lintcode:Copy Books

最新推荐文章于 2024-04-21 14:14:37 发布

martin_liang

最新推荐文章于 2024-04-21 14:14:37 发布

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本文探讨了一种分配连续编号书籍给不同人数以完成复制任务的算法，目标是在最短时间内完成所有书籍的复制。通过动态规划方法，该文提出了在给定书籍数量和人数的情况下，如何最优地分配书籍以最小化总复制时间。实例分析展示了算法的有效性和实用性。

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Given an array A of integer with size of n( means n books and number of pages of each book) and k people to copy the book. You must distribute the continuous id books to one people to copy. (You can give book A[1],A[2] to one people, but you cannot give book A[1], A[3] to one people, because book A[1] and A[3] is not continuous.) Each person have can copy one page per minute. Return the number of smallest minutes need to copy all the books.

Have you met this question in a real interview?

Yes

Example

Given array A = [3,2,4], k = 2.

Return 5( First person spends 5 minutes to copy book 1 and book 2 and second person spends 4 minutes to copy book 3. )

Challenge

Could you do this in O(n*k) time ?

class Solution {
public:
    /**
     * @param pages: a vector of integers
     * @param k: an integer
     * @return: an integer
     */

int copyBooks(vector<int> &pages, int k) {
    // write your code here
    if (pages.size() == 0 || k == 0)
        return 0;
        
    if (k > pages.size())
        k = pages.size();
    
    vector<vector<int>> dp(k, vector<int>(pages.size()));
    vector<int> auxPages(pages.size());

    auxPages[0] = pages[0]; //页面的累加和
    dp[0][0] = pages[0];
    for (int i=1; i<pages.size(); i++)
    {
        auxPages[i] = auxPages[i-1]+pages[i];
        dp[0][i] = auxPages[i];
    }
    
    //dp[i][j] i个人抄写书，抄到以j下标结尾的书时的最少耗时
    
    for (int i=1; i<k; i++)
    {
        for (int j=i; j<pages.size(); j++)
        {
            int targetValue = INT_MAX;
            for (int m=i; m<=j; m++)
            {
                //i-1人抄到下标为m-1的书，剩下的一人抄从m到j的书
                targetValue = min(max(dp[i-1][m-1], auxPages[j]-auxPages[m-1]), targetValue);
            }
            dp[i][j] = targetValue;
        }
    }
    
    return dp[k-1][pages.size()-1];
}

};