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2026-07-07 13:14:19 +00:00 · 2023-11-09 05:13:54 +08:00
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-<p><a class="glightbox" href="../backtracking_algorithm.assets/preorder_find_nodes.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="在前序遍历中搜索节点" src="../backtracking_algorithm.assets/preorder_find_nodes.png" /></a></p>
+<p><a class="glightbox" href="../backtracking_algorithm.assets/preorder_find_nodes.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="在前序遍历中搜索节点" class="animation-figure" src="../backtracking_algorithm.assets/preorder_find_nodes.png" /></a></p>
 <p align="center"> 图 13-1 &nbsp; 在前序遍历中搜索节点 </p>

 <h2 id="1311">13.1.1 &nbsp; 尝试与回退<a class="headerlink" href="#1311" title="Permanent link">&para;</a></h2>
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-<p><a class="glightbox" href="../backtracking_algorithm.assets/preorder_find_paths_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="尝试与回退" src="../backtracking_algorithm.assets/preorder_find_paths_step1.png" /></a></p>
+<p><a class="glightbox" href="../backtracking_algorithm.assets/preorder_find_paths_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="尝试与回退" class="animation-figure" src="../backtracking_algorithm.assets/preorder_find_paths_step1.png" /></a></p>
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+<p><a class="glightbox" href="../backtracking_algorithm.assets/preorder_find_paths_step3.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="preorder_find_paths_step3" class="animation-figure" src="../backtracking_algorithm.assets/preorder_find_paths_step3.png" /></a></p>
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+<p><a class="glightbox" href="../backtracking_algorithm.assets/preorder_find_paths_step4.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="preorder_find_paths_step4" class="animation-figure" src="../backtracking_algorithm.assets/preorder_find_paths_step4.png" /></a></p>
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+<p><a class="glightbox" href="../backtracking_algorithm.assets/preorder_find_paths_step5.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="preorder_find_paths_step5" class="animation-figure" src="../backtracking_algorithm.assets/preorder_find_paths_step5.png" /></a></p>
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+<p><a class="glightbox" href="../backtracking_algorithm.assets/preorder_find_paths_step6.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="preorder_find_paths_step6" class="animation-figure" src="../backtracking_algorithm.assets/preorder_find_paths_step6.png" /></a></p>
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+<p><a class="glightbox" href="../backtracking_algorithm.assets/preorder_find_paths_step7.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="preorder_find_paths_step7" class="animation-figure" src="../backtracking_algorithm.assets/preorder_find_paths_step7.png" /></a></p>
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+<p><a class="glightbox" href="../backtracking_algorithm.assets/preorder_find_paths_step9.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="preorder_find_paths_step9" class="animation-figure" src="../backtracking_algorithm.assets/preorder_find_paths_step9.png" /></a></p>
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+<p><a class="glightbox" href="../backtracking_algorithm.assets/preorder_find_paths_step10.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="preorder_find_paths_step10" class="animation-figure" src="../backtracking_algorithm.assets/preorder_find_paths_step10.png" /></a></p>
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+<p><a class="glightbox" href="../backtracking_algorithm.assets/preorder_find_paths_step11.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="preorder_find_paths_step11" class="animation-figure" src="../backtracking_algorithm.assets/preorder_find_paths_step11.png" /></a></p>
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@@ -4114,7 +4114,7 @@
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 <p>剪枝是一个非常形象的名词。如图 13-3 所示，在搜索过程中，<strong>我们“剪掉”了不满足约束条件的搜索分支</strong>，避免许多无意义的尝试，从而提高了搜索效率。</p>
-<p><a class="glightbox" href="../backtracking_algorithm.assets/preorder_find_constrained_paths.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="根据约束条件剪枝" src="../backtracking_algorithm.assets/preorder_find_constrained_paths.png" /></a></p>
+<p><a class="glightbox" href="../backtracking_algorithm.assets/preorder_find_constrained_paths.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="根据约束条件剪枝" class="animation-figure" src="../backtracking_algorithm.assets/preorder_find_constrained_paths.png" /></a></p>
 <p align="center"> 图 13-3 &nbsp; 根据约束条件剪枝 </p>

