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chapter_dynamic_programming/dp_problem_features/index.html
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10.2 &nbsp; 二分查找插入点
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10.3 &nbsp; 二分查找边界
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第 12 章 &nbsp; 分治
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12.2 &nbsp; 分治搜索策略
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第 14 章 &nbsp; 动态规划
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<p>给定一个楼梯，你每步可以上 <span class="arithmatex">\(1\)</span> 阶或者 <span class="arithmatex">\(2\)</span> 阶，每一阶楼梯上都贴有一个非负整数，表示你在该台阶所需要付出的代价。给定一个非负整数数组 <span class="arithmatex">\(cost\)</span> ，其中 <span class="arithmatex">\(cost[i]\)</span> 表示在第 <span class="arithmatex">\(i\)</span> 个台阶需要付出的代价，<span class="arithmatex">\(cost[0]\)</span> 为地面起始点。请计算最少需要付出多少代价才能到达顶部？</p>
</div>
<p>如图 14-6 所示，若第 <span class="arithmatex">\(1\)</span>、<span class="arithmatex">\(2\)</span>、<span class="arithmatex">\(3\)</span> 阶的代价分别为 <span class="arithmatex">\(1\)</span>、<span class="arithmatex">\(10\)</span>、<span class="arithmatex">\(1\)</span> ，则从地面爬到第 <span class="arithmatex">\(3\)</span> 阶的最小代价为 <span class="arithmatex">\(2\)</span> 。</p>
<p><img alt="爬到第 3 阶的最小代价" src="../dp_problem_features.assets/min_cost_cs_example.png" /></p>
<p><a class="glightbox" href="../dp_problem_features.assets/min_cost_cs_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="爬到第 3 阶的最小代价" src="../dp_problem_features.assets/min_cost_cs_example.png" /></a></p>
<p align="center"> 图 14-6 &nbsp; 爬到第 3 阶的最小代价 </p>
<p>设 <span class="arithmatex">\(dp[i]\)</span> 为爬到第 <span class="arithmatex">\(i\)</span> 阶累计付出的代价，由于第 <span class="arithmatex">\(i\)</span> 阶只可能从 <span class="arithmatex">\(i - 1\)</span> 阶或 <span class="arithmatex">\(i - 2\)</span> 阶走来，因此 <span class="arithmatex">\(dp[i]\)</span> 只可能等于 <span class="arithmatex">\(dp[i - 1] + cost[i]\)</span> 或 <span class="arithmatex">\(dp[i - 2] + cost[i]\)</span> 。为了尽可能减少代价，我们应该选择两者中较小的那一个：</p>
@@ -3725,7 +3552,7 @@ dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
</div>
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<p>图 14-7 展示了以上代码的动态规划过程。</p>
<p><img alt="爬楼梯最小代价的动态规划过程" src="../dp_problem_features.assets/min_cost_cs_dp.png" /></p>
<p><a class="glightbox" href="../dp_problem_features.assets/min_cost_cs_dp.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="爬楼梯最小代价的动态规划过程" src="../dp_problem_features.assets/min_cost_cs_dp.png" /></a></p>
<p align="center"> 图 14-7 &nbsp; 爬楼梯最小代价的动态规划过程 </p>
<p>本题也可以进行空间优化，将一维压缩至零维，使得空间复杂度从 <span class="arithmatex">\(O(n)\)</span> 降低至 <span class="arithmatex">\(O(1)\)</span> 。</p>
@@ -3925,7 +3752,7 @@ dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
<p>给定一个共有 <span class="arithmatex">\(n\)</span> 阶的楼梯，你每步可以上 <span class="arithmatex">\(1\)</span> 阶或者 <span class="arithmatex">\(2\)</span> 阶，<strong>但不能连续两轮跳 <span class="arithmatex">\(1\)</span> 阶</strong>，请问有多少种方案可以爬到楼顶。</p>
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<p>例如图 14-8 ，爬上第 <span class="arithmatex">\(3\)</span> 阶仅剩 <span class="arithmatex">\(2\)</span> 种可行方案，其中连续三次跳 <span class="arithmatex">\(1\)</span> 阶的方案不满足约束条件，因此被舍弃。</p>
<p><img alt="带约束爬到第 3 阶的方案数量" src="../dp_problem_features.assets/climbing_stairs_constraint_example.png" /></p>
<p><a class="glightbox" href="../dp_problem_features.assets/climbing_stairs_constraint_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="带约束爬到第 3 阶的方案数量" src="../dp_problem_features.assets/climbing_stairs_constraint_example.png" /></a></p>
<p align="center"> 图 14-8 &nbsp; 带约束爬到第 3 阶的方案数量 </p>
<p>在该问题中，如果上一轮是跳 <span class="arithmatex">\(1\)</span> 阶上来的，那么下一轮就必须跳 <span class="arithmatex">\(2\)</span> 阶。这意味着，<strong>下一步选择不能由当前状态（当前楼梯阶数）独立决定，还和前一个状态（上轮楼梯阶数）有关</strong>。</p>
@@ -3942,7 +3769,7 @@ dp[i, 1] = dp[i-1, 2] \\
dp[i, 2] = dp[i-2, 1] + dp[i-2, 2]
\end{cases}
\]</div>
<p><img alt="考虑约束下的递推关系" src="../dp_problem_features.assets/climbing_stairs_constraint_state_transfer.png" /></p>
<p><a class="glightbox" href="../dp_problem_features.assets/climbing_stairs_constraint_state_transfer.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="考虑约束下的递推关系" src="../dp_problem_features.assets/climbing_stairs_constraint_state_transfer.png" /></a></p>
<p align="center"> 图 14-9 &nbsp; 考虑约束下的递推关系 </p>
<p>最终，返回 <span class="arithmatex">\(dp[n, 1] + dp[n, 2]\)</span> 即可，两者之和代表爬到第 <span class="arithmatex">\(n\)</span> 阶的方案总数。</p>
