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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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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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14.3 &nbsp; DP 解题思路
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14.4 &nbsp; 0-1 背包问题
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14.5 &nbsp; 完全背包问题
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14.6 &nbsp; 编辑距离问题
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14.7 &nbsp; 小结
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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 <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">
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<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">
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<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]
</div>
<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]
</div>
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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>
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<div class="tabbed-set tabbed-alternate" data-tabs="4:12"><input checked="checked" id="__tabbed_4_1" name="__tabbed_4" type="radio" /><input id="__tabbed_4_2" name="__tabbed_4" type="radio" /><input id="__tabbed_4_3" name="__tabbed_4" type="radio" /><input id="__tabbed_4_4" name="__tabbed_4" type="radio" /><input id="__tabbed_4_5" name="__tabbed_4" type="radio" /><input id="__tabbed_4_6" name="__tabbed_4" type="radio" /><input id="__tabbed_4_7" name="__tabbed_4" type="radio" /><input id="__tabbed_4_8" name="__tabbed_4" type="radio" /><input id="__tabbed_4_9" name="__tabbed_4" type="radio" /><input id="__tabbed_4_10" name="__tabbed_4" type="radio" /><input id="__tabbed_4_11" name="__tabbed_4" type="radio" /><input id="__tabbed_4_12" name="__tabbed_4" type="radio" /><div class="tabbed-labels"><label for="__tabbed_4_1">&lt;1&gt;</label><label for="__tabbed_4_2">&lt;2&gt;</label><label for="__tabbed_4_3">&lt;3&gt;</label><label for="__tabbed_4_4">&lt;4&gt;</label><label for="__tabbed_4_5">&lt;5&gt;</label><label for="__tabbed_4_6">&lt;6&gt;</label><label for="__tabbed_4_7">&lt;7&gt;</label><label for="__tabbed_4_8">&lt;8&gt;</label><label for="__tabbed_4_9">&lt;9&gt;</label><label for="__tabbed_4_10">&lt;10&gt;</label><label for="__tabbed_4_11">&lt;11&gt;</label><label for="__tabbed_4_12">&lt;12&gt;</label></div>
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@ -4906,10 +4733,15 @@ aria-label="页脚"
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