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<h1 id="_1"><a class="headerlink" href="#_1" title="Permanent link">&para;</a></h1>
<p><em>图建模了关系和连接——从社交网络到道路地图再到依赖链。本文件涵盖图的表示、BFS、DFS、最短路径、拓扑排序和连通分量,包括遍历和寻路模式,这些是图面试题中的核心。</em></p>
<ul>
<li>
<p>我们在第12章(邻接矩阵、拉普拉斯矩阵、谱性质)和第13章(树、平面性、着色)中已经介绍了图论。这里我们专注于<strong>算法模式</strong>:如何在代码中遍历、搜索和优化图。</p>
</li>
<li>
<p>两种基本的图算法是 <strong>BFS</strong><strong>DFS</strong>。几乎所有图问题都可以归结为其中一种,可能带有修改。掌握这两种算法,你就能解决绝大多数图问题。</p>
</li>
</ul>
<h2 id="_2">图的表示<a class="headerlink" href="#_2" title="Permanent link">&para;</a></h2>
<ul>
<li><strong>邻接表</strong>:对于每个节点,存储一个邻居列表。空间:<span class="arithmatex">\(O(|V| + |E|)\)</span>。最适合稀疏图(大多数现实世界的图)。</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a><span class="c1"># 无向图</span>
<a id="__codelineno-0-2" name="__codelineno-0-2" href="#__codelineno-0-2"></a><span class="n">graph</span> <span class="o">=</span> <span class="p">{</span>
<a id="__codelineno-0-3" name="__codelineno-0-3" href="#__codelineno-0-3"></a> <span class="mi">0</span><span class="p">:</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">],</span>
<a id="__codelineno-0-4" name="__codelineno-0-4" href="#__codelineno-0-4"></a> <span class="mi">1</span><span class="p">:</span> <span class="p">[</span><span class="mi">0</span><span class="p">,</span> <span class="mi">3</span><span class="p">],</span>
<a id="__codelineno-0-5" name="__codelineno-0-5" href="#__codelineno-0-5"></a> <span class="mi">2</span><span class="p">:</span> <span class="p">[</span><span class="mi">0</span><span class="p">,</span> <span class="mi">3</span><span class="p">],</span>
<a id="__codelineno-0-6" name="__codelineno-0-6" href="#__codelineno-0-6"></a> <span class="mi">3</span><span class="p">:</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">]</span>
<a id="__codelineno-0-7" name="__codelineno-0-7" href="#__codelineno-0-7"></a><span class="p">}</span>
<a id="__codelineno-0-8" name="__codelineno-0-8" href="#__codelineno-0-8"></a>
<a id="__codelineno-0-9" name="__codelineno-0-9" href="#__codelineno-0-9"></a><span class="c1"># 从边列表构建</span>
<a id="__codelineno-0-10" name="__codelineno-0-10" href="#__codelineno-0-10"></a><span class="k">def</span><span class="w"> </span><span class="nf">build_graph</span><span class="p">(</span><span class="n">n</span><span class="p">,</span> <span class="n">edges</span><span class="p">):</span>
<a id="__codelineno-0-11" name="__codelineno-0-11" href="#__codelineno-0-11"></a> <span class="n">graph</span> <span class="o">=</span> <span class="p">{</span><span class="n">i</span><span class="p">:</span> <span class="p">[]</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span><span class="p">)}</span>
<a id="__codelineno-0-12" name="__codelineno-0-12" href="#__codelineno-0-12"></a> <span class="k">for</span> <span class="n">u</span><span class="p">,</span> <span class="n">v</span> <span class="ow">in</span> <span class="n">edges</span><span class="p">:</span>
<a id="__codelineno-0-13" name="__codelineno-0-13" href="#__codelineno-0-13"></a> <span class="n">graph</span><span class="p">[</span><span class="n">u</span><span class="p">]</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">v</span><span class="p">)</span>
<a id="__codelineno-0-14" name="__codelineno-0-14" href="#__codelineno-0-14"></a> <span class="n">graph</span><span class="p">[</span><span class="n">v</span><span class="p">]</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">u</span><span class="p">)</span> <span class="c1"># 有向图省略这一行</span>
<a id="__codelineno-0-15" name="__codelineno-0-15" href="#__codelineno-0-15"></a> <span class="k">return</span> <span class="n">graph</span>
</code></pre></div>
<ul>
<li>
<p><strong>邻接矩阵</strong><span class="arithmatex">\(n \times n\)</span> 矩阵,其中 <span class="arithmatex">\(A[i][j] = 1\)</span> 如果边 <span class="arithmatex">\((i, j)\)</span> 存在。空间:<span class="arithmatex">\(O(|V|^2)\)</span>。最适合稠密图或需要 <span class="arithmatex">\(O(1)\)</span> 边查找时。</p>
</li>
<li>
<p><strong>何时使用哪种</strong>:绝大多数情况使用邻接表。只有当图很稠密(<span class="arithmatex">\(|E| \approx |V|^2\)</span>)或需要常数时间边存在性检查时才使用矩阵。</p>
