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chapter_divide_and_conquer/hanota_problem/index.html
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<span class="md-ellipsis">
8.3 &nbsp; Top-K 问题
8.3 &nbsp; Top-k 问题
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<div class="tabbed-set tabbed-alternate" data-tabs="1:2"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">&lt;1&gt;</label><label for="__tabbed_1_2">&lt;2&gt;</label></div>
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<p><a class="glightbox" href="../hanota_problem.assets/hanota_f1_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="规模为 1 问题的解" class="animation-figure" src="../hanota_problem.assets/hanota_f1_step1.png" /></a></p>
<p><a class="glightbox" href="../hanota_problem.assets/hanota_f1_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="规模为 1 的问题的解" class="animation-figure" src="../hanota_problem.assets/hanota_f1_step1.png" /></a></p>
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<p><a class="glightbox" href="../hanota_problem.assets/hanota_f1_step2.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="hanota_f1_step2" class="animation-figure" src="../hanota_problem.assets/hanota_f1_step2.png" /></a></p>
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<p align="center"> 图 12-11 &nbsp; 规模为 1 问题的解 </p>
<p align="center"> 图 12-11 &nbsp; 规模为 1 的问题的解 </p>
<p>如图 12-12 所示，对于问题 <span class="arithmatex">\(f(2)\)</span> ，即当有两个圆盘时，<strong>由于要时刻满足小圆盘在大圆盘之上，因此需要借助 <code>B</code> 来完成移动</strong>。</p>
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<div class="tabbed-set tabbed-alternate" data-tabs="2:4"><input checked="checked" id="__tabbed_2_1" name="__tabbed_2" type="radio" /><input id="__tabbed_2_2" name="__tabbed_2" type="radio" /><input id="__tabbed_2_3" name="__tabbed_2" type="radio" /><input id="__tabbed_2_4" name="__tabbed_2" type="radio" /><div class="tabbed-labels"><label for="__tabbed_2_1">&lt;1&gt;</label><label for="__tabbed_2_2">&lt;2&gt;</label><label for="__tabbed_2_3">&lt;3&gt;</label><label for="__tabbed_2_4">&lt;4&gt;</label></div>
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<p><a class="glightbox" href="../hanota_problem.assets/hanota_f2_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="规模为 2 问题的解" class="animation-figure" src="../hanota_problem.assets/hanota_f2_step1.png" /></a></p>
<p><a class="glightbox" href="../hanota_problem.assets/hanota_f2_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="规模为 2 的问题的解" class="animation-figure" src="../hanota_problem.assets/hanota_f2_step1.png" /></a></p>
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<p><a class="glightbox" href="../hanota_problem.assets/hanota_f2_step2.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="hanota_f2_step2" class="animation-figure" src="../hanota_problem.assets/hanota_f2_step2.png" /></a></p>
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<p align="center"> 图 12-12 &nbsp; 规模为 2 问题的解 </p>
<p align="center"> 图 12-12 &nbsp; 规模为 2 的问题的解 </p>
<p>解决问题 <span class="arithmatex">\(f(2)\)</span> 的过程可总结为：<strong>将两个圆盘借助 <code>B</code> 从 <code>A</code> 移至 <code>C</code></strong> 。其中，<code>C</code> 称为目标柱、<code>B</code> 称为缓冲柱。</p>
<h3 id="2">2. &nbsp; 子问题分解<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
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<div class="tabbed-set tabbed-alternate" data-tabs="3:4"><input checked="checked" id="__tabbed_3_1" name="__tabbed_3" type="radio" /><input id="__tabbed_3_2" name="__tabbed_3" type="radio" /><input id="__tabbed_3_3" name="__tabbed_3" type="radio" /><input id="__tabbed_3_4" name="__tabbed_3" type="radio" /><div class="tabbed-labels"><label for="__tabbed_3_1">&lt;1&gt;</label><label for="__tabbed_3_2">&lt;2&gt;</label><label for="__tabbed_3_3">&lt;3&gt;</label><label for="__tabbed_3_4">&lt;4&gt;</label></div>
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<p><a class="glightbox" href="../hanota_problem.assets/hanota_f3_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="规模为 3 问题的解" class="animation-figure" src="../hanota_problem.assets/hanota_f3_step1.png" /></a></p>
<p><a class="glightbox" href="../hanota_problem.assets/hanota_f3_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="规模为 3 的问题的解" class="animation-figure" src="../hanota_problem.assets/hanota_f3_step1.png" /></a></p>
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<p><a class="glightbox" href="../hanota_problem.assets/hanota_f3_step2.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="hanota_f3_step2" class="animation-figure" src="../hanota_problem.assets/hanota_f3_step2.png" /></a></p>
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<p align="center"> 图 12-13 &nbsp; 规模为 3 问题的解 </p>
<p align="center"> 图 12-13 &nbsp; 规模为 3 的问题的解 </p>
<p>从本质上看，<strong>我们将问题 <span class="arithmatex">\(f(3)\)</span> 划分为两个子问题 <span class="arithmatex">\(f(2)\)</span> 和子问题 <span class="arithmatex">\(f(1)\)</span></strong> 。按顺序解决这三个子问题之后，原问题随之得到解决。这说明子问题是独立的，而且解可以合并。</p>
<p>从本质上看，<strong>我们将问题 <span class="arithmatex">\(f(3)\)</span> 划分为两个子问题 <span class="arithmatex">\(f(2)\)</span> 和一个子问题 <span class="arithmatex">\(f(1)\)</span></strong> 。按顺序解决这三个子问题之后，原问题随之得到解决。这说明子问题是独立的，而且解可以合并。</p>
<p>至此，我们可总结出图 12-14 所示的解决汉诺塔问题的分治策略：将原问题 <span class="arithmatex">\(f(n)\)</span> 划分为两个子问题 <span class="arithmatex">\(f(n-1)\)</span> 和一个子问题 <span class="arithmatex">\(f(1)\)</span> ，并按照以下顺序解决这三个子问题。</p>
<ol>
<li>将 <span class="arithmatex">\(n-1\)</span> 个圆盘借助 <code>C</code> 从 <code>A</code> 移至 <code>B</code> 。</li>