 <h2 id="1313">13.1.3 &nbsp; 框架代码<a class="headerlink" href="#1313" title="Permanent link">&para;</a></h2>
@@ -4944,7 +4944,7 @@
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 <p>根据题意，我们在找到值为 <span class="arithmatex">\(7\)</span> 的节点后应该继续搜索，<strong>因此需要将记录解之后的 <code>return</code> 语句删除</strong>。图 13-4 对比了保留或删除 <code>return</code> 语句的搜索过程。</p>
-<p><a class="glightbox" href="../backtracking_algorithm.assets/backtrack_remove_return_or_not.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="保留与删除 return 的搜索过程对比" src="../backtracking_algorithm.assets/backtrack_remove_return_or_not.png" /></a></p>
+<p><a class="glightbox" href="../backtracking_algorithm.assets/backtrack_remove_return_or_not.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="保留与删除 return 的搜索过程对比" class="animation-figure" src="../backtracking_algorithm.assets/backtrack_remove_return_or_not.png" /></a></p>
 <p align="center"> 图 13-4 &nbsp; 保留与删除 return 的搜索过程对比 </p>

 <p>相比基于前序遍历的代码实现，基于回溯算法框架的代码实现虽然显得啰嗦，但通用性更好。实际上，<strong>许多回溯问题都可以在该框架下解决</strong>。我们只需根据具体问题来定义 <code>state</code> 和 <code>choices</code> ，并实现框架中的各个方法即可。</p>
@@ -3292,7 +3292,7 @@
 <!-- Page content -->
 <h1 id="13">第 13 章 &nbsp; 回溯<a class="headerlink" href="#13" title="Permanent link">&para;</a></h1>
 <div class="center-table">
-<p><a class="glightbox" href="../assets/covers/chapter_backtracking.jpg" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="回溯" src="../assets/covers/chapter_backtracking.jpg" width="600" /></a></p>
+<p><a class="glightbox" href="../assets/covers/chapter_backtracking.jpg" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="回溯" class="cover-image" src="../assets/covers/chapter_backtracking.jpg" /></a></p>
 </div>
 <div class="admonition abstract">
 <p class="admonition-title">Abstract</p>
@@ -3371,18 +3371,18 @@
 <p>根据国际象棋的规则，皇后可以攻击与之处在同一行或同一列或同一斜线上的棋子。给定 <span class="arithmatex">\(n\)</span> 个皇后和一个 <span class="arithmatex">\(n \times n\)</span> 大小的棋盘，寻找使得所有皇后之间无法相互攻击的摆放方案。</p>
 </div>
 <p>如图 13-15 所示，当 <span class="arithmatex">\(n = 4\)</span> 时，共可以找到两个解。从回溯算法的角度看，<span class="arithmatex">\(n \times n\)</span> 大小的棋盘共有 <span class="arithmatex">\(n^2\)</span> 个格子，给出了所有的选择 <code>choices</code> 。在逐个放置皇后的过程中，棋盘状态在不断地变化，每个时刻的棋盘就是状态 <code>state</code> 。</p>
-<p><a class="glightbox" href="../n_queens_problem.assets/solution_4_queens.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="4 皇后问题的解" src="../n_queens_problem.assets/solution_4_queens.png" /></a></p>
+<p><a class="glightbox" href="../n_queens_problem.assets/solution_4_queens.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="4 皇后问题的解" class="animation-figure" src="../n_queens_problem.assets/solution_4_queens.png" /></a></p>
 <p align="center"> 图 13-15 &nbsp; 4 皇后问题的解 </p>

 <p>图 13-16 展示了本题的三个约束条件：<strong>多个皇后不能在同一行、同一列、同一对角线</strong>。值得注意的是，对角线分为主对角线 <code>\</code> 和次对角线 <code>/</code> 两种。</p>
-<p><a class="glightbox" href="../n_queens_problem.assets/n_queens_constraints.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="n 皇后问题的约束条件" src="../n_queens_problem.assets/n_queens_constraints.png" /></a></p>
+<p><a class="glightbox" href="../n_queens_problem.assets/n_queens_constraints.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="n 皇后问题的约束条件" class="animation-figure" src="../n_queens_problem.assets/n_queens_constraints.png" /></a></p>
 <p align="center"> 图 13-16 &nbsp; n 皇后问题的约束条件 </p>

