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第 8 章 &nbsp; ヒープ
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第 9 章 &nbsp; グラフ
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第 10 章 &nbsp; 探索
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第 11 章 &nbsp; ソート
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第 12 章 &nbsp; 分割統治
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第 13 章 &nbsp; バックトラッキング
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13.2 &nbsp; 順列問題
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13.3 &nbsp; 部分集合和問題
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第 14 章 &nbsp; 動的プログラミング
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14.1 &nbsp; 動的プログラミング入門
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14.2 &nbsp; DP問題の特性
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14.3 &nbsp; DP問題解決アプローチ
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14.5 &nbsp; 無制限ナップサック問題
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14.6 &nbsp; 編集距離問題
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<h1 id="153">15.3 &nbsp; 最大容量問題<a class="headerlink" href="#153" title="Permanent link">&para;</a></h1>
<div class="admonition question">
<p class="admonition-title">Question</p>
<p>配列 <span class="arithmatex">\(ht\)</span> を入力します。各要素は垂直仕切りの高さを表します。配列内の任意の2つの仕切りと、それらの間のスペースによってコンテナを形成できます。</p>
<p>コンテナの容量は高さと幅の積面積で、高さは短い方の仕切りによって決定され、幅は2つの仕切りの配列インデックスの差です。</p>
<p>コンテナの容量を最大化する2つの仕切りを配列から選択し、この最大容量を返してください。例を下の図に示します。</p>
</div>
<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="最大容量問題の例データ" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_example.png" /></a></p>
<p align="center"> 図 15-7 &nbsp; 最大容量問題の例データ </p>
<p>コンテナは任意の2つの仕切りによって形成されるため、<strong>この問題の状態は2つの仕切りのインデックスで表現され、<span class="arithmatex">\([i, j]\)</span> と表記されます</strong></p>
<p>問題の記述によれば、容量は高さと幅の積に等しく、高さは短い方の仕切りによって決定され、幅は2つの仕切りの配列インデックスの差です。容量 <span class="arithmatex">\(cap[i, j]\)</span> の式は:</p>
<div class="arithmatex">\[
cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
\]</div>
<p>配列の長さを <span class="arithmatex">\(n\)</span> と仮定すると、2つの仕切りの組み合わせ数状態の総数<span class="arithmatex">\(C_n^2 = \frac{n(n - 1)}{2}\)</span> です。最も直接的なアプローチは**すべての可能な状態を列挙する**ことで、時間計算量は <span class="arithmatex">\(O(n^2)\)</span> になります。</p>
<h3 id="1">1. &nbsp; 貪欲戦略の決定<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
<p>この問題にはより効率的な解法があります。下の図に示すように、インデックス <span class="arithmatex">\(i &lt; j\)</span> かつ高さ <span class="arithmatex">\(ht[i] &lt; ht[j]\)</span> の状態 <span class="arithmatex">\([i, j]\)</span> を選択します。つまり、<span class="arithmatex">\(i\)</span> は短い仕切り、<span class="arithmatex">\(j\)</span> は高い仕切りです。</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="初期状態" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_initial_state.png" /></a></p>
<p align="center"> 図 15-8 &nbsp; 初期状態 </p>
<p>下の図に示すように、<strong>高い仕切り <span class="arithmatex">\(j\)</span> を短い仕切り <span class="arithmatex">\(i\)</span> に近づけて移動すると、容量は確実に減少します</strong></p>
<p>これは、高い仕切り <span class="arithmatex">\(j\)</span> を移動すると、幅 <span class="arithmatex">\(j-i\)</span> が確実に減少するためです。高さは短い仕切りによって決定されるため、高さは同じまま(<span class="arithmatex">\(i\)</span> が短い仕切りのまま)か減少(移動した <span class="arithmatex">\(j\)</span> が短い仕切りになる)しかありません。</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="高い仕切りを内側に移動した後の状態" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_moving_long_board.png" /></a></p>
<p align="center"> 図 15-9 &nbsp; 高い仕切りを内側に移動した後の状態 </p>
<p>逆に、<strong>短い仕切り <span class="arithmatex">\(i\)</span> を内側に移動することによってのみ容量を増加させることが可能です</strong>。幅は確実に減少しますが、<strong>高さが増加する可能性があります</strong>(移動した短い仕切り <span class="arithmatex">\(i\)</span> が高くなる場合)。例えば、下の図では、短い仕切りを移動した後に面積が増加しています。</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="短い仕切りを内側に移動した後の状態" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_moving_short_board.png" /></a></p>
<p align="center"> 図 15-10 &nbsp; 短い仕切りを内側に移動した後の状態 </p>
<p>これにより、この問題の貪欲戦略が導かれますコンテナの両端に2つのポインタを初期化し、各ラウンドで短い仕切りに対応するポインタを内側に移動し、2つのポインタが出会うまで続けます。</p>
<p>下の図は貪欲戦略の実行を示しています。</p>
<ol>
<li>最初に、ポインタ <span class="arithmatex">\(i\)</span><span class="arithmatex">\(j\)</span> が配列の両端に配置されます。</li>
<li>現在の状態の容量 <span class="arithmatex">\(cap[i, j]\)</span> を計算し、最大容量を更新します。</li>
<li>仕切り <span class="arithmatex">\(i\)</span><span class="arithmatex">\(j\)</span> の高さを比較し、短い仕切りを1ステップ内側に移動します。</li>
