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第 8 章 &nbsp; 堆
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第 9 章 &nbsp; 图
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9.1 &nbsp; 图
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第 10 章 &nbsp; 搜索
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10.1 &nbsp; 二分查找
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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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10.4 &nbsp; 哈希优化策略
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10.5 &nbsp; 重识搜索算法
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第 11 章 &nbsp; 排序
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11.2 &nbsp; 选择排序
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11.3 &nbsp; 冒泡排序
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11.4 &nbsp; 插入排序
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11.5 &nbsp; 快速排序
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11.6 &nbsp; 归并排序
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11.7 &nbsp; 堆排序
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11.8 &nbsp; 桶排序
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11.9 &nbsp; 计数排序
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11.10 &nbsp; 基数排序
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11.11 &nbsp; 小结
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11.12 &nbsp; 练习
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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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第 13 章 &nbsp; 回溯
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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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14.6 &nbsp; 编辑距离问题
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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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<h1 id="156">15.6 &nbsp; 练习<a class="headerlink" href="#156" title="Permanent link">&para;</a></h1>
<h2 id="1561">15.6.1 &nbsp; 知识巩固<a class="headerlink" href="#1561" title="Permanent link">&para;</a></h2>
<h3 id="1">1. &nbsp; 每次选最大硬币一定最好吗<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
<p>硬币面值为 <code>[1, 7, 10]</code>，目标金额为 14。</p>
<!-- numbered-subquestions -->
<ol>
<li>按“每次取不超过剩余金额的最大面值”执行，写出所取硬币；</li>
<li>是否存在使用更少硬币的方案？如果存在，请写出一种；如果不存在，请说明理由；</li>
<li>这能否说明该贪心策略对任意硬币面值都正确？</li>
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<summary>参考答案</summary>
<ol>
<li>
<p>贪心依次选择 <code>10 + 1 + 1 + 1 + 1</code>，共 5 枚硬币。</p>
</li>
<li>
<p>存在使用更少硬币的方案：<code>7 + 7</code>，只需 2 枚硬币。</p>
</li>
<li>
<p>不能。这个反例说明，对任意硬币面值，每次选择当前能取的最大面值并不一定得到最少硬币数；
眼前最大的选择可能破坏后续更好的组合。</p>
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<h3 id="2">2. &nbsp; 背包应该先装哪件物品<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
<p>一个容量为 4 千克的背包中可以装入以下物品，物品允许只取一部分，
所取价值与重量成正比：</p>
<ul>
<li>物品 A：重量 4 千克，价值 20；</li>
<li>物品 B：重量 3 千克，价值 18。</li>
</ul>
<!-- numbered-subquestions -->
<ol>
<li>两件物品每千克的价值分别是多少？应先装哪一件？</li>
<li>按分数背包的贪心策略装满背包，最终价值是多少？</li>
<li>在物品可以切分、背包限制总重量的情况下，选择物品时应该比较总价值还是每千克价值？为什么？</li>
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<summary>参考答案</summary>
<ol>
<li>
<p>A 每千克价值为 <code>20 ÷ 4 = 5</code>，B 每千克价值为 <code>18 ÷ 3 = 6</code>，
因此应先装单位价值更高的 B。</p>
</li>
<li>
<p>先装入全部 B，占用 3 千克、获得价值 18；剩余 1 千克容量，
再装入 1 千克 A，获得价值 5。最终价值为 <code>18 + 5 = 23</code>。</p>
</li>
<li>
<p>背包限制的是总重量，而且物品可以切分，所以应比较单位重量的价值。
A 的总价值虽然更高，但每千克价值低于 B；若先装满 A，只能得到价值 20。</p>
</li>
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</details>
<h3 id="3">3. &nbsp; 两个指针下一步怎样移动<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
<p>隔板高度为 <code>[1, 8, 6, 2, 5]</code>，用首尾双指针寻找最大容量。
容量等于“两块隔板中较短者的高度 × 两块隔板的下标距离”。</p>
<!-- numbered-subquestions -->
<ol>
<li>初始时左指针在下标 0、右指针在下标 4，当前容量是多少？下一步应移动哪个指针？</li>
<li>按照第 1 问的选择移动一次后，两个指针分别位于哪个下标？此时容量是多少？下一步又应移动哪个指针？</li>
<li>在当前一对隔板中，可以移动较短隔板对应的指针，也可以移动较长隔板对应的指针。哪一种移动仍有可能得到更大的容量？为什么？</li>
</ol>
<details class="success">
<summary>参考答案</summary>
<ol>
<li>
<p>当前容量为 <code>min(1, 5) × (4 - 0) = 4</code>。左侧隔板较短，因此移动左指针。</p>
</li>
<li>
<p>移动左指针后，两个指针分别位于下标 1 和 4。当前容量为
<code>min(8, 5) × (4 - 1) = 15</code>。右侧隔板较短，因此下一步移动右指针。</p>
</li>
<li>
<p>移动较短隔板对应的指针，才仍有可能得到更大的容量。移动较长隔板后，两块隔板的距离一定缩短，而容器高度仍受未移动的短板限制，
只可能不变或变小。因此容量不可能超过移动前；只有移动短板，才有机会遇到更高的隔板。</p>
</li>
</ol>
</details>
<h2 id="1562">15.6.2 &nbsp; 编程练习<a class="headerlink" href="#1562" title="Permanent link">&para;</a></h2>
<h3 id="1_1">1. &nbsp; 分数背包<a class="headerlink" href="#1_1" title="Permanent link">&para;</a></h3>
<p>给定等长数组 <code>wgt</code> 和 <code>val</code>，其中 <code>wgt[i] &gt; 0</code>、<code>val[i] &gt;= 0</code>，背包容量 <code>cap &gt;= 0</code>。
每种物品只有一件，但允许只装入其中一部分，
获得的价值按所装重量占该物品总重量的比例计算。请使用贪心算法，
并以实数返回背包能够获得的最大总价值。</p>
<details class="tip">
<summary>解题提示</summary>
<ol>
<li>先计算每件物品的单位重量价值 val[i] / wgt[i]，除法结果要保留小数</li>
<li>单位价值越高的物品越先装入背包</li>
<li>若剩余容量小于当前物品重量，就装入恰好填满背包的部分并结束</li>
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