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chapter_dynamic_programming/dp_solution_pipeline/index.html
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<p>给定一个 <span class="arithmatex">\(n \times m\)</span> 的二维网格 <code>grid</code> ，网格中的每个单元格包含一个非负整数，表示该单元格的代价。机器人以左上角单元格为起始点，每次只能向下或者向右移动一步，直至到达右下角单元格。请返回从左上角到右下角的最小路径和。</p>
</div>
<p>图 14-10 展示了一个例子，给定网格的最小路径和为 <span class="arithmatex">\(13\)</span> 。</p>
<p><img alt="最小路径和示例数据" src="../dp_solution_pipeline.assets/min_path_sum_example.png" /></p>
<p><a class="glightbox" href="../dp_solution_pipeline.assets/min_path_sum_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="最小路径和示例数据" src="../dp_solution_pipeline.assets/min_path_sum_example.png" /></a></p>
<p align="center"> 图 14-10 &nbsp; 最小路径和示例数据 </p>
<p><strong>第一步：思考每轮的决策，定义状态，从而得到 <span class="arithmatex">\(dp\)</span> 表</strong></p>
<p>本题的每一轮的决策就是从当前格子向下或向右一步。设当前格子的行列索引为 <span class="arithmatex">\([i, j]\)</span> ，则向下或向右走一步后，索引变为 <span class="arithmatex">\([i+1, j]\)</span> 或 <span class="arithmatex">\([i, j+1]\)</span> 。因此，状态应包含行索引和列索引两个变量，记为 <span class="arithmatex">\([i, j]\)</span> 。</p>
<p>状态 <span class="arithmatex">\([i, j]\)</span> 对应的子问题为：从起始点 <span class="arithmatex">\([0, 0]\)</span> 走到 <span class="arithmatex">\([i, j]\)</span> 的最小路径和，解记为 <span class="arithmatex">\(dp[i, j]\)</span> 。</p>
<p>至此，我们就得到了图 14-11 所示的二维 <span class="arithmatex">\(dp\)</span> 矩阵，其尺寸与输入网格 <span class="arithmatex">\(grid\)</span> 相同。</p>
<p><img alt="状态定义与 dp 表" src="../dp_solution_pipeline.assets/min_path_sum_solution_step1.png" /></p>
<p><a class="glightbox" href="../dp_solution_pipeline.assets/min_path_sum_solution_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="状态定义与 dp 表" src="../dp_solution_pipeline.assets/min_path_sum_solution_step1.png" /></a></p>
<p align="center"> 图 14-11 &nbsp; 状态定义与 dp 表 </p>
<div class="admonition note">
@@ -3599,7 +3426,7 @@
<div class="arithmatex">\[
dp[i, j] = \min(dp[i-1, j], dp[i, j-1]) + grid[i, j]
\]</div>
<p><img alt="最优子结构与状态转移方程" src="../dp_solution_pipeline.assets/min_path_sum_solution_step2.png" /></p>
<p><a class="glightbox" href="../dp_solution_pipeline.assets/min_path_sum_solution_step2.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="最优子结构与状态转移方程" src="../dp_solution_pipeline.assets/min_path_sum_solution_step2.png" /></a></p>
<p align="center"> 图 14-12 &nbsp; 最优子结构与状态转移方程 </p>
<div class="admonition note">
@@ -3610,7 +3437,7 @@ dp[i, j] = \min(dp[i-1, j], dp[i, j-1]) + grid[i, j]
<p><strong>第三步：确定边界条件和状态转移顺序</strong></p>
<p>在本题中，首行的状态只能从其左边的状态得来，首列的状态只能从其上边的状态得来，因此首行 <span class="arithmatex">\(i = 0\)</span> 和首列 <span class="arithmatex">\(j = 0\)</span> 是边界条件。</p>
<p>如图 14-13 所示，由于每个格子是由其左方格子和上方格子转移而来，因此我们使用采用循环来遍历矩阵，外循环遍历各行、内循环遍历各列。</p>
<p><img alt="边界条件与状态转移顺序" src="../dp_solution_pipeline.assets/min_path_sum_solution_step3.png" /></p>
<p><a class="glightbox" href="../dp_solution_pipeline.assets/min_path_sum_solution_step3.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="边界条件与状态转移顺序" src="../dp_solution_pipeline.assets/min_path_sum_solution_step3.png" /></a></p>
<p align="center"> 图 14-13 &nbsp; 边界条件与状态转移顺序 </p>
<div class="admonition note">
@@ -3848,7 +3675,7 @@ dp[i, j] = \min(dp[i-1, j], dp[i, j-1]) + grid[i, j]
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<p>图 14-14 给出了以 <span class="arithmatex">\(dp[2, 1]\)</span> 为根节点的递归树，其中包含一些重叠子问题，其数量会随着网格 <code>grid</code> 的尺寸变大而急剧增多。</p>
<p>本质上看，造成重叠子问题的原因为：<strong>存在多条路径可以从左上角到达某一单元格</strong>。</p>
<p><img alt="暴力搜索递归树" src="../dp_solution_pipeline.assets/min_path_sum_dfs.png" /></p>
<p><a class="glightbox" href="../dp_solution_pipeline.assets/min_path_sum_dfs.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="暴力搜索递归树" src="../dp_solution_pipeline.assets/min_path_sum_dfs.png" /></a></p>
<p align="center"> 图 14-14 &nbsp; 暴力搜索递归树 </p>
<p>每个状态都有向下和向右两种选择，从左上角走到右下角总共需要 <span class="arithmatex">\(m + n - 2\)</span> 步，所以最差时间复杂度为 <span class="arithmatex">\(O(2^{m + n})\)</span> 。请注意，这种计算方式未考虑临近网格边界的情况，当到达网络边界时只剩下一种选择。因此实际的路径数量会少一些。</p>
@@ -4132,7 +3959,7 @@ dp[i, j] = \min(dp[i-1, j], dp[i, j-1]) + grid[i, j]
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<p>如图 14-15 所示，在引入记忆化后，所有子问题的解只需计算一次，因此时间复杂度取决于状态总数，即网格尺寸 <span class="arithmatex">\(O(nm)\)</span> 。</p>
<p><img alt="记忆化搜索递归树" src="../dp_solution_pipeline.assets/min_path_sum_dfs_mem.png" /></p>