<li><span class="arithmatex">\(i\)</span><span class="arithmatex">\(j\)</span> が出会うまでステップ <code>2.</code><code>3.</code> を繰り返します。</li>
</ol>
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<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="最大容量問題の貪欲プロセス" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step1.png" /></a></p>
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<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step2.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step2" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step2.png" /></a></p>
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<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step3.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step3" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step3.png" /></a></p>
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<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step4.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step4" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step4.png" /></a></p>
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<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step5.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step5" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step5.png" /></a></p>
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<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step6.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step6" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step6.png" /></a></p>
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<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step7.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step7" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step7.png" /></a></p>
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<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step8.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step8" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step8.png" /></a></p>
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<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step9.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step9" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step9.png" /></a></p>
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<p align="center"> 図 15-11 &nbsp; 最大容量問題の貪欲プロセス </p>
<h3 id="2">2. &nbsp; 実装<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
<p>コードは最大 <span class="arithmatex">\(n\)</span> 回ループするため、<strong>時間計算量は <span class="arithmatex">\(O(n)\)</span></strong> です。</p>
<p>変数 <span class="arithmatex">\(i\)</span><span class="arithmatex">\(j\)</span><span class="arithmatex">\(res\)</span> は一定量の追加スペースを使用するため、<strong>空間計算量は <span class="arithmatex">\(O(1)\)</span></strong> です。</p>
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<div class="highlight"><span class="filename">max_capacity.py</span><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">max_capacity</span><span class="p">(</span><span class="n">ht</span><span class="p">:</span> <span class="nb">list</span><span class="p">[</span><span class="nb">int</span><span class="p">])</span> <span class="o">-&gt;</span> <span class="nb">int</span><span class="p">:</span>
<a id="__codelineno-0-2" name="__codelineno-0-2" href="#__codelineno-0-2"></a><span class="w"> </span><span class="sd">&quot;&quot;&quot;最大容量:貪欲法&quot;&quot;&quot;</span>
<a id="__codelineno-0-3" name="__codelineno-0-3" href="#__codelineno-0-3"></a> <span class="c1"># i、j を初期化、配列の両端で分割させる</span>
<a id="__codelineno-0-4" name="__codelineno-0-4" href="#__codelineno-0-4"></a> <span class="n">i</span><span class="p">,</span> <span class="n">j</span> <span class="o">=</span> <span class="mi">0</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">ht</span><span class="p">)</span> <span class="o">-</span> <span class="mi">1</span>
<a id="__codelineno-0-5" name="__codelineno-0-5" href="#__codelineno-0-5"></a> <span class="c1"># 初期最大容量は 0</span>
<a id="__codelineno-0-6" name="__codelineno-0-6" href="#__codelineno-0-6"></a> <span class="n">res</span> <span class="o">=</span> <span class="mi">0</span>
<a id="__codelineno-0-7" name="__codelineno-0-7" href="#__codelineno-0-7"></a> <span class="c1"># 2 つの板が出会うまで貪欲選択をループ</span>