 <h3 id="1">1. &nbsp; 逐行放置策略<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
 <p>皇后的数量和棋盘的行数都为 <span class="arithmatex">\(n\)</span> ，因此我们容易得到一个推论：<strong>棋盘每行都允许且只允许放置一个皇后</strong>。</p>
 <p>也就是说，我们可以采取逐行放置策略：从第一行开始，在每行放置一个皇后，直至最后一行结束。</p>
 <p>如图 13-17 所示，为 <span class="arithmatex">\(4\)</span> 皇后问题的逐行放置过程。受画幅限制，图 13-17 仅展开了第一行的其中一个搜索分支，并且将不满足列约束和对角线约束的方案都进行了剪枝。</p>
-<p><a class="glightbox" href="../n_queens_problem.assets/n_queens_placing.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="逐行放置策略" src="../n_queens_problem.assets/n_queens_placing.png" /></a></p>
+<p><a class="glightbox" href="../n_queens_problem.assets/n_queens_placing.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="逐行放置策略" class="animation-figure" src="../n_queens_problem.assets/n_queens_placing.png" /></a></p>
 <p align="center"> 图 13-17 &nbsp; 逐行放置策略 </p>

 <p>本质上看，<strong>逐行放置策略起到了剪枝的作用</strong>，它避免了同一行出现多个皇后的所有搜索分支。</p>
@@ -3391,7 +3391,7 @@
 <p>那么，如何处理对角线约束呢？设棋盘中某个格子的行列索引为 <span class="arithmatex">\((row, col)\)</span> ，选定矩阵中的某条主对角线，我们发现该对角线上所有格子的行索引减列索引都相等，<strong>即对角线上所有格子的 <span class="arithmatex">\(row - col\)</span> 为恒定值</strong>。</p>
 <p>也就是说，如果两个格子满足 <span class="arithmatex">\(row_1 - col_1 = row_2 - col_2\)</span> ，则它们一定处在同一条主对角线上。利用该规律，我们可以借助图 13-18 所示的数组 <code>diags1</code> ，记录每条主对角线上是否有皇后。</p>
 <p>同理，<strong>次对角线上的所有格子的 <span class="arithmatex">\(row + col\)</span> 是恒定值</strong>。我们同样也可以借助数组 <code>diags2</code> 来处理次对角线约束。</p>
-<p><a class="glightbox" href="../n_queens_problem.assets/n_queens_cols_diagonals.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="处理列约束和对角线约束" src="../n_queens_problem.assets/n_queens_cols_diagonals.png" /></a></p>
+<p><a class="glightbox" href="../n_queens_problem.assets/n_queens_cols_diagonals.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="处理列约束和对角线约束" class="animation-figure" src="../n_queens_problem.assets/n_queens_cols_diagonals.png" /></a></p>
 <p align="center"> 图 13-18 &nbsp; 处理列约束和对角线约束 </p>

 <h3 id="3">3. &nbsp; 代码实现<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
@@ -3482,7 +3482,7 @@
 <p>从回溯算法的角度看，<strong>我们可以把生成排列的过程想象成一系列选择的结果</strong>。假设输入数组为 <span class="arithmatex">\([1, 2, 3]\)</span> ，如果我们先选择 <span class="arithmatex">\(1\)</span>、再选择 <span class="arithmatex">\(3\)</span>、最后选择 <span class="arithmatex">\(2\)</span> ，则获得排列 <span class="arithmatex">\([1, 3, 2]\)</span> 。回退表示撤销一个选择，之后继续尝试其他选择。</p>
 <p>从回溯代码的角度看，候选集合 <code>choices</code> 是输入数组中的所有元素，状态 <code>state</code> 是直至目前已被选择的元素。请注意，每个元素只允许被选择一次，<strong>因此 <code>state</code> 中的所有元素都应该是唯一的</strong>。</p>
 <p>如图 13-5 所示，我们可以将搜索过程展开成一个递归树，树中的每个节点代表当前状态 <code>state</code> 。从根节点开始，经过三轮选择后到达叶节点，每个叶节点都对应一个排列。</p>
-<p><a class="glightbox" href="../permutations_problem.assets/permutations_i.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="全排列的递归树" src="../permutations_problem.assets/permutations_i.png" /></a></p>
+<p><a class="glightbox" href="../permutations_problem.assets/permutations_i.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="全排列的递归树" class="animation-figure" src="../permutations_problem.assets/permutations_i.png" /></a></p>
 <p align="center"> 图 13-5 &nbsp; 全排列的递归树 </p>