</li>
</ul>
<h2 id="bfs">模式:BFS(广度优先搜索)<a class="headerlink" href="#bfs" title="Permanent link">&para;</a></h2>
<ul>
<li>BFS 使用队列<strong>逐层</strong>探索节点。它是以下问题的首选算法:<ul>
<li><strong>无权</strong>图中的最短路径</li>
<li>层序遍历</li>
<li>寻找连通分量</li>
<li>任何询问"最小步数"的问题</li>
</ul>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-1-1" name="__codelineno-1-1" href="#__codelineno-1-1"></a><span class="kn">from</span><span class="w"> </span><span class="nn">collections</span><span class="w"> </span><span class="kn">import</span> <span class="n">deque</span>
<a id="__codelineno-1-2" name="__codelineno-1-2" href="#__codelineno-1-2"></a>
<a id="__codelineno-1-3" name="__codelineno-1-3" href="#__codelineno-1-3"></a><span class="k">def</span><span class="w"> </span><span class="nf">bfs</span><span class="p">(</span><span class="n">graph</span><span class="p">,</span> <span class="n">start</span><span class="p">):</span>
<a id="__codelineno-1-4" name="__codelineno-1-4" href="#__codelineno-1-4"></a> <span class="n">visited</span> <span class="o">=</span> <span class="p">{</span><span class="n">start</span><span class="p">}</span>
<a id="__codelineno-1-5" name="__codelineno-1-5" href="#__codelineno-1-5"></a> <span class="n">queue</span> <span class="o">=</span> <span class="n">deque</span><span class="p">([</span><span class="n">start</span><span class="p">])</span>
<a id="__codelineno-1-6" name="__codelineno-1-6" href="#__codelineno-1-6"></a>
<a id="__codelineno-1-7" name="__codelineno-1-7" href="#__codelineno-1-7"></a> <span class="k">while</span> <span class="n">queue</span><span class="p">:</span>
<a id="__codelineno-1-8" name="__codelineno-1-8" href="#__codelineno-1-8"></a> <span class="n">node</span> <span class="o">=</span> <span class="n">queue</span><span class="o">.</span><span class="n">popleft</span><span class="p">()</span>
<a id="__codelineno-1-9" name="__codelineno-1-9" href="#__codelineno-1-9"></a> <span class="k">for</span> <span class="n">neighbour</span> <span class="ow">in</span> <span class="n">graph</span><span class="p">[</span><span class="n">node</span><span class="p">]:</span>
<a id="__codelineno-1-10" name="__codelineno-1-10" href="#__codelineno-1-10"></a> <span class="k">if</span> <span class="n">neighbour</span> <span class="ow">not</span> <span class="ow">in</span> <span class="n">visited</span><span class="p">:</span>
<a id="__codelineno-1-11" name="__codelineno-1-11" href="#__codelineno-1-11"></a> <span class="n">visited</span><span class="o">.</span><span class="n">add</span><span class="p">(</span><span class="n">neighbour</span><span class="p">)</span>
<a id="__codelineno-1-12" name="__codelineno-1-12" href="#__codelineno-1-12"></a> <span class="n">queue</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">neighbour</span><span class="p">)</span>
</code></pre></div>
<ul>
<li><strong>关键</strong>:在<strong>入队时</strong>添加到 <code>visited</code>,而不是在出队时。如果你在出队时标记已访问,同一个节点可能被不同前驱多次入队,浪费时间并可能导致错误结果。</li>
</ul>
<h3 id="_3">简单:岛屿数量<a class="headerlink" href="#_3" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p><strong>问题</strong>:给定一个由 '1'(陆地)和 '0'(水)组成的 2D 网格,计算岛屿的数量。</p>
</li>
<li>
<p><strong>模式</strong>:遍历网格。当找到一个 '1' 时,使用 BFS/DFS 将所有连通的陆地单元格标记为已访问。每次开始 BFS 就是一个岛屿。</p>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-2-1" name="__codelineno-2-1" href="#__codelineno-2-1"></a><span class="kn">from</span><span class="w"> </span><span class="nn">collections</span><span class="w"> </span><span class="kn">import</span> <span class="n">deque</span>
<a id="__codelineno-2-2" name="__codelineno-2-2" href="#__codelineno-2-2"></a>
<a id="__codelineno-2-3" name="__codelineno-2-3" href="#__codelineno-2-3"></a><span class="k">def</span><span class="w"> </span><span class="nf">num_islands</span><span class="p">(</span><span class="n">grid</span><span class="p">):</span>
<a id="__codelineno-2-4" name="__codelineno-2-4" href="#__codelineno-2-4"></a> <span class="k">if</span> <span class="ow">not</span> <span class="n">grid</span><span class="p">:</span>
