价值的计算。通过对比二维数组和一维数组的实现方式，阐述了如何避免中间状态的遗漏，确保算法正确性和效率。

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Bone Collector

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

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

Sample Output

附链接：
http://acm.hdu.edu.cn/showproblem.php?pid=2602

【碰到问题】一开始我用二维数组去做，但遇到背包体积为0时，面对（2，0）（1， 1），（3， 0）本来应该是5，可是费用为1时不执行内循环，那么最后结果是3.
【错误代码】

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <map>

using namespace std;
int dp[1010][1010;
int main()
{
    freopen("input.txt", "r", stdin);
    int c[1010], w[1010];
    int t, N, V;
    cin>>t;
    while(t--)
    {
        cin>>N>>V;
        memset(c, 0, sizeof(c));
        memset(w, 0, sizeof(w));
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i<=N; i++)
        {
            cin>>w[i];
        }
        for(int i = 1; i<=N; i++)
        {
            cin>>c[i];
        }
        int i, j;
        for( i = 1; i<=N; i++)
        {
            for( j = V; j>=c[i]; j--)
            {
                dp[i][j] = max(dp[i-1][j], dp[i-1][j - c[i]] + w[i]);
            }
        }
        cout<<dp[N][V]<<endl;
    }
    return 0;
}

看了discuss，发现大家是用if语句做判断，并不是直接用状态转移方程
因为跳过了中间一行，所以考虑一维数组

【AC代码】

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <map>

using namespace std;
int dp[1010];
int main()
{
    freopen("input.txt", "r", stdin);
    int c[1010], w[1010];
    int t, N, V;
    cin>>t;
    while(t--)
    {
        cin>>N>>V;
        memset(c, 0, sizeof(c));
        memset(w, 0, sizeof(w));
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i<=N; i++)
        {
            cin>>w[i];
        }
        for(int i = 1; i<=N; i++)
        {
            cin>>c[i];
        }
        int i, j;
        for( i = 1; i<=N; i++)
        {
            for( j = V; j>=c[i]; j--)
            {
                dp[j] = max(dp[j], dp[j - c[i]] + w[i]);
            }
        }
        cout<<dp[V]<<endl;
    }
    return 0;
}

以上是我第一次写背包的感受