 <h3 id="1">1. &nbsp; 重复选择剪枝<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
@@ -3492,7 +3492,7 @@
 <li>遍历选择列表 <code>choices</code> 时，跳过所有已被选择过的节点，即剪枝。</li>
 </ul>
 <p>如图 13-6 所示，假设我们第一轮选择 1 ，第二轮选择 3 ，第三轮选择 2 ，则需要在第二轮剪掉元素 1 的分支，在第三轮剪掉元素 1 和元素 3 的分支。</p>
-<p><a class="glightbox" href="../permutations_problem.assets/permutations_i_pruning.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="全排列剪枝示例" src="../permutations_problem.assets/permutations_i_pruning.png" /></a></p>
+<p><a class="glightbox" href="../permutations_problem.assets/permutations_i_pruning.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="全排列剪枝示例" class="animation-figure" src="../permutations_problem.assets/permutations_i_pruning.png" /></a></p>
 <p align="center"> 图 13-6 &nbsp; 全排列剪枝示例 </p>

 <p>观察图 13-6 发现，该剪枝操作将搜索空间大小从 <span class="arithmatex">\(O(n^n)\)</span> 降低至 <span class="arithmatex">\(O(n!)\)</span> 。</p>
@@ -3901,7 +3901,7 @@
 </div>
 <p>假设输入数组为 <span class="arithmatex">\([1, 1, 2]\)</span> 。为了方便区分两个重复元素 <span class="arithmatex">\(1\)</span> ，我们将第二个 <span class="arithmatex">\(1\)</span> 记为 <span class="arithmatex">\(\hat{1}\)</span> 。</p>
 <p>如图 13-7 所示，上述方法生成的排列有一半都是重复的。</p>
-<p><a class="glightbox" href="../permutations_problem.assets/permutations_ii.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="重复排列" src="../permutations_problem.assets/permutations_ii.png" /></a></p>
+<p><a class="glightbox" href="../permutations_problem.assets/permutations_ii.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="重复排列" class="animation-figure" src="../permutations_problem.assets/permutations_ii.png" /></a></p>
 <p align="center"> 图 13-7 &nbsp; 重复排列 </p>

 <p>那么如何去除重复的排列呢？最直接地，考虑借助一个哈希表，直接对排列结果进行去重。然而这样做不够优雅，<strong>因为生成重复排列的搜索分支是没有必要的，应当被提前识别并剪枝</strong>，这样可以进一步提升算法效率。</p>
@@ -3909,7 +3909,7 @@
 <p>观察图 13-8 ，在第一轮中，选择 <span class="arithmatex">\(1\)</span> 或选择 <span class="arithmatex">\(\hat{1}\)</span> 是等价的，在这两个选择之下生成的所有排列都是重复的。因此应该把 <span class="arithmatex">\(\hat{1}\)</span> 剪枝掉。</p>
 <p>同理，在第一轮选择 <span class="arithmatex">\(2\)</span> 之后，第二轮选择中的 <span class="arithmatex">\(1\)</span> 和 <span class="arithmatex">\(\hat{1}\)</span> 也会产生重复分支，因此也应将第二轮的 <span class="arithmatex">\(\hat{1}\)</span> 剪枝。</p>
 <p>本质上看，<strong>我们的目标是在某一轮选择中，保证多个相等的元素仅被选择一次</strong>。</p>
-<p><a class="glightbox" href="../permutations_problem.assets/permutations_ii_pruning.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="重复排列剪枝" src="../permutations_problem.assets/permutations_ii_pruning.png" /></a></p>
+<p><a class="glightbox" href="../permutations_problem.assets/permutations_ii_pruning.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="重复排列剪枝" class="animation-figure" src="../permutations_problem.assets/permutations_ii_pruning.png" /></a></p>
 <p align="center"> 图 13-8 &nbsp; 重复排列剪枝 </p>