<a id="__codelineno-2-5" name="__codelineno-2-5" href="#__codelineno-2-5"></a> <span class="k">return</span> <span class="mi">0</span>
<a id="__codelineno-2-6" name="__codelineno-2-6" href="#__codelineno-2-6"></a>
<a id="__codelineno-2-7" name="__codelineno-2-7" href="#__codelineno-2-7"></a> <span class="n">rows</span><span class="p">,</span> <span class="n">cols</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">grid</span><span class="p">),</span> <span class="nb">len</span><span class="p">(</span><span class="n">grid</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span>
<a id="__codelineno-2-8" name="__codelineno-2-8" href="#__codelineno-2-8"></a> <span class="n">count</span> <span class="o">=</span> <span class="mi">0</span>
<a id="__codelineno-2-9" name="__codelineno-2-9" href="#__codelineno-2-9"></a>
<a id="__codelineno-2-10" name="__codelineno-2-10" href="#__codelineno-2-10"></a> <span class="k">for</span> <span class="n">r</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">rows</span><span class="p">):</span>
<a id="__codelineno-2-11" name="__codelineno-2-11" href="#__codelineno-2-11"></a> <span class="k">for</span> <span class="n">c</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">cols</span><span class="p">):</span>
<a id="__codelineno-2-12" name="__codelineno-2-12" href="#__codelineno-2-12"></a> <span class="k">if</span> <span class="n">grid</span><span class="p">[</span><span class="n">r</span><span class="p">][</span><span class="n">c</span><span class="p">]</span> <span class="o">==</span> <span class="s1">&#39;1&#39;</span><span class="p">:</span>
<a id="__codelineno-2-13" name="__codelineno-2-13" href="#__codelineno-2-13"></a> <span class="n">count</span> <span class="o">+=</span> <span class="mi">1</span>
<a id="__codelineno-2-14" name="__codelineno-2-14" href="#__codelineno-2-14"></a> <span class="c1"># BFS 标记整个岛屿</span>
<a id="__codelineno-2-15" name="__codelineno-2-15" href="#__codelineno-2-15"></a> <span class="n">queue</span> <span class="o">=</span> <span class="n">deque</span><span class="p">([(</span><span class="n">r</span><span class="p">,</span> <span class="n">c</span><span class="p">)])</span>
<a id="__codelineno-2-16" name="__codelineno-2-16" href="#__codelineno-2-16"></a> <span class="n">grid</span><span class="p">[</span><span class="n">r</span><span class="p">][</span><span class="n">c</span><span class="p">]</span> <span class="o">=</span> <span class="s1">&#39;0&#39;</span> <span class="c1"># 标记已访问</span>
<a id="__codelineno-2-17" name="__codelineno-2-17" href="#__codelineno-2-17"></a> <span class="k">while</span> <span class="n">queue</span><span class="p">:</span>
<a id="__codelineno-2-18" name="__codelineno-2-18" href="#__codelineno-2-18"></a> <span class="n">cr</span><span class="p">,</span> <span class="n">cc</span> <span class="o">=</span> <span class="n">queue</span><span class="o">.</span><span class="n">popleft</span><span class="p">()</span>
<a id="__codelineno-2-19" name="__codelineno-2-19" href="#__codelineno-2-19"></a> <span class="k">for</span> <span class="n">dr</span><span class="p">,</span> <span class="n">dc</span> <span class="ow">in</span> <span class="p">[(</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">),(</span><span class="mi">0</span><span class="p">,</span><span class="o">-</span><span class="mi">1</span><span class="p">),(</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">),(</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">)]:</span>
<a id="__codelineno-2-20" name="__codelineno-2-20" href="#__codelineno-2-20"></a> <span class="n">nr</span><span class="p">,</span> <span class="n">nc</span> <span class="o">=</span> <span class="n">cr</span> <span class="o">+</span> <span class="n">dr</span><span class="p">,</span> <span class="n">cc</span> <span class="o">+</span> <span class="n">dc</span>
