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chapter_dynamic_programming/intro_to_dynamic_programming/index.html

@@ -3387,7 +3387,7 @@
 <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><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><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" class="animation-figure" 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>
@@ -3748,7 +3748,7 @@ 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><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><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" class="animation-figure" 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>
@@ -3959,7 +3959,7 @@ dp[i] = dp[i-1] + dp[i-2]
 </div>
 </div>
 <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><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><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" class="animation-figure" 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>
@@ -4269,7 +4269,7 @@ dp[i] = dp[i-1] + dp[i-2]
 </div>
 </div>
 <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><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><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" class="animation-figure" 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>
@@ -4497,7 +4497,7 @@ 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><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><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" class="animation-figure" 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>