<p><a class="glightbox" href="../dp_solution_pipeline.assets/min_path_sum_dfs_mem.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="记忆化搜索递归树" src="../dp_solution_pipeline.assets/min_path_sum_dfs_mem.png" /></a></p>
<p align="center"> 图 14-15 &nbsp; 记忆化搜索递归树 </p>
<h3 id="3">3. &nbsp; 方法三：动态规划<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
@@ -4431,40 +4258,40 @@ dp[i, j] = \min(dp[i-1, j], dp[i, j-1]) + grid[i, j]
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chapter_dynamic_programming/edit_distance_problem/index.html
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10.2 &nbsp; 二分查找插入点
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10.3 &nbsp; 二分查找边界
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第 12 章 &nbsp; 分治
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12.1 &nbsp; 分治算法
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12.2 &nbsp; 分治搜索策略
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12.3 &nbsp; 构建树问题
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12.4 &nbsp; 汉诺塔问题
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第 14 章 &nbsp; 动态规划
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14.1 &nbsp; 初探动态规划
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<p>你可以在一个字符串中进行三种编辑操作：插入一个字符、删除一个字符、替换字符为任意一个字符。</p>
</div>
<p>如图 14-27 所示，将 <code>kitten</code> 转换为 <code>sitting</code> 需要编辑 3 步，包括 2 次替换操作与 1 次添加操作；将 <code>hello</code> 转换为 <code>algo</code> 需要 3 步，包括 2 次替换操作和 1 次删除操作。</p>
<p><img alt="编辑距离的示例数据" src="../edit_distance_problem.assets/edit_distance_example.png" /></p>
<p><a class="glightbox" href="../edit_distance_problem.assets/edit_distance_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="编辑距离的示例数据" src="../edit_distance_problem.assets/edit_distance_example.png" /></a></p>
<p align="center"> 图 14-27 &nbsp; 编辑距离的示例数据 </p>
<p><strong>编辑距离问题可以很自然地用决策树模型来解释</strong>。字符串对应树节点，一轮决策（一次编辑操作）对应树的一条边。</p>
<p>如图 14-28 所示，在不限制操作的情况下，每个节点都可以派生出许多条边，每条边对应一种操作，这意味着从 <code>hello</code> 转换到 <code>algo</code> 有许多种可能的路径。</p>
<p>从决策树的角度看，本题的目标是求解节点 <code>hello</code> 和节点 <code>algo</code> 之间的最短路径。</p>
<p><img alt="基于决策树模型表示编辑距离问题" src="../edit_distance_problem.assets/edit_distance_decision_tree.png" /></p>
<p><a class="glightbox" href="../edit_distance_problem.assets/edit_distance_decision_tree.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="基于决策树模型表示编辑距离问题" src="../edit_distance_problem.assets/edit_distance_decision_tree.png" /></a></p>
<p align="center"> 图 14-28 &nbsp; 基于决策树模型表示编辑距离问题 </p>
<h3 id="1">1. &nbsp; 动态规划思路<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
@@ -3530,7 +3357,7 @@
<li>删除 <span class="arithmatex">\(s[i-1]\)</span> ，则剩余子问题 <span class="arithmatex">\(dp[i-1, j]\)</span> 。</li>
<li>将 <span class="arithmatex">\(s[i-1]\)</span> 替换为 <span class="arithmatex">\(t[j-1]\)</span> ，则剩余子问题 <span class="arithmatex">\(dp[i-1, j-1]\)</span> 。</li>
</ol>
<p><img alt="编辑距离的状态转移" src="../edit_distance_problem.assets/edit_distance_state_transfer.png" /></p>
<p><a class="glightbox" href="../edit_distance_problem.assets/edit_distance_state_transfer.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="编辑距离的状态转移" src="../edit_distance_problem.assets/edit_distance_state_transfer.png" /></a></p>
<p align="center"> 图 14-29 &nbsp; 编辑距离的状态转移 </p>
<p>根据以上分析，可得最优子结构：<span class="arithmatex">\(dp[i, j]\)</span> 的最少编辑步数等于 <span class="arithmatex">\(dp[i, j-1]\)</span>、<span class="arithmatex">\(dp[i-1, j]\)</span>、<span class="arithmatex">\(dp[i-1, j-1]\)</span> 三者中的最少编辑步数，再加上本次的编辑步数 <span class="arithmatex">\(1\)</span> 。对应的状态转移方程为：</p>
@@ -3877,49 +3704,49 @@ dp[i, j] = dp[i-1, j-1]
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<div class="tabbed-block">
<p><img alt="编辑距离的动态规划过程" src="../edit_distance_problem.assets/edit_distance_dp_step1.png" /></p>
<p><a class="glightbox" href="../edit_distance_problem.assets/edit_distance_dp_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="编辑距离的动态规划过程" src="../edit_distance_problem.assets/edit_distance_dp_step1.png" /></a></p>
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@@ -4440,10 +4267,15 @@ aria-label="页脚"
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chapter_dynamic_programming/index.html
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<link href="../assets/stylesheets/glightbox.min.css" rel="stylesheet"/><style>
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10.2 &nbsp; 二分查找插入点
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10.3 &nbsp; 二分查找边界
