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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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14.4 &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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15.3 &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>容器的容量等于高度和宽度的乘积（即面积），其中高度由较短的隔板决定，宽度是两个隔板的数组索引之差。</p>
<p>请在数组中选择两个隔板，使得组成的容器的容量最大，返回最大容量。</p>
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<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><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="最大容量问题的示例数据" src="../max_capacity_problem.assets/max_capacity_example.png" /></a></p>
<p align="center"> 图 15-7 &nbsp; 最大容量问题的示例数据 </p>
<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>
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<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>
<h3 id="1">1. &nbsp; 贪心策略确定<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
<p>这道题还有更高效率的解法。如图 15-8 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><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_initial_state.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="初始状态" src="../max_capacity_problem.assets/max_capacity_initial_state.png" /></a></p>
<p align="center"> 图 15-8 &nbsp; 初始状态 </p>
<p>如图 15-9 所示，<strong>若此时将长板 <span class="arithmatex">\(j\)</span> 向短板 <span class="arithmatex">\(i\)</span> 靠近，则容量一定变小</strong>。</p>
<p>这是因为在移动长板 <span class="arithmatex">\(j\)</span> 后，宽度 <span class="arithmatex">\(j-i\)</span> 肯定变小；而高度由短板决定，因此高度只可能不变（ <span class="arithmatex">\(i\)</span> 仍为短板）或变小（移动后的 <span class="arithmatex">\(j\)</span> 成为短板）。</p>
<p><img alt="向内移动长板后的状态" src="../max_capacity_problem.assets/max_capacity_moving_long_board.png" /></p>
<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_moving_long_board.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="向内移动长板后的状态" src="../max_capacity_problem.assets/max_capacity_moving_long_board.png" /></a></p>
<p align="center"> 图 15-9 &nbsp; 向内移动长板后的状态 </p>
<p>反向思考，<strong>我们只有向内收缩短板 <span class="arithmatex">\(i\)</span> ，才有可能使容量变大</strong>。因为虽然宽度一定变小，<strong>但高度可能会变大</strong>（移动后的短板 <span class="arithmatex">\(i\)</span> 可能会变长）。例如在图 15-10 中，移动短板后面积变大。</p>
<p><img alt="向内移动短板后的状态" src="../max_capacity_problem.assets/max_capacity_moving_short_board.png" /></p>
<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_moving_short_board.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="向内移动短板后的状态" src="../max_capacity_problem.assets/max_capacity_moving_short_board.png" /></a></p>
<p align="center"> 图 15-10 &nbsp; 向内移动短板后的状态 </p>
<p>由此便可推出本题的贪心策略：初始化两指针分裂容器两端，每轮向内收缩短板对应的指针，直至两指针相遇。</p>
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<div class="arithmatex">\[
cap[i, i+1], cap[i, i+2], \dots, cap[i, j-2], cap[i, j-1]
\]</div>
<p><img alt="移动短板导致被跳过的状态" src="../max_capacity_problem.assets/max_capacity_skipped_states.png" /></p>
<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_skipped_states.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="移动短板导致被跳过的状态" src="../max_capacity_problem.assets/max_capacity_skipped_states.png" /></a></p>
<p align="center"> 图 15-12 &nbsp; 移动短板导致被跳过的状态 </p>
<p>观察发现，<strong>这些被跳过的状态实际上就是将长板 <span class="arithmatex">\(j\)</span> 向内移动的所有状态</strong>。而在第二步中，我们已经证明内移长板一定会导致容量变小。也就是说，被跳过的状态都不可能是最优解，<strong>跳过它们不会导致错过最优解</strong>。</p>
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