<a id="__codelineno-2-21" name="__codelineno-2-21" href="#__codelineno-2-21"></a> <span class="k">if</span> <span class="mi">0</span> <span class="o">&lt;=</span> <span class="n">nr</span> <span class="o">&lt;</span> <span class="n">rows</span> <span class="ow">and</span> <span class="mi">0</span> <span class="o">&lt;=</span> <span class="n">nc</span> <span class="o">&lt;</span> <span class="n">cols</span> <span class="ow">and</span> <span class="n">grid</span><span class="p">[</span><span class="n">nr</span><span class="p">][</span><span class="n">nc</span><span class="p">]</span> <span class="o">==</span> <span class="s1">&#39;1&#39;</span><span class="p">:</span>
<a id="__codelineno-2-22" name="__codelineno-2-22" href="#__codelineno-2-22"></a> <span class="n">grid</span><span class="p">[</span><span class="n">nr</span><span class="p">][</span><span class="n">nc</span><span class="p">]</span> <span class="o">=</span> <span class="s1">&#39;0&#39;</span>
<a id="__codelineno-2-23" name="__codelineno-2-23" href="#__codelineno-2-23"></a> <span class="n">queue</span><span class="o">.</span><span class="n">append</span><span class="p">((</span><span class="n">nr</span><span class="p">,</span> <span class="n">nc</span><span class="p">))</span>
<a id="__codelineno-2-24" name="__codelineno-2-24" href="#__codelineno-2-24"></a>
<a id="__codelineno-2-25" name="__codelineno-2-25" href="#__codelineno-2-25"></a> <span class="k">return</span> <span class="n">count</span>
</code></pre></div>
<ul>
<li>
<p><strong>陷阱</strong><code>directions = [(0,1),(0,-1),(1,0),(-1,0)]</code> 模式用于四连通网格邻居,几乎每个网格问题都会用到。记住它。对于八连通,加上对角线。</p>
</li>
<li>
<p><strong>陷阱</strong>:修改输入网格(<code>grid[r][c] = '0'</code>)避免了需要单独的 <code>visited</code> 集合。在面试中这是可以接受的,但要明确说明权衡(改变了输入)。</p>
</li>
</ul>
<h3 id="_4">中等:腐烂的橘子<a class="headerlink" href="#_4" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p><strong>问题</strong>:新鲜橘子如果与腐烂橘子相邻则会腐烂。返回所有橘子都腐烂的最短时间(如果不可能则返回 -1)。</p>
</li>
<li>
<p><strong>模式</strong>:多源 BFS。将所有初始腐烂的橘子同时放入队列。每层 BFS 就是一个时间步。</p>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-3-1" name="__codelineno-3-1" href="#__codelineno-3-1"></a><span class="kn">from</span><span class="w"> </span><span class="nn">collections</span><span class="w"> </span><span class="kn">import</span> <span class="n">deque</span>
<a id="__codelineno-3-2" name="__codelineno-3-2" href="#__codelineno-3-2"></a>
<a id="__codelineno-3-3" name="__codelineno-3-3" href="#__codelineno-3-3"></a><span class="k">def</span><span class="w"> </span><span class="nf">oranges_rotting</span><span class="p">(</span><span class="n">grid</span><span class="p">):</span>
<a id="__codelineno-3-4" name="__codelineno-3-4" href="#__codelineno-3-4"></a> <span class="n">rows</span><span class="p">,</span> <span class="n">cols</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">grid</span><span class="p">),</span> <span class="nb">len</span><span class="p">(</span><span class="n">grid</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span>
<a id="__codelineno-3-5" name="__codelineno-3-5" href="#__codelineno-3-5"></a> <span class="n">queue</span> <span class="o">=</span> <span class="n">deque</span><span class="p">()</span>
<a id="__codelineno-3-6" name="__codelineno-3-6" href="#__codelineno-3-6"></a> <span class="n">fresh</span> <span class="o">=</span> <span class="mi">0</span>
<a id="__codelineno-3-7" name="__codelineno-3-7" href="#__codelineno-3-7"></a>
<a id="__codelineno-3-8" name="__codelineno-3-8" href="#__codelineno-3-8"></a> <span class="k">for</span> <span class="n">r</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">rows</span><span class="p">):</span>
<a id="__codelineno-3-9" name="__codelineno-3-9" href="#__codelineno-3-9"></a> <span class="k">for</span> <span class="n">c</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">cols</span><span class="p">):</span>
<a id="__codelineno-3-10" name="__codelineno-3-10" href="#__codelineno-3-10"></a> <span class="k">if</span> <span class="n">grid</span><span class="p">[</span><span class="n">r</span><span class="p">][</span><span class="n">c</span><span class="p">]</span> <span class="o">==</span> <span class="mi">2</span><span class="p">:</span>