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第 12 章 &nbsp; 分治
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12.1 &nbsp; 分治算法
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12.3 &nbsp; 构建树问题
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12.4 &nbsp; 汉诺塔问题
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12.5 &nbsp; 小结
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第 14 章 &nbsp; 动态规划
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14.1 &nbsp; 初探动态规划
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14.2 &nbsp; DP 问题特性
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14.3 &nbsp; DP 解题思路
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14.6 &nbsp; 编辑距离问题
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15.1 &nbsp; 贪心算法
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15.2 &nbsp; 分数背包问题
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15.3 &nbsp; 最大容量问题
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15.4 &nbsp; 最大切分乘积问题
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15.5 &nbsp; 小结
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<!-- Page content -->
<h1 id="14">第 14 章 &nbsp; 动态规划<a class="headerlink" href="#14" title="Permanent link">&para;</a></h1>
<div class="center-table">
<p><img alt="动态规划" src="../assets/covers/chapter_dynamic_programming.jpg" width="600" /></p>
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@@ -3594,10 +3429,15 @@ aria-label="页脚"
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chapter_dynamic_programming/intro_to_dynamic_programming/index.html
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@@ -60,7 +60,18 @@
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html.glightbox-open { overflow: initial; height: 100%; }
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@@ -1875,14 +1886,6 @@
10.2 &nbsp; 二分查找插入点
</span>
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10.3 &nbsp; 二分查找边界
</span>
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</a>
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第 12 章 &nbsp; 分治
</span>
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12.1 &nbsp; 分治算法
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12.2 &nbsp; 分治搜索策略
</span>
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12.3 &nbsp; 构建树问题
</span>
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12.4 &nbsp; 汉诺塔问题
</span>
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12.5 &nbsp; 小结
</span>
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第 14 章 &nbsp; 动态规划
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14.1 &nbsp; 初探动态规划
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14.2 &nbsp; DP 问题特性
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14.4 &nbsp; 0-1 背包问题
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第 15 章 &nbsp; 贪心
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15.1 &nbsp; 贪心算法
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15.2 &nbsp; 分数背包问题
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15.3 &nbsp; 最大容量问题
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15.4 &nbsp; 最大切分乘积问题
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15.5 &nbsp; 小结
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<p>给定一个共有 <span class="arithmatex">\(n\)</span> 阶的楼梯，你每步可以上 <span class="arithmatex">\(1\)</span> 阶或者 <span class="arithmatex">\(2\)</span> 阶，请问有多少种方案可以爬到楼顶。</p>
</div>
<p>如图 14-1 所示，对于一个 <span class="arithmatex">\(3\)</span> 阶楼梯，共有 <span class="arithmatex">\(3\)</span> 种方案可以爬到楼顶。</p>
<p><img alt="爬到第 3 阶的方案数量" src="../intro_to_dynamic_programming.assets/climbing_stairs_example.png" /></p>
<p><a class="glightbox" href="../intro_to_dynamic_programming.assets/climbing_stairs_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="爬到第 3 阶的方案数量" src="../intro_to_dynamic_programming.assets/climbing_stairs_example.png" /></a></p>
<p align="center"> 图 14-1 &nbsp; 爬到第 3 阶的方案数量 </p>
<p>本题的目标是求解方案数量，<strong>我们可以考虑通过回溯来穷举所有可能性</strong>。具体来说，将爬楼梯想象为一个多轮选择的过程：从地面出发，每轮选择上 <span class="arithmatex">\(1\)</span> 阶或 <span class="arithmatex">\(2\)</span> 阶，每当到达楼梯顶部时就将方案数量加 <span class="arithmatex">\(1\)</span> ，当越过楼梯顶部时就将其剪枝。</p>
@@ -3852,7 +3679,7 @@ dp[i-1], dp[i-2], \dots, dp[2], dp[1]
dp[i] = dp[i-1] + dp[i-2]
\]</div>
<p>这意味着在爬楼梯问题中，各个子问题之间存在递推关系，<strong>原问题的解可以由子问题的解构建得来</strong>。图 14-2 展示了该递推关系。</p>
<p><img alt="方案数量递推关系" src="../intro_to_dynamic_programming.assets/climbing_stairs_state_transfer.png" /></p>
