動態規劃解C++解Leetcode 1547難題 Minimum Cost to Cut a Stick

這個問題被視為困難的可能原因是，使用暴力解法的時間複雜度為O(m!)，其中m代表切割點的數量。然而，我們可以使用動態規劃來有效地解決這個問題。

直接的遞迴方法的時間複雜度是指數級的。通過使用記憶化的技巧，我們可以將其降低到O(m^3)。記憶化技巧能夠幫助我們儲存先前計算過的結果，避免重複計算。

在這段程式碼中，我們使用一個二維向量dp來儲存計算過的結果。dp的維度為mxm，其中m代表切割點的數量。這個矩陣可以幫助我們追蹤切割點之間的最小成本。

透過運用記憶化的遞迴函式，我們可以使用迴圈的方式來迭代地解決這個問題。這進一步優化了解決方案，將記憶體複雜度降低為O(m^2)。

總而言之，這個問題可能看似困難，但我們可以使用動態規劃來有效率地解決。所提供的程式碼利用了記憶化和迭代的方法，時間複雜度O(m^3)和記憶體複雜度O(m^2)。

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The problem at hand can be considered difficult because the brute-force solution would have a time complexity of O(m!), where m is the number of cuts. However, we can solve it efficiently using dynamic programming. 

 The straightforward recursive approach has an exponential time complexity. By incorporating memoization, we can reduce it to O(m^3). 

The memoization technique helps us store previously computed results, avoiding redundant calculations. In this code, we use a 2D vector, dp, to store the computed results. The dimensions of dp are m x m, where m is the size of the cuts vector. This matrix allows us to keep track of the minimum cost for cutting the rod between different positions. 

 By leveraging the memoized recursive function, we can solve the problem iteratively using for loops. This further optimizes the solution by reducing the memory complexity to O(m^2). 

 In summary, although the problem may initially appear challenging, we can efficiently solve it using dynamic programming. The code provided employs memoization and an iterative approach to achieve a time complexity of O(m^3) and a memory complexity of O(m^2).