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chapter_dynamic_programming/dp_problem_features/index.html
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@@ -3438,7 +3438,32 @@
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<h1 id="142">14.2 &nbsp; 动态规划问题特性<a class="headerlink" href="#142" title="Permanent link">&para;</a></h1>
<p>在上节中，我们学习了动态规划是如何通过子问题分解来求解问题的。实际上，子问题分解是一种通用的算法思路，在分治、动态规划、回溯中的侧重点不同。</p>
<ul>
@@ -4165,11 +4197,73 @@ dp[i, 2] = dp[i-2, 1] + dp[i-2, 2]
<p>在这个问题中，下次跳跃依赖于过去所有的状态，因为每一次跳跃都会在更高的阶梯上设置障碍，并影响未来的跳跃。对于这类问题，动态规划往往难以解决。</p>
<p>实际上，许多复杂的组合优化问题（例如旅行商问题）都不满足无后效性。对于这类问题，我们通常会选择使用其他方法，例如启发式搜索、遗传算法、强化学习等，从而在有限时间内得到可用的局部最优解。</p>
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<h1 id="143">14.3 &nbsp; 动态规划解题思路<a class="headerlink" href="#143" title="Permanent link">&para;</a></h1>
<p>上两节介绍了动态规划问题的主要特征，接下来我们一起探究两个更加实用的问题。</p>
<ol>
@@ -4712,11 +4744,73 @@ dp[i, j] = \min(dp[i-1, j], dp[i, j-1]) + grid[i, j]
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<h1 id="146">14.6 &nbsp; 编辑距离问题<a class="headerlink" href="#146" title="Permanent link">&para;</a></h1>
<p>编辑距离，也被称为 Levenshtein 距离，指两个字符串之间互相转换的最小修改次数，通常用于在信息检索和自然语言处理中度量两个序列的相似度。</p>
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<h1 id="141">14.1 &nbsp; 初探动态规划<a class="headerlink" href="#141" title="Permanent link">&para;</a></h1>
<p>「动态规划 dynamic programming」是一个重要的算法范式，它将一个问题分解为一系列更小的子问题，并通过存储子问题的解来避免重复计算，从而大幅提升时间效率。</p>
<p>在本节中，我们从一个经典例题入手，先给出它的暴力回溯解法，观察其中包含的重叠子问题，再逐步导出更高效的动态规划解法。</p>
@@ -4703,11 +4735,73 @@ dp[i] = dp[i-1] + dp[i-2]
<p>观察以上代码，由于省去了数组 <code>dp</code> 占用的空间，因此空间复杂度从 <span class="arithmatex">\(O(n)\)</span> 降低至 <span class="arithmatex">\(O(1)\)</span> 。</p>
<p>在动态规划问题中，当前状态往往仅与前面有限个状态有关，这时我们可以只保留必要的状态，通过“降维”来节省内存空间。<strong>这种空间优化技巧被称为“滚动变量”或“滚动数组”</strong>。</p>
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<h1 id="144-0-1">14.4 &nbsp; 0-1 背包问题<a class="headerlink" href="#144-0-1" title="Permanent link">&para;</a></h1>
<p>背包问题是一个非常好的动态规划入门题目，是动态规划中最常见的问题形式。其具有很多变种，例如 0-1 背包问题、完全背包问题、多重背包问题等。</p>
<p>在本节中，我们先来求解最常见的 0-1 背包问题。</p>
@@ -4615,11 +4647,73 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
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@@ -3400,11 +3432,73 @@
<li>在编辑距离中，状态依赖于其正上方、正左方、左上方的状态，因此空间优化后正序或倒序遍历都无法正确地进行状态转移。为此，我们利用一个变量暂存左上方状态，从而转化到与完全背包等价的情况，可以在空间优化后进行正序遍历。</li>
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<h1 id="145">14.5 &nbsp; 完全背包问题<a class="headerlink" href="#145" title="Permanent link">&para;</a></h1>
<p>在本节中，我们先求解另一个常见的背包问题：完全背包，再了解它的一种特例：零钱兑换。</p>
<h2 id="1451">14.5.1 &nbsp; 完全背包<a class="headerlink" href="#1451" title="Permanent link">&para;</a></h2>
@@ -5454,11 +5486,73 @@ dp[i, a] = dp[i-1, a] + dp[i, a - coins[i-1]]
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