<p><a class="glightbox" href="../intro_to_dynamic_programming.assets/climbing_stairs_state_transfer.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="方案数量递推关系" src="../intro_to_dynamic_programming.assets/climbing_stairs_state_transfer.png" /></a></p>
<p align="center"> 图 14-2 &nbsp; 方案数量递推关系 </p>
<p>我们可以根据递推公式得到暴力搜索解法。以 <span class="arithmatex">\(dp[n]\)</span> 为起始点，<strong>递归地将一个较大问题拆解为两个较小问题的和</strong>，直至到达最小子问题 <span class="arithmatex">\(dp[1]\)</span> 和 <span class="arithmatex">\(dp[2]\)</span> 时返回。其中，最小子问题的解是已知的，即 <span class="arithmatex">\(dp[1] = 1\)</span>、<span class="arithmatex">\(dp[2] = 2\)</span> ，表示爬到第 <span class="arithmatex">\(1\)</span>、<span class="arithmatex">\(2\)</span> 阶分别有 <span class="arithmatex">\(1\)</span>、<span class="arithmatex">\(2\)</span> 种方案。</p>
@@ -4052,7 +3879,7 @@ dp[i] = dp[i-1] + dp[i-2]
</div>
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<p>图 14-3 展示了暴力搜索形成的递归树。对于问题 <span class="arithmatex">\(dp[n]\)</span> ，其递归树的深度为 <span class="arithmatex">\(n\)</span> ，时间复杂度为 <span class="arithmatex">\(O(2^n)\)</span> 。指数阶属于爆炸式增长，如果我们输入一个比较大的 <span class="arithmatex">\(n\)</span> ，则会陷入漫长的等待之中。</p>
<p><img alt="爬楼梯对应递归树" src="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_tree.png" /></p>
<p><a class="glightbox" href="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_tree.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="爬楼梯对应递归树" src="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_tree.png" /></a></p>
<p align="center"> 图 14-3 &nbsp; 爬楼梯对应递归树 </p>
<p>观察图 14-3 ，<strong>指数阶的时间复杂度是由于“重叠子问题”导致的</strong>。例如 <span class="arithmatex">\(dp[9]\)</span> 被分解为 <span class="arithmatex">\(dp[8]\)</span> 和 <span class="arithmatex">\(dp[7]\)</span> ，<span class="arithmatex">\(dp[8]\)</span> 被分解为 <span class="arithmatex">\(dp[7]\)</span> 和 <span class="arithmatex">\(dp[6]\)</span> ，两者都包含子问题 <span class="arithmatex">\(dp[7]\)</span> 。</p>
@@ -4339,7 +4166,7 @@ dp[i] = dp[i-1] + dp[i-2]
</div>
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<p>观察图 14-4 ，<strong>经过记忆化处理后，所有重叠子问题都只需被计算一次，时间复杂度被优化至 <span class="arithmatex">\(O(n)\)</span></strong> ，这是一个巨大的飞跃。</p>
<p><img alt="记忆化搜索对应递归树" src="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_memo_tree.png" /></p>
<p><a class="glightbox" href="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_memo_tree.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="记忆化搜索对应递归树" src="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_memo_tree.png" /></a></p>
<p align="center"> 图 14-4 &nbsp; 记忆化搜索对应递归树 </p>
<h2 id="1413">14.1.3 &nbsp; 方法三：动态规划<a class="headerlink" href="#1413" title="Permanent link">&para;</a></h2>
@@ -4551,7 +4378,7 @@ dp[i] = dp[i-1] + dp[i-2]
</div>
</div>
<p>图 14-5 模拟了以上代码的执行过程。</p>
<p><img alt="爬楼梯的动态规划过程" src="../intro_to_dynamic_programming.assets/climbing_stairs_dp.png" /></p>
<p><a class="glightbox" href="../intro_to_dynamic_programming.assets/climbing_stairs_dp.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="爬楼梯的动态规划过程" src="../intro_to_dynamic_programming.assets/climbing_stairs_dp.png" /></a></p>
<p align="center"> 图 14-5 &nbsp; 爬楼梯的动态规划过程 </p>
<p>与回溯算法一样，动态规划也使用“状态”概念来表示问题求解的某个特定阶段，每个状态都对应一个子问题以及相应的局部最优解。例如，爬楼梯问题的状态定义为当前所在楼梯阶数 <span class="arithmatex">\(i\)</span> 。</p>
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<p>给定 <span class="arithmatex">\(n\)</span> 个物品，第 <span class="arithmatex">\(i\)</span> 个物品的重量为 <span class="arithmatex">\(wgt[i-1]\)</span>、价值为 <span class="arithmatex">\(val[i-1]\)</span> ，和一个容量为 <span class="arithmatex">\(cap\)</span> 的背包。每个物品只能选择一次，问在不超过背包容量下能放入物品的最大价值。</p>
</div>
<p>观察图 14-17 ，由于物品编号 <span class="arithmatex">\(i\)</span> 从 <span class="arithmatex">\(1\)</span> 开始计数，数组索引从 <span class="arithmatex">\(0\)</span> 开始计数，因此物品 <span class="arithmatex">\(i\)</span> 对应重量 <span class="arithmatex">\(wgt[i-1]\)</span> 和价值 <span class="arithmatex">\(val[i-1]\)</span> 。</p>
<p><img alt="0-1 背包的示例数据" src="../knapsack_problem.assets/knapsack_example.png" /></p>
<p><a class="glightbox" href="../knapsack_problem.assets/knapsack_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="0-1 背包的示例数据" src="../knapsack_problem.assets/knapsack_example.png" /></a></p>
<p align="center"> 图 14-17 &nbsp; 0-1 背包的示例数据 </p>
<p>我们可以将 0-1 背包问题看作是一个由 <span class="arithmatex">\(n\)</span> 轮决策组成的过程，每个物体都有不放入和放入两种决策，因此该问题是满足决策树模型的。</p>
