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chapter_greedy/max_capacity_problem/index.html

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+        <label class="md-nav__link md-nav__link--active" for="__toc">
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+    15.3. &nbsp; 最大容量问题
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+        <li class="md-nav__item">
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+    第一步：问题分析
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+  
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+      
+        <li class="md-nav__item">
+  <a href="#_2" class="md-nav__link">
+    第二步：贪心策略确定
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+  
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+      
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+  <a href="#_3" class="md-nav__link">
+    代码实现
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+  <a href="#_4" class="md-nav__link">
+    第三步：正确性证明
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+    15.4. &nbsp; 最大切分乘积问题
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+      目录
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+      
+        <li class="md-nav__item">
+  <a href="#_1" class="md-nav__link">
+    第一步：问题分析
+  </a>
+  
+</li>
+      
+        <li class="md-nav__item">
+  <a href="#_2" class="md-nav__link">
+    第二步：贪心策略确定
+  </a>
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+    代码实现
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+  <a href="#_4" class="md-nav__link">
+    第三步：正确性证明
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@@ -3256,14 +3388,14 @@
 <p><img alt="最大容量问题的示例数据" src="../max_capacity_problem.assets/max_capacity_example.png" /></p>
 <p align="center"> Fig. 最大容量问题的示例数据 </p>
 
-<p><strong>第一步：问题分析</strong></p>
+<h3 id="_1">第一步：问题分析<a class="headerlink" href="#_1" title="Permanent link">&para;</a></h3>
 <p>容器由任意两个隔板围成，<strong>因此本题的状态为两个隔板的索引，记为 <span class="arithmatex">\([i, j]\)</span></strong> 。</p>
 <p>根据定义，容量等于高度乘以宽度，其中高度由短板决定，宽度是两隔板的索引之差。设容量为 <span class="arithmatex">\(cap[i, j]\)</span> ，可得计算公式：</p>
 <div class="arithmatex">\[
 cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
 \]</div>
 <p>设数组长度为 <span class="arithmatex">\(n\)</span> ，两个隔板的组合数量（即状态总数）为 <span class="arithmatex">\(C_n^2 = \frac{n(n - 1)}{2}\)</span> 个。最直接地，<strong>我们可以穷举所有状态</strong>，从而求得最大容量，时间复杂度为 <span class="arithmatex">\(O(n^2)\)</span> 。</p>
-<p><strong>第二步：贪心策略确定</strong></p>
+<h3 id="_2">第二步：贪心策略确定<a class="headerlink" href="#_2" title="Permanent link">&para;</a></h3>
 <p>当然，这道题还有更高效率的解法。如下图所示，现选取一个状态 <span class="arithmatex">\([i, j]\)</span> ，其满足索引 <span class="arithmatex">\(i &lt; j\)</span> 且高度 <span class="arithmatex">\(ht[i] &lt; ht[j]\)</span> ，即 <span class="arithmatex">\(i\)</span> 为短板、 <span class="arithmatex">\(j\)</span> 为长板。</p>
 <p><img alt="初始状态" src="../max_capacity_problem.assets/max_capacity_initial_state.png" /></p>
 <p align="center"> Fig. 初始状态 </p>
@@ -3318,7 +3450,8 @@ cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
 </div>
 </div>
 </div>
-<p>代码实现如下所示。最多循环 <span class="arithmatex">\(n\)</span> 轮，<strong>因此时间复杂度为 <span class="arithmatex">\(O(n)\)</span></strong> 。变量 <span class="arithmatex">\(i\)</span> , <span class="arithmatex">\(j\)</span> , <span class="arithmatex">\(res\)</span> 使用常数大小额外空间，<strong>因此空间复杂度为 <span class="arithmatex">\(O(1)\)</span></strong> 。</p>
+<h3 id="_3">代码实现<a class="headerlink" href="#_3" title="Permanent link">&para;</a></h3>
+<p>如下代码所示，循环最多 <span class="arithmatex">\(n\)</span> 轮，<strong>因此时间复杂度为 <span class="arithmatex">\(O(n)\)</span></strong> 。变量 <span class="arithmatex">\(i\)</span> , <span class="arithmatex">\(j\)</span> , <span class="arithmatex">\(res\)</span> 使用常数大小额外空间，<strong>因此空间复杂度为 <span class="arithmatex">\(O(1)\)</span></strong> 。</p>
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@@ -3421,7 +3554,7 @@ cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
 </div>
 </div>
 </div>
-<p><strong>第三步：正确性证明</strong></p>
+<h3 id="_4">第三步：正确性证明<a class="headerlink" href="#_4" title="Permanent link">&para;</a></h3>
 <p>之所以贪心比穷举更快，是因为每轮的贪心选择都会“跳过”一些状态。</p>
 <p>比如在状态 <span class="arithmatex">\(cap[i, j]\)</span> 下，<span class="arithmatex">\(i\)</span> 为短板、<span class="arithmatex">\(j\)</span> 为长板。若贪心地将短板 <span class="arithmatex">\(i\)</span> 向内移动一格，会导致以下状态被“跳过”，<strong>意味着之后无法验证这些状态的容量大小</strong>。</p>
 <div class="arithmatex">\[
@@ -3528,13 +3661,13 @@ cap[i, i+1], cap[i, i+2], \cdots, cap[i, j-2], cap[i, j-1]
         
         
           
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