<a id="__codelineno-0-8" name="__codelineno-0-8" href="#__codelineno-0-8"></a> <span class="k">while</span> <span class="n">i</span> <span class="o">&lt;</span> <span class="n">j</span><span class="p">:</span>
<a id="__codelineno-0-9" name="__codelineno-0-9" href="#__codelineno-0-9"></a> <span class="c1"># 最大容量を更新</span>
<a id="__codelineno-0-10" name="__codelineno-0-10" href="#__codelineno-0-10"></a> <span class="n">cap</span> <span class="o">=</span> <span class="nb">min</span><span class="p">(</span><span class="n">ht</span><span class="p">[</span><span class="n">i</span><span class="p">],</span> <span class="n">ht</span><span class="p">[</span><span class="n">j</span><span class="p">])</span> <span class="o">*</span> <span class="p">(</span><span class="n">j</span> <span class="o">-</span> <span class="n">i</span><span class="p">)</span>
<a id="__codelineno-0-11" name="__codelineno-0-11" href="#__codelineno-0-11"></a> <span class="n">res</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="n">res</span><span class="p">,</span> <span class="n">cap</span><span class="p">)</span>
<a id="__codelineno-0-12" name="__codelineno-0-12" href="#__codelineno-0-12"></a> <span class="c1"># 短い板を内側に移動</span>
<a id="__codelineno-0-13" name="__codelineno-0-13" href="#__codelineno-0-13"></a> <span class="k">if</span> <span class="n">ht</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">&lt;</span> <span class="n">ht</span><span class="p">[</span><span class="n">j</span><span class="p">]:</span>
<a id="__codelineno-0-14" name="__codelineno-0-14" href="#__codelineno-0-14"></a> <span class="n">i</span> <span class="o">+=</span> <span class="mi">1</span>
<a id="__codelineno-0-15" name="__codelineno-0-15" href="#__codelineno-0-15"></a> <span class="k">else</span><span class="p">:</span>
<a id="__codelineno-0-16" name="__codelineno-0-16" href="#__codelineno-0-16"></a> <span class="n">j</span> <span class="o">-=</span> <span class="mi">1</span>
<a id="__codelineno-0-17" name="__codelineno-0-17" href="#__codelineno-0-17"></a> <span class="k">return</span> <span class="n">res</span>
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><span class="filename">max_capacity.cpp</span><pre><span></span><code><a id="__codelineno-1-1" name="__codelineno-1-1" href="#__codelineno-1-1"></a><span class="cm">/* 最大容量:貪欲法 */</span>
<a id="__codelineno-1-2" name="__codelineno-1-2" href="#__codelineno-1-2"></a><span class="kt">int</span><span class="w"> </span><span class="nf">maxCapacity</span><span class="p">(</span><span class="n">vector</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;</span><span class="w"> </span><span class="o">&amp;</span><span class="n">ht</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-1-3" name="__codelineno-1-3" href="#__codelineno-1-3"></a><span class="w"> </span><span class="c1">// i、j を初期化し、配列の両端で分割させる</span>
<a id="__codelineno-1-4" name="__codelineno-1-4" href="#__codelineno-1-4"></a><span class="w"> </span><span class="kt">int</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mi">0</span><span class="p">,</span><span class="w"> </span><span class="n">j</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">ht</span><span class="p">.</span><span class="n">size</span><span class="p">()</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="mi">1</span><span class="p">;</span>
<a id="__codelineno-1-5" name="__codelineno-1-5" href="#__codelineno-1-5"></a><span class="w"> </span><span class="c1">// 初期最大容量は 0</span>
<a id="__codelineno-1-6" name="__codelineno-1-6" href="#__codelineno-1-6"></a><span class="w"> </span><span class="kt">int</span><span class="w"> </span><span class="n">res</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mi">0</span><span class="p">;</span>
<a id="__codelineno-1-7" name="__codelineno-1-7" href="#__codelineno-1-7"></a><span class="w"> </span><span class="c1">// 2つの板が出会うまで貪欲選択をループ</span>
<a id="__codelineno-1-8" name="__codelineno-1-8" href="#__codelineno-1-8"></a><span class="w"> </span><span class="k">while</span><span class="w"> </span><span class="p">(</span><span class="n">i</span><span class="w"> </span><span class="o">&lt;</span><span class="w"> </span><span class="n">j</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-1-9" name="__codelineno-1-9" href="#__codelineno-1-9"></a><span class="w"> </span><span class="c1">// 最大容量を更新</span>