@@ -3770,7 +3597,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
</div>
<p>如图 14-18 所示，由于每个物品都会产生不选和选两条搜索分支，因此时间复杂度为 <span class="arithmatex">\(O(2^n)\)</span> 。</p>
<p>观察递归树，容易发现其中存在重叠子问题，例如 <span class="arithmatex">\(dp[1, 10]\)</span> 等。而当物品较多、背包容量较大，尤其是相同重量的物品较多时，重叠子问题的数量将会大幅增多。</p>
<p><img alt="0-1 背包的暴力搜索递归树" src="../knapsack_problem.assets/knapsack_dfs.png" /></p>
<p><a class="glightbox" href="../knapsack_problem.assets/knapsack_dfs.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="0-1 背包的暴力搜索递归树" src="../knapsack_problem.assets/knapsack_dfs.png" /></a></p>
<p align="center"> 图 14-18 &nbsp; 0-1 背包的暴力搜索递归树 </p>
<h3 id="2">2. &nbsp; 方法二：记忆化搜索<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
@@ -4061,7 +3888,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
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<p>图 14-19 展示了在记忆化递归中被剪掉的搜索分支。</p>
<p><img alt="0-1 背包的记忆化搜索递归树" src="../knapsack_problem.assets/knapsack_dfs_mem.png" /></p>
<p><a class="glightbox" href="../knapsack_problem.assets/knapsack_dfs_mem.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="0-1 背包的记忆化搜索递归树" src="../knapsack_problem.assets/knapsack_dfs_mem.png" /></a></p>
<p align="center"> 图 14-19 &nbsp; 0-1 背包的记忆化搜索递归树 </p>
<h3 id="3">3. &nbsp; 方法三：动态规划<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
@@ -4333,46 +4160,46 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
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<p class="admonition-title">Question</p>
<p>给定 <span class="arithmatex">\(n\)</span> 个物品，第 <span class="arithmatex">\(i\)</span> 个物品的重量为 <span class="arithmatex">\(wgt[i-1]\)</span>、价值为 <span class="arithmatex">\(val[i-1]\)</span> ，和一个容量为 <span class="arithmatex">\(cap\)</span> 的背包。<strong>每个物品可以重复选取</strong>，问在不超过背包容量下能放入物品的最大价值。</p>
</div>
<p><img alt="完全背包问题的示例数据" src="../unbounded_knapsack_problem.assets/unbounded_knapsack_example.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/unbounded_knapsack_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="完全背包问题的示例数据" src="../unbounded_knapsack_problem.assets/unbounded_knapsack_example.png" /></a></p>
<p align="center"> 图 14-22 &nbsp; 完全背包问题的示例数据 </p>
<h3 id="1">1. &nbsp; 动态规划思路<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
@@ -3953,22 +3780,22 @@ dp[i, c] = \max(dp[i-1, c], dp[i, c - wgt[i-1]] + val[i-1])
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<p><img alt="unbounded_knapsack_dp_comp_step3" src="../unbounded_knapsack_problem.assets/unbounded_knapsack_dp_comp_step3.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/unbounded_knapsack_dp_comp_step3.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="unbounded_knapsack_dp_comp_step3" src="../unbounded_knapsack_problem.assets/unbounded_knapsack_dp_comp_step3.png" /></a></p>
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<p><img alt="unbounded_knapsack_dp_comp_step4" src="../unbounded_knapsack_problem.assets/unbounded_knapsack_dp_comp_step4.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/unbounded_knapsack_dp_comp_step4.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="unbounded_knapsack_dp_comp_step4" src="../unbounded_knapsack_problem.assets/unbounded_knapsack_dp_comp_step4.png" /></a></p>
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<p><img alt="unbounded_knapsack_dp_comp_step5" src="../unbounded_knapsack_problem.assets/unbounded_knapsack_dp_comp_step5.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/unbounded_knapsack_dp_comp_step5.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="unbounded_knapsack_dp_comp_step5" src="../unbounded_knapsack_problem.assets/unbounded_knapsack_dp_comp_step5.png" /></a></p>
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<p><img alt="unbounded_knapsack_dp_comp_step6" src="../unbounded_knapsack_problem.assets/unbounded_knapsack_dp_comp_step6.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/unbounded_knapsack_dp_comp_step6.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="unbounded_knapsack_dp_comp_step6" src="../unbounded_knapsack_problem.assets/unbounded_knapsack_dp_comp_step6.png" /></a></p>
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</div>
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@@ -4232,7 +4059,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i, c - wgt[i-1]] + val[i-1])
<p class="admonition-title">Question</p>
