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第 1 章 &nbsp; 初識演算法
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第 4 章 &nbsp; 陣列與鏈結串列
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4.1 &nbsp; 陣列
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4.3 &nbsp; 串列
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4.4 &nbsp; 記憶體與快取 *
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第 5 章 &nbsp; 堆疊與佇列
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第 6 章 &nbsp; 雜湊表
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6.1 &nbsp; 雜湊表
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6.3 &nbsp; 雜湊演算法
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第 7 章 &nbsp; 樹
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7.5 &nbsp; AVL *
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第 8 章 &nbsp; 堆積
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第 8 章 &nbsp; 堆積
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第 9 章 &nbsp; 圖
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第 11 章 &nbsp; 排序
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第 15 章 &nbsp; 貪婪
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15.2 &nbsp; 分數背包問題
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<h1 id="156">15.6 &nbsp; 練習<a class="headerlink" href="#156" title="Permanent link">&para;</a></h1>
<h2 id="1561">15.6.1 &nbsp; 知識鞏固<a class="headerlink" href="#1561" title="Permanent link">&para;</a></h2>
<h3 id="1">1. &nbsp; 每次選最大硬幣一定最好嗎<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
<p>硬幣面值為 <code>[1, 7, 10]</code>，目標金額為 14。</p>
<!-- numbered-subquestions -->
<ol>
<li>按“每次取不超過剩餘金額的最大面值”執行，寫出所取硬幣；</li>
<li>是否存在使用更少硬幣的方案？如果存在，請寫出一種；如果不存在，請說明理由；</li>
<li>這能否說明該貪婪策略對任意硬幣面值都正確？</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>貪婪依次選擇 <code>10 + 1 + 1 + 1 + 1</code>，共 5 枚硬幣。</p>
</li>
<li>
<p>存在使用更少硬幣的方案：<code>7 + 7</code>，只需 2 枚硬幣。</p>
</li>
<li>
<p>不能。這個反例說明，對任意硬幣面值，每次選擇當前能取的最大面值並不一定得到最少硬幣數；
眼前最大的選擇可能破壞後續更好的組合。</p>
</li>
</ol>
</details>
<h3 id="2">2. &nbsp; 背包應該先裝哪件物品<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
<p>一個容量為 4 千克的背包中可以裝入以下物品，物品允許只取一部分，
所取價值與重量成正比：</p>
<ul>
<li>物品 A：重量 4 千克，價值 20；</li>
<li>物品 B：重量 3 千克，價值 18。</li>
</ul>
<!-- numbered-subquestions -->
<ol>
<li>兩件物品每千克的價值分別是多少？應先裝哪一件？</li>
<li>按分數背包的貪婪策略裝滿背包，最終價值是多少？</li>
<li>在物品可以切分、背包限制總重量的情況下，選擇物品時應該比較總價值還是每千克價值？為什麼？</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>A 每千克價值為 <code>20 ÷ 4 = 5</code>，B 每千克價值為 <code>18 ÷ 3 = 6</code>，
因此應先裝單位價值更高的 B。</p>
</li>
<li>
<p>先裝入全部 B，佔用 3 千克、獲得價值 18；剩餘 1 千克容量，
再裝入 1 千克 A，獲得價值 5。最終價值為 <code>18 + 5 = 23</code>。</p>
</li>
<li>
<p>背包限制的是總重量，而且物品可以切分，所以應比較單位重量的價值。
A 的總價值雖然更高，但每千克價值低於 B；若先裝滿 A，只能得到價值 20。</p>
</li>
</ol>
</details>
<h3 id="3">3. &nbsp; 兩個指標下一步怎樣移動<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
<p>隔板高度為 <code>[1, 8, 6, 2, 5]</code>，用首尾雙指標尋找最大容量。
容量等於“兩塊隔板中較短者的高度 × 兩塊隔板的下標距離”。</p>
<!-- numbered-subquestions -->
<ol>
<li>初始時左指標在下標 0、右指標在下標 4，當前容量是多少？下一步應移動哪個指標？</li>
<li>按照第 1 問的選擇移動一次後，兩個指標分別位於哪個下標？此時容量是多少？下一步又應移動哪個指標？</li>
<li>在當前一對隔板中，可以移動較短隔板對應的指標，也可以移動較長隔板對應的指標。哪一種移動仍有可能得到更大的容量？為什麼？</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>當前容量為 <code>min(1, 5) × (4 - 0) = 4</code>。左側隔板較短，因此移動左指標。</p>
</li>
<li>
<p>移動左指標後，兩個指標分別位於下標 1 和 4。當前容量為
<code>min(8, 5) × (4 - 1) = 15</code>。右側隔板較短，因此下一步移動右指標。</p>
</li>
<li>
<p>移動較短隔板對應的指標，才仍有可能得到更大的容量。移動較長隔板後，兩塊隔板的距離一定縮短，而容器高度仍受未移動的短板限制，
只可能不變或變小。因此容量不可能超過移動前；只有移動短板，才有機會遇到更高的隔板。</p>
</li>
</ol>
</details>
<h2 id="1562">15.6.2 &nbsp; 程式設計練習<a class="headerlink" href="#1562" title="Permanent link">&para;</a></h2>
<h3 id="1_1">1. &nbsp; 分數背包<a class="headerlink" href="#1_1" title="Permanent link">&para;</a></h3>
<p>給定等長陣列 <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>。
每種物品只有一件，但允許只裝入其中一部分，
獲得的價值按所裝重量佔該物品總重量的比例計算。請使用貪婪演算法，
並以實數返回背包能夠獲得的最大總價值。</p>
<details class="tip">
<summary>解題提示</summary>
<ol>
<li>先計算每件物品的單位重量價值 val[i] / wgt[i]，除法結果要保留小數</li>
<li>單位價值越高的物品越先裝入背包</li>
<li>若剩餘容量小於當前物品重量，就裝入恰好填滿背包的部分並結束</li>
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