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1. &nbsp; 무방향 그래프에 경로가 존재하는지 판단하기
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<h1 id="95">9.5 &nbsp; 연습 문제<a class="headerlink" href="#95" title="고유 링크">&para;</a></h1>
<h2 id="951">9.5.1 &nbsp; 개념 확인<a class="headerlink" href="#951" title="고유 링크">&para;</a></h2>
<h3 id="1">1. &nbsp; 같은 그래프를 두 가지 방식으로 표현하기<a class="headerlink" href="#1" title="고유 링크">&para;</a></h3>
<p>무방향 그래프의 정점은 <code>A, B, C, D</code> 4개이며, 간선은
<code>A-B, A-C, B-C, C-D</code>다.</p>
<!-- numbered-subquestions -->
<ol>
<li>이 그래프의 인접 리스트를 적어 보자.</li>
<li>0과 1만 들어가는 인접 행렬을 완성해 보자.</li>
<li><code>A</code>와 <code>D</code>가 직접 연결되어 있는지 판단할 때, 저장된 항목 하나만 확인하면 되는 표현 방식은 무엇인가?</li>
<li>정점이 많고 간선이 적다면 일반적으로 어느 표현 방식이 공간을 더 절약하는가?</li>
</ol>
<details class="success">
<summary>참고 답안</summary>
<ol>
<li>
<p>인접 리스트는 다음과 같다.</p>
<div class="highlight"><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a>A: B, C
<a id="__codelineno-0-2" name="__codelineno-0-2" href="#__codelineno-0-2"></a>B: A, C
<a id="__codelineno-0-3" name="__codelineno-0-3" href="#__codelineno-0-3"></a>C: A, B, D
<a id="__codelineno-0-4" name="__codelineno-0-4" href="#__codelineno-0-4"></a>D: C
</code></pre></div>
</li>
<li>
<p>인접 행렬은 다음과 같다.</p>
<table>
<thead>
<tr>
<th></th>
<th>A</th>
<th>B</th>
<th>C</th>
<th>D</th>
</tr>
</thead>
<tbody>
<tr>
<td>A</td>
<td>0</td>
<td>1</td>
<td>1</td>
<td>0</td>
</tr>
<tr>
<td>B</td>
<td>1</td>
<td>0</td>
<td>1</td>
<td>0</td>
</tr>
<tr>
<td>C</td>
<td>1</td>
<td>1</td>
<td>0</td>
<td>1</td>
</tr>
<tr>
<td>D</td>
<td>0</td>
<td>0</td>
<td>1</td>
<td>0</td>
</tr>
</tbody>
</table>
</li>
<li>
<p>인접 행렬에서는 <code>A</code>행과 <code>D</code>열이 만나는 항목을 바로 확인할 수 있다. 따라서 임의의 두 정점이 직접 연결되어 있는지 판단하기에 적합하다.</p>
</li>
<li>
<p>정점이 많고 간선이 적을 때 인접 리스트는 실제로 존재하는 간선만 기록한다. 따라서 모든 정점 쌍에 대해 자리를 마련하는 인접 행렬보다 일반적으로 공간을 더 절약한다.</p>
</li>
</ol>
</details>
<h3 id="2">2. &nbsp; 너비 우선 탐색과 깊이 우선 탐색의 방문 순서<a class="headerlink" href="#2" title="고유 링크">&para;</a></h3>
<p>무방향 그래프의 정점은 <code>A, B, C, D, E</code>이며, 간선은
<code>A-B, A-C, B-D, C-D, D-E</code>다.</p>
<p>A에서 출발하고, 방문하지 않은 인접 정점이 여러 개면 알파벳순으로 선택한다고 하자.</p>
<!-- numbered-subquestions -->
<ol>
<li>너비 우선 탐색(BFS)의 방문 순서를 적어 보자.</li>
<li>재귀 깊이 우선 탐색(DFS)의 방문 순서를 적어 보자.</li>
<li>두 탐색 모두 이미 방문한 정점을 기록해야 하는 이유는 무엇인가?</li>
</ol>
<details class="success">
<summary>참고 답안</summary>
<ol>
<li>
<p>BFS의 방문 순서는 <code>A, B, C, D, E</code>다. A에서 간선 하나로 도달하는 B와 C를 먼저 방문하고,