<p>给定 <span class="arithmatex">\(n\)</span> 种硬币，第 <span class="arithmatex">\(i\)</span> 种硬币的面值为 <span class="arithmatex">\(coins[i - 1]\)</span> ，目标金额为 <span class="arithmatex">\(amt\)</span> ，<strong>每种硬币可以重复选取</strong>，问能够凑出目标金额的最少硬币个数。如果无法凑出目标金额则返回 <span class="arithmatex">\(-1\)</span> 。</p>
</div>
<p><img alt="零钱兑换问题的示例数据" src="../unbounded_knapsack_problem.assets/coin_change_example.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="零钱兑换问题的示例数据" src="../unbounded_knapsack_problem.assets/coin_change_example.png" /></a></p>
<p align="center"> 图 14-24 &nbsp; 零钱兑换问题的示例数据 </p>
<h3 id="1_1">1. &nbsp; 动态规划思路<a class="headerlink" href="#1_1" title="Permanent link">&para;</a></h3>
@@ -4579,49 +4406,49 @@ dp[i, a] = \min(dp[i-1, a], dp[i, a - coins[i-1]] + 1)
<div class="tabbed-set tabbed-alternate" data-tabs="5:15"><input checked="checked" id="__tabbed_5_1" name="__tabbed_5" type="radio" /><input id="__tabbed_5_2" name="__tabbed_5" type="radio" /><input id="__tabbed_5_3" name="__tabbed_5" type="radio" /><input id="__tabbed_5_4" name="__tabbed_5" type="radio" /><input id="__tabbed_5_5" name="__tabbed_5" type="radio" /><input id="__tabbed_5_6" name="__tabbed_5" type="radio" /><input id="__tabbed_5_7" name="__tabbed_5" type="radio" /><input id="__tabbed_5_8" name="__tabbed_5" type="radio" /><input id="__tabbed_5_9" name="__tabbed_5" type="radio" /><input id="__tabbed_5_10" name="__tabbed_5" type="radio" /><input id="__tabbed_5_11" name="__tabbed_5" type="radio" /><input id="__tabbed_5_12" name="__tabbed_5" type="radio" /><input id="__tabbed_5_13" name="__tabbed_5" type="radio" /><input id="__tabbed_5_14" name="__tabbed_5" type="radio" /><input id="__tabbed_5_15" name="__tabbed_5" type="radio" /><div class="tabbed-labels"><label for="__tabbed_5_1">&lt;1&gt;</label><label for="__tabbed_5_2">&lt;2&gt;</label><label for="__tabbed_5_3">&lt;3&gt;</label><label for="__tabbed_5_4">&lt;4&gt;</label><label for="__tabbed_5_5">&lt;5&gt;</label><label for="__tabbed_5_6">&lt;6&gt;</label><label for="__tabbed_5_7">&lt;7&gt;</label><label for="__tabbed_5_8">&lt;8&gt;</label><label for="__tabbed_5_9">&lt;9&gt;</label><label for="__tabbed_5_10">&lt;10&gt;</label><label for="__tabbed_5_11">&lt;11&gt;</label><label for="__tabbed_5_12">&lt;12&gt;</label><label for="__tabbed_5_13">&lt;13&gt;</label><label for="__tabbed_5_14">&lt;14&gt;</label><label for="__tabbed_5_15">&lt;15&gt;</label></div>
<div class="tabbed-content">
<div class="tabbed-block">
<p><img alt="零钱兑换问题的动态规划过程" src="../unbounded_knapsack_problem.assets/coin_change_dp_step1.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_dp_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="零钱兑换问题的动态规划过程" src="../unbounded_knapsack_problem.assets/coin_change_dp_step1.png" /></a></p>
</div>
<div class="tabbed-block">
<p><img alt="coin_change_dp_step2" src="../unbounded_knapsack_problem.assets/coin_change_dp_step2.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_dp_step2.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="coin_change_dp_step2" src="../unbounded_knapsack_problem.assets/coin_change_dp_step2.png" /></a></p>
</div>
<div class="tabbed-block">
<p><img alt="coin_change_dp_step3" src="../unbounded_knapsack_problem.assets/coin_change_dp_step3.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_dp_step3.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="coin_change_dp_step3" src="../unbounded_knapsack_problem.assets/coin_change_dp_step3.png" /></a></p>
</div>
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<p><img alt="coin_change_dp_step4" src="../unbounded_knapsack_problem.assets/coin_change_dp_step4.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_dp_step4.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="coin_change_dp_step4" src="../unbounded_knapsack_problem.assets/coin_change_dp_step4.png" /></a></p>
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<p><img alt="coin_change_dp_step5" src="../unbounded_knapsack_problem.assets/coin_change_dp_step5.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_dp_step5.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="coin_change_dp_step5" src="../unbounded_knapsack_problem.assets/coin_change_dp_step5.png" /></a></p>
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<p><img alt="coin_change_dp_step6" src="../unbounded_knapsack_problem.assets/coin_change_dp_step6.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_dp_step6.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="coin_change_dp_step6" src="../unbounded_knapsack_problem.assets/coin_change_dp_step6.png" /></a></p>