 <h3 id="2_1">2. &nbsp; 代码实现<a class="headerlink" href="#2_1" title="Permanent link">&para;</a></h3>
@@ -4342,7 +4342,7 @@
 <li><strong>相等元素剪枝</strong>：每轮选择（即每个调用的 <code>backtrack</code> 函数）都包含一个 <code>duplicated</code> 。它记录的是在本轮遍历（即 <code>for</code> 循环）中哪些元素已被选择过，作用是保证相等的元素只被选择一次。</li>
 </ul>
 <p>图 13-9 展示了两个剪枝条件的生效范围。注意，树中的每个节点代表一个选择，从根节点到叶节点的路径上的各个节点构成一个排列。</p>
-<p><a class="glightbox" href="../permutations_problem.assets/permutations_ii_pruning_summary.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="两种剪枝条件的作用范围" src="../permutations_problem.assets/permutations_ii_pruning_summary.png" /></a></p>
+<p><a class="glightbox" href="../permutations_problem.assets/permutations_ii_pruning_summary.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="两种剪枝条件的作用范围" class="animation-figure" src="../permutations_problem.assets/permutations_ii_pruning_summary.png" /></a></p>
 <p align="center"> 图 13-9 &nbsp; 两种剪枝条件的作用范围 </p>

 <!-- Source file information -->
@@ -3848,7 +3848,7 @@
 </div>
 <p>向以上代码输入数组 <span class="arithmatex">\([3, 4, 5]\)</span> 和目标元素 <span class="arithmatex">\(9\)</span> ，输出结果为 <span class="arithmatex">\([3, 3, 3], [4, 5], [5, 4]\)</span> 。<strong>虽然成功找出了所有和为 <span class="arithmatex">\(9\)</span> 的子集，但其中存在重复的子集 <span class="arithmatex">\([4, 5]\)</span> 和 <span class="arithmatex">\([5, 4]\)</span></strong> 。</p>
 <p>这是因为搜索过程是区分选择顺序的，然而子集不区分选择顺序。如图 13-10 所示，先选 <span class="arithmatex">\(4\)</span> 后选 <span class="arithmatex">\(5\)</span> 与先选 <span class="arithmatex">\(5\)</span> 后选 <span class="arithmatex">\(4\)</span> 是两个不同的分支，但两者对应同一个子集。</p>
-<p><a class="glightbox" href="../subset_sum_problem.assets/subset_sum_i_naive.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="子集搜索与越界剪枝" src="../subset_sum_problem.assets/subset_sum_i_naive.png" /></a></p>
+<p><a class="glightbox" href="../subset_sum_problem.assets/subset_sum_i_naive.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="子集搜索与越界剪枝" class="animation-figure" src="../subset_sum_problem.assets/subset_sum_i_naive.png" /></a></p>
 <p align="center"> 图 13-10 &nbsp; 子集搜索与越界剪枝 </p>

 <p>为了去除重复子集，<strong>一种直接的思路是对结果列表进行去重</strong>。但这个方法效率很低，有两方面原因。</p>
@@ -3868,7 +3868,7 @@
 <li>前两轮选择 <span class="arithmatex">\(4\)</span> 和 <span class="arithmatex">\(5\)</span> ，生成子集 <span class="arithmatex">\([4, 5, \dots]\)</span> 。</li>
 <li>若第一轮选择 <span class="arithmatex">\(5\)</span> ，<strong>则第二轮应该跳过 <span class="arithmatex">\(3\)</span> 和 <span class="arithmatex">\(4\)</span></strong> ，因为子集 <span class="arithmatex">\([5, 3, \dots]\)</span> 和 <span class="arithmatex">\([5, 4, \dots]\)</span> 与第 <code>1.</code> 和 <code>2.</code> 步中描述的子集完全重复。</li>
 </ol>
-<p><a class="glightbox" href="../subset_sum_problem.assets/subset_sum_i_pruning.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="不同选择顺序导致的重复子集" src="../subset_sum_problem.assets/subset_sum_i_pruning.png" /></a></p>
+<p><a class="glightbox" href="../subset_sum_problem.assets/subset_sum_i_pruning.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="不同选择顺序导致的重复子集" class="animation-figure" src="../subset_sum_problem.assets/subset_sum_i_pruning.png" /></a></p>
 <p align="center"> 图 13-11 &nbsp; 不同选择顺序导致的重复子集 </p>