그다음 더 멀리 있는 D와 E를 방문한다.</p>
</li>
<li>
<p>DFS의 방문 순서는 <code>A, B, D, C, E</code>다. 현재 정점에서 방문하지 않은 인접 정점으로 차례대로 들어가므로
먼저 <code>A → B → D → C</code>를 따라간다. C에서 새로 방문할 인접 정점이 없으면 D로 돌아와 E를 방문한다.</p>
</li>
<li>
<p>그래프에는 <code>A-B-D-C-A</code>와 같은 사이클이 있다. 이미 방문한 정점을 기록하지 않으면
같은 정점들을 사이클을 따라 반복해서 방문하여 탐색이 정상적으로 끝나지 않을 수 있다.</p>
</li>
</ol>
</details>
<h3 id="3-bfs">3. &nbsp; BFS 한 번으로 그래프 전체를 방문할 수 있는가<a class="headerlink" href="#3-bfs" title="고유 링크">&para;</a></h3>
<p>무방향 그래프의 정점은 <code>A, B, C, D, E, F</code>이며, 간선은
<code>A-B, B-C, D-E</code>뿐이다.</p>
<!-- numbered-subquestions -->
<ol>
<li>A에서 BFS를 한 번 시작하면 어떤 정점을 방문할 수 있는가?</li>
<li>1번의 결과를 바탕으로, 이 BFS가 그래프의 모든 정점을 방문했는지 판단하고 이유를 설명하라.</li>
<li>모든 정점을 알파벳순으로 확인하면서 방문하지 않은 정점을 만날 때마다 BFS를 새로 시작한다면,
각 BFS의 시작 정점은 무엇인가? 이 그래프는 서로 연결되지 않은 부분인 연결 요소 몇 개로 이루어져 있는가?</li>
</ol>
<details class="success">
<summary>참고 답안</summary>
<ol>
<li>
<p>A에서 출발하면 <code>A, B, C</code>만 방문할 수 있다.</p>
</li>
<li>
<p>모든 정점을 방문한 것은 아니다. <code>D, E</code>는 서로 연결된 별개의 부분을 이루고, F는 홀로 떨어진 정점이다.
이 정점들과 A 사이에는 경로가 없으므로 A에서 출발해서는 도달할 수 없다.</p>
</li>
<li>
<p>세 번의 BFS는 차례대로 <code>A, D, F</code>에서 시작하여 각각
<code>{A, B, C}</code>, <code>{D, E}</code>, <code>{F}</code>를 방문한다. 따라서 이 그래프의 연결 요소는 3개다.</p>
</li>
</ol>
</details>
<h2 id="952">9.5.2 &nbsp; 프로그래밍 연습<a class="headerlink" href="#952" title="고유 링크">&para;</a></h2>
<h3 id="1_1">1. &nbsp; 무방향 그래프에 경로가 존재하는지 판단하기<a class="headerlink" href="#1_1" title="고유 링크">&para;</a></h3>
<p>정점이 <span class="arithmatex">\(n\)</span>개인 무방향 그래프가 주어지며, 정점 번호는 <span class="arithmatex">\(0\)</span>부터 <span class="arithmatex">\(n-1\)</span>까지다. 배열 <code>edges</code>의 각 항목 <code>[u, v]</code>는 정점 <code>u</code>와 <code>v</code> 사이에 무방향 간선이 있음을 나타낸다.</p>
<p>시작 정점 <code>source</code>와 도착 정점 <code>destination</code>도 주어진다. 먼저 <code>edges</code>로 인접 리스트를 만든 뒤 BFS 또는 DFS를 사용하여
<code>source</code>에서 <code>destination</code>까지의 경로가 있는지 판단하라. 있으면 <code>true</code>를, 없으면 <code>false</code>를 반환하라.
그래프에는 사이클이 있을 수 있으며, 모든 정점이 서로 연결되어 있지 않을 수도 있다.</p>
<details class="tip">
<summary>풀이 힌트</summary>
<ol>
<li>무방향 간선은 양쪽 방향을 모두 추가해야 한다.</li>
<li>그래프에는 사이클이 있을 수 있으므로 이미 방문한 노드를 반드시 기록해야 한다.</li>
<li>source에서 출발하여 destination을 만나면 true를 반환한다. 탐색을 마칠 때까지 만나지 못하면 false를 반환한다.</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.com/problems/find-if-path-exists-in-graph/" rel="noopener noreferrer" target="_blank">LeetCode</a></p>
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