<a id="__codelineno-1-10" name="__codelineno-1-10" href="#__codelineno-1-10"></a><span class="w"> </span><span class="kt">int</span><span class="w"> </span><span class="n">cap</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">min</span><span class="p">(</span><span class="n">ht</span><span class="p">[</span><span class="n">i</span><span class="p">],</span><span class="w"> </span><span class="n">ht</span><span class="p">[</span><span class="n">j</span><span class="p">])</span><span class="w"> </span><span class="o">*</span><span class="w"> </span><span class="p">(</span><span class="n">j</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="n">i</span><span class="p">);</span>
<a id="__codelineno-1-11" name="__codelineno-1-11" href="#__codelineno-1-11"></a><span class="w"> </span><span class="n">res</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">max</span><span class="p">(</span><span class="n">res</span><span class="p">,</span><span class="w"> </span><span class="n">cap</span><span class="p">);</span>
<a id="__codelineno-1-12" name="__codelineno-1-12" href="#__codelineno-1-12"></a><span class="w"> </span><span class="c1">// より短い板を内側に移動</span>
<a id="__codelineno-1-13" name="__codelineno-1-13" href="#__codelineno-1-13"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">ht</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="w"> </span><span class="o">&lt;</span><span class="w"> </span><span class="n">ht</span><span class="p">[</span><span class="n">j</span><span class="p">])</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-1-14" name="__codelineno-1-14" href="#__codelineno-1-14"></a><span class="w"> </span><span class="n">i</span><span class="o">++</span><span class="p">;</span>
<a id="__codelineno-1-15" name="__codelineno-1-15" href="#__codelineno-1-15"></a><span class="w"> </span><span class="p">}</span><span class="w"> </span><span class="k">else</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-1-16" name="__codelineno-1-16" href="#__codelineno-1-16"></a><span class="w"> </span><span class="n">j</span><span class="o">--</span><span class="p">;</span>
<a id="__codelineno-1-17" name="__codelineno-1-17" href="#__codelineno-1-17"></a><span class="w"> </span><span class="p">}</span>
<a id="__codelineno-1-18" name="__codelineno-1-18" href="#__codelineno-1-18"></a><span class="w"> </span><span class="p">}</span>
<a id="__codelineno-1-19" name="__codelineno-1-19" href="#__codelineno-1-19"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">res</span><span class="p">;</span>
<a id="__codelineno-1-20" name="__codelineno-1-20" href="#__codelineno-1-20"></a><span class="p">}</span>
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><span class="filename">max_capacity.java</span><pre><span></span><code><a id="__codelineno-2-1" name="__codelineno-2-1" href="#__codelineno-2-1"></a><span class="cm">/* 最大容量:貪欲法 */</span>
<a id="__codelineno-2-2" name="__codelineno-2-2" href="#__codelineno-2-2"></a><span class="kt">int</span><span class="w"> </span><span class="nf">maxCapacity</span><span class="p">(</span><span class="kt">int</span><span class="o">[]</span><span class="w"> </span><span class="n">ht</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-2-3" name="__codelineno-2-3" href="#__codelineno-2-3"></a><span class="w"> </span><span class="c1">// i、j を初期化し、配列の両端で分割させる</span>
<a id="__codelineno-2-4" name="__codelineno-2-4" href="#__codelineno-2-4"></a><span class="w"> </span><span class="kt">int</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mi">0</span><span class="p">,</span><span class="w"> </span><span class="n">j</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">ht</span><span class="p">.</span><span class="na">length</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="mi">1</span><span class="p">;</span>
<a id="__codelineno-2-5" name="__codelineno-2-5" href="#__codelineno-2-5"></a><span class="w"> </span><span class="c1">// 初期最大容量は 0</span>
<a id="__codelineno-2-6" name="__codelineno-2-6" href="#__codelineno-2-6"></a><span class="w"> </span><span class="kt">int</span><span class="w"> </span><span class="n">res</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mi">0</span><span class="p">;</span>
<a id="__codelineno-2-7" name="__codelineno-2-7" href="#__codelineno-2-7"></a><span class="w"> </span><span class="c1">// 2つの板が出会うまで貪欲選択をループ</span>