 <p>总结来看，给定输入数组 <span class="arithmatex">\([x_1, x_2, \dots, x_n]\)</span> ，设搜索过程中的选择序列为 <span class="arithmatex">\([x_{i_1}, x_{i_2}, \dots, x_{i_m}]\)</span> ，则该选择序列需要满足 <span class="arithmatex">\(i_1 \leq i_2 \leq \dots \leq i_m\)</span> ，<strong>不满足该条件的选择序列都会造成重复，应当剪枝</strong>。</p>
@@ -4297,7 +4297,7 @@
 </div>
 </div>
 <p>如图 13-12 所示，为将数组 <span class="arithmatex">\([3, 4, 5]\)</span> 和目标元素 <span class="arithmatex">\(9\)</span> 输入到以上代码后的整体回溯过程。</p>
-<p><a class="glightbox" href="../subset_sum_problem.assets/subset_sum_i.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="子集和 I 回溯过程" src="../subset_sum_problem.assets/subset_sum_i.png" /></a></p>
+<p><a class="glightbox" href="../subset_sum_problem.assets/subset_sum_i.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="子集和 I 回溯过程" class="animation-figure" src="../subset_sum_problem.assets/subset_sum_i.png" /></a></p>
 <p align="center"> 图 13-12 &nbsp; 子集和 I 回溯过程 </p>

 <h2 id="1332">13.3.2 &nbsp; 考虑重复元素的情况<a class="headerlink" href="#1332" title="Permanent link">&para;</a></h2>
@@ -4307,7 +4307,7 @@
 </div>
 <p>相比于上题，<strong>本题的输入数组可能包含重复元素</strong>，这引入了新的问题。例如，给定数组 <span class="arithmatex">\([4, \hat{4}, 5]\)</span> 和目标元素 <span class="arithmatex">\(9\)</span> ，则现有代码的输出结果为 <span class="arithmatex">\([4, 5], [\hat{4}, 5]\)</span> ，出现了重复子集。</p>
 <p><strong>造成这种重复的原因是相等元素在某轮中被多次选择</strong>。在图 13-13 中，第一轮共有三个选择，其中两个都为 <span class="arithmatex">\(4\)</span> ，会产生两个重复的搜索分支，从而输出重复子集；同理，第二轮的两个 <span class="arithmatex">\(4\)</span> 也会产生重复子集。</p>
-<p><a class="glightbox" href="../subset_sum_problem.assets/subset_sum_ii_repeat.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="相等元素导致的重复子集" src="../subset_sum_problem.assets/subset_sum_ii_repeat.png" /></a></p>
+<p><a class="glightbox" href="../subset_sum_problem.assets/subset_sum_ii_repeat.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="相等元素导致的重复子集" class="animation-figure" src="../subset_sum_problem.assets/subset_sum_ii_repeat.png" /></a></p>
 <p align="center"> 图 13-13 &nbsp; 相等元素导致的重复子集 </p>

 <h3 id="1_1">1. &nbsp; 相等元素剪枝<a class="headerlink" href="#1_1" title="Permanent link">&para;</a></h3>
@@ -4786,7 +4786,7 @@
 </div>
 </div>
 <p>图 13-14 展示了数组 <span class="arithmatex">\([4, 4, 5]\)</span> 和目标元素 <span class="arithmatex">\(9\)</span> 的回溯过程，共包含四种剪枝操作。请你将图示与代码注释相结合，理解整个搜索过程，以及每种剪枝操作是如何工作的。</p>
-<p><a class="glightbox" href="../subset_sum_problem.assets/subset_sum_ii.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="子集和 II 回溯过程" src="../subset_sum_problem.assets/subset_sum_ii.png" /></a></p>
+<p><a class="glightbox" href="../subset_sum_problem.assets/subset_sum_ii.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="子集和 II 回溯过程" class="animation-figure" src="../subset_sum_problem.assets/subset_sum_ii.png" /></a></p>
 <p align="center"> 图 13-14 &nbsp; 子集和 II 回溯过程 </p>

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