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<p><img alt="coin_change_dp_step7" src="../unbounded_knapsack_problem.assets/coin_change_dp_step7.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_dp_step7.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="coin_change_dp_step7" src="../unbounded_knapsack_problem.assets/coin_change_dp_step7.png" /></a></p>
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<p><img alt="coin_change_dp_step8" src="../unbounded_knapsack_problem.assets/coin_change_dp_step8.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_dp_step8.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="coin_change_dp_step8" src="../unbounded_knapsack_problem.assets/coin_change_dp_step8.png" /></a></p>
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<p><img alt="coin_change_dp_step9" src="../unbounded_knapsack_problem.assets/coin_change_dp_step9.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_dp_step9.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="coin_change_dp_step9" src="../unbounded_knapsack_problem.assets/coin_change_dp_step9.png" /></a></p>
</div>
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<p><img alt="coin_change_dp_step10" src="../unbounded_knapsack_problem.assets/coin_change_dp_step10.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_dp_step10.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="coin_change_dp_step10" src="../unbounded_knapsack_problem.assets/coin_change_dp_step10.png" /></a></p>
</div>
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<p><img alt="coin_change_dp_step11" src="../unbounded_knapsack_problem.assets/coin_change_dp_step11.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_dp_step11.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="coin_change_dp_step11" src="../unbounded_knapsack_problem.assets/coin_change_dp_step11.png" /></a></p>
</div>
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<p><img alt="coin_change_dp_step12" src="../unbounded_knapsack_problem.assets/coin_change_dp_step12.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_dp_step12.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="coin_change_dp_step12" src="../unbounded_knapsack_problem.assets/coin_change_dp_step12.png" /></a></p>
</div>
<div class="tabbed-block">
<p><img alt="coin_change_dp_step13" src="../unbounded_knapsack_problem.assets/coin_change_dp_step13.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_dp_step13.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="coin_change_dp_step13" src="../unbounded_knapsack_problem.assets/coin_change_dp_step13.png" /></a></p>
</div>
<div class="tabbed-block">
<p><img alt="coin_change_dp_step14" src="../unbounded_knapsack_problem.assets/coin_change_dp_step14.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_dp_step14.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="coin_change_dp_step14" src="../unbounded_knapsack_problem.assets/coin_change_dp_step14.png" /></a></p>
</div>
<div class="tabbed-block">
<p><img alt="coin_change_dp_step15" src="../unbounded_knapsack_problem.assets/coin_change_dp_step15.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_dp_step15.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="coin_change_dp_step15" src="../unbounded_knapsack_problem.assets/coin_change_dp_step15.png" /></a></p>
</div>
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</div>
@@ -4917,7 +4744,7 @@ dp[i, a] = \min(dp[i-1, a], dp[i, a - coins[i-1]] + 1)
<p class="admonition-title">Question</p>
<p>给定 <span class="arithmatex">\(n\)</span> 种硬币，第 <span class="arithmatex">\(i\)</span> 种硬币的面值为 <span class="arithmatex">\(coins[i - 1]\)</span> ，目标金额为 <span class="arithmatex">\(amt\)</span> ，每种硬币可以重复选取，<strong>问在凑出目标金额的硬币组合数量</strong>。</p>
</div>
<p><img alt="零钱兑换问题 II 的示例数据" src="../unbounded_knapsack_problem.assets/coin_change_ii_example.png" /></p>
<p><a class="glightbox" href="../unbounded_knapsack_problem.assets/coin_change_ii_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="零钱兑换问题 II 的示例数据" src="../unbounded_knapsack_problem.assets/coin_change_ii_example.png" /></a></p>
<p align="center"> 图 14-26 &nbsp; 零钱兑换问题 II 的示例数据 </p>
<h3 id="1_2">1. &nbsp; 动态规划思路<a class="headerlink" href="#1_2" title="Permanent link">&para;</a></h3>
@@ -5648,10 +5475,15 @@ aria-label="页脚"
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