<a id="__codelineno-2-8" name="__codelineno-2-8" href="#__codelineno-2-8"></a><span class="w"> </span><span class="k">while</span><span class="w"> </span><span class="p">(</span><span class="n">i</span><span class="w"> </span><span class="o">&lt;</span><span class="w"> </span><span class="n">j</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-2-9" name="__codelineno-2-9" href="#__codelineno-2-9"></a><span class="w"> </span><span class="c1">// 最大容量を更新</span>
<a id="__codelineno-2-10" name="__codelineno-2-10" href="#__codelineno-2-10"></a><span class="w"> </span><span class="kt">int</span><span class="w"> </span><span class="n">cap</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">Math</span><span class="p">.</span><span class="na">min</span><span class="p">(</span><span class="n">ht</span><span class="o">[</span><span class="n">i</span><span class="o">]</span><span class="p">,</span><span class="w"> </span><span class="n">ht</span><span class="o">[</span><span class="n">j</span><span class="o">]</span><span class="p">)</span><span class="w"> </span><span class="o">*</span><span class="w"> </span><span class="p">(</span><span class="n">j</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="n">i</span><span class="p">);</span>
<a id="__codelineno-2-11" name="__codelineno-2-11" href="#__codelineno-2-11"></a><span class="w"> </span><span class="n">res</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">Math</span><span class="p">.</span><span class="na">max</span><span class="p">(</span><span class="n">res</span><span class="p">,</span><span class="w"> </span><span class="n">cap</span><span class="p">);</span>
<a id="__codelineno-2-12" name="__codelineno-2-12" href="#__codelineno-2-12"></a><span class="w"> </span><span class="c1">// より短い板を内側に移動</span>
<a id="__codelineno-2-13" name="__codelineno-2-13" href="#__codelineno-2-13"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">ht</span><span class="o">[</span><span class="n">i</span><span class="o">]</span><span class="w"> </span><span class="o">&lt;</span><span class="w"> </span><span class="n">ht</span><span class="o">[</span><span class="n">j</span><span class="o">]</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-2-14" name="__codelineno-2-14" href="#__codelineno-2-14"></a><span class="w"> </span><span class="n">i</span><span class="o">++</span><span class="p">;</span>
<a id="__codelineno-2-15" name="__codelineno-2-15" href="#__codelineno-2-15"></a><span class="w"> </span><span class="p">}</span><span class="w"> </span><span class="k">else</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-2-16" name="__codelineno-2-16" href="#__codelineno-2-16"></a><span class="w"> </span><span class="n">j</span><span class="o">--</span><span class="p">;</span>
<a id="__codelineno-2-17" name="__codelineno-2-17" href="#__codelineno-2-17"></a><span class="w"> </span><span class="p">}</span>
<a id="__codelineno-2-18" name="__codelineno-2-18" href="#__codelineno-2-18"></a><span class="w"> </span><span class="p">}</span>
<a id="__codelineno-2-19" name="__codelineno-2-19" href="#__codelineno-2-19"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">res</span><span class="p">;</span>
<a id="__codelineno-2-20" name="__codelineno-2-20" href="#__codelineno-2-20"></a><span class="p">}</span>
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><span class="filename">max_capacity.cs</span><pre><span></span><code><a id="__codelineno-3-1" name="__codelineno-3-1" href="#__codelineno-3-1"></a><span class="na">[class]</span><span class="p">{</span><span class="n">max_capacity</span><span class="p">}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">MaxCapacity</span><span class="p">}</span>
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><span class="filename">max_capacity.go</span><pre><span></span><code><a id="__codelineno-4-1" name="__codelineno-4-1" href="#__codelineno-4-1"></a><span class="p">[</span><span class="nx">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="kd">func</span><span class="p">]{</span><span class="nx">maxCapacity</span><span class="p">}</span>
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><span class="filename">max_capacity.swift</span><pre><span></span><code><a id="__codelineno-5-1" name="__codelineno-5-1" href="#__codelineno-5-1"></a><span class="p">[</span><span class="kd">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="kd">func</span><span class="p">]{</span><span class="n">maxCapacity</span><span class="p">}</span>
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><span class="filename">max_capacity.js</span><pre><span></span><code><a id="__codelineno-6-1" name="__codelineno-6-1" href="#__codelineno-6-1"></a><span class="p">[</span><span class="kd">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="nx">func</span><span class="p">]{</span><span class="nx">maxCapacity</span><span class="p">}</span>
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><span class="filename">max_capacity.ts</span><pre><span></span><code><a id="__codelineno-7-1" name="__codelineno-7-1" href="#__codelineno-7-1"></a><span class="p">[</span><span class="kd">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="nx">func</span><span class="p">]{</span><span class="nx">maxCapacity</span><span class="p">}</span>
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><span class="filename">max_capacity.dart</span><pre><span></span><code><a id="__codelineno-8-1" name="__codelineno-8-1" href="#__codelineno-8-1"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">maxCapacity</span><span class="p">}</span>
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><span class="filename">max_capacity.rs</span><pre><span></span><code><a id="__codelineno-9-1" name="__codelineno-9-1" href="#__codelineno-9-1"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">max_capacity</span><span class="p">}</span>
</code></pre></div>
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<div class="tabbed-block">
<div class="highlight"><span class="filename">max_capacity.c</span><pre><span></span><code><a id="__codelineno-10-1" name="__codelineno-10-1" href="#__codelineno-10-1"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">maxCapacity</span><span class="p">}</span>
</code></pre></div>
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<div class="tabbed-block">
<div class="highlight"><span class="filename">max_capacity.kt</span><pre><span></span><code><a id="__codelineno-11-1" name="__codelineno-11-1" href="#__codelineno-11-1"></a><span class="o">[</span><span class="n">class</span><span class="o">]</span><span class="p">{}</span><span class="o">-[</span><span class="n">func</span><span class="o">]</span><span class="p">{</span><span class="n">maxCapacity</span><span class="p">}</span>
</code></pre></div>
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<div class="tabbed-block">
<div class="highlight"><span class="filename">max_capacity.rb</span><pre><span></span><code><a id="__codelineno-12-1" name="__codelineno-12-1" href="#__codelineno-12-1"></a><span class="o">[</span><span class="n">class</span><span class="o">]</span><span class="p">{}</span><span class="o">-[</span><span class="n">func</span><span class="o">]</span><span class="p">{</span><span class="n">max_capacity</span><span class="p">}</span>
</code></pre></div>
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<div class="tabbed-block">
<div class="highlight"><span class="filename">max_capacity.zig</span><pre><span></span><code><a id="__codelineno-13-1" name="__codelineno-13-1" href="#__codelineno-13-1"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">maxCapacity</span><span class="p">}</span>
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<h3 id="3">3. &nbsp; 正しさの証明<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
<p>貪欲法が列挙よりも高速である理由は、各ラウンドの貪欲選択が一部の状態を「スキップ」するからです。</p>
<p>例えば、<span class="arithmatex">\(i\)</span> が短い仕切りで <span class="arithmatex">\(j\)</span> が高い仕切りである状態 <span class="arithmatex">\(cap[i, j]\)</span> の下で、短い仕切り <span class="arithmatex">\(i\)</span> を貪欲に1ステップ内側に移動すると、下の図に示す「スキップされた」状態につながります。<strong>これは、これらの状態の容量を後で検証できないことを意味します</strong></p>
<div class="arithmatex">\[
cap[i, i+1], cap[i, i+2], \dots, cap[i, j-2], cap[i, j-1]
\]</div>
<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="短い仕切りの移動によってスキップされる状態" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_skipped_states.png" /></a></p>
<p align="center"> 図 15-12 &nbsp; 短い仕切りの移動によってスキップされる状態 </p>
<p>観察すると、<strong>これらのスキップされた状態は実際には高い仕切り <span class="arithmatex">\(j\)</span> が内側に移動したすべての状態**です。高い仕切りを内側に移動すると容量が確実に減少することをすでに証明しました。したがって、スキップされた状態は最適解である可能性がなく、**それらをスキップしても最適解を見逃すことはありません</strong></p>
<p>分析により、短い仕切りを移動する操作は「安全」であり、貪欲戦略が効果的であることが示されます。</p>
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