<a id="__codelineno-3-11" name="__codelineno-3-11" href="#__codelineno-3-11"></a> <span class="n">queue</span><span class="o">.</span><span class="n">append</span><span class="p">((</span><span class="n">r</span><span class="p">,</span> <span class="n">c</span><span class="p">))</span>
<a id="__codelineno-3-12" name="__codelineno-3-12" href="#__codelineno-3-12"></a> <span class="k">elif</span> <span class="n">grid</span><span class="p">[</span><span class="n">r</span><span class="p">][</span><span class="n">c</span><span class="p">]</span> <span class="o">==</span> <span class="mi">1</span><span class="p">:</span>
<a id="__codelineno-3-13" name="__codelineno-3-13" href="#__codelineno-3-13"></a> <span class="n">fresh</span> <span class="o">+=</span> <span class="mi">1</span>
<a id="__codelineno-3-14" name="__codelineno-3-14" href="#__codelineno-3-14"></a>
<a id="__codelineno-3-15" name="__codelineno-3-15" href="#__codelineno-3-15"></a> <span class="k">if</span> <span class="n">fresh</span> <span class="o">==</span> <span class="mi">0</span><span class="p">:</span>
<a id="__codelineno-3-16" name="__codelineno-3-16" href="#__codelineno-3-16"></a> <span class="k">return</span> <span class="mi">0</span>
<a id="__codelineno-3-17" name="__codelineno-3-17" href="#__codelineno-3-17"></a>
<a id="__codelineno-3-18" name="__codelineno-3-18" href="#__codelineno-3-18"></a> <span class="n">time</span> <span class="o">=</span> <span class="mi">0</span>
<a id="__codelineno-3-19" name="__codelineno-3-19" href="#__codelineno-3-19"></a> <span class="k">while</span> <span class="n">queue</span> <span class="ow">and</span> <span class="n">fresh</span> <span class="o">&gt;</span> <span class="mi">0</span><span class="p">:</span>
<a id="__codelineno-3-20" name="__codelineno-3-20" href="#__codelineno-3-20"></a> <span class="n">time</span> <span class="o">+=</span> <span class="mi">1</span>
<a id="__codelineno-3-21" name="__codelineno-3-21" href="#__codelineno-3-21"></a> <span class="k">for</span> <span class="n">_</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">queue</span><span class="p">)):</span>
<a id="__codelineno-3-22" name="__codelineno-3-22" href="#__codelineno-3-22"></a> <span class="n">cr</span><span class="p">,</span> <span class="n">cc</span> <span class="o">=</span> <span class="n">queue</span><span class="o">.</span><span class="n">popleft</span><span class="p">()</span>
<a id="__codelineno-3-23" name="__codelineno-3-23" href="#__codelineno-3-23"></a> <span class="k">for</span> <span class="n">dr</span><span class="p">,</span> <span class="n">dc</span> <span class="ow">in</span> <span class="p">[(</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">),(</span><span class="mi">0</span><span class="p">,</span><span class="o">-</span><span class="mi">1</span><span class="p">),(</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">),(</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">)]:</span>
<a id="__codelineno-3-24" name="__codelineno-3-24" href="#__codelineno-3-24"></a> <span class="n">nr</span><span class="p">,</span> <span class="n">nc</span> <span class="o">=</span> <span class="n">cr</span> <span class="o">+</span> <span class="n">dr</span><span class="p">,</span> <span class="n">cc</span> <span class="o">+</span> <span class="n">dc</span>
<a id="__codelineno-3-25" name="__codelineno-3-25" href="#__codelineno-3-25"></a> <span class="k">if</span> <span class="mi">0</span> <span class="o">&lt;=</span> <span class="n">nr</span> <span class="o">&lt;</span> <span class="n">rows</span> <span class="ow">and</span> <span class="mi">0</span> <span class="o">&lt;=</span> <span class="n">nc</span> <span class="o">&lt;</span> <span class="n">cols</span> <span class="ow">and</span> <span class="n">grid</span><span class="p">[</span><span class="n">nr</span><span class="p">][</span><span class="n">nc</span><span class="p">]</span> <span class="o">==</span> <span class="mi">1</span><span class="p">:</span>
<a id="__codelineno-3-26" name="__codelineno-3-26" href="#__codelineno-3-26"></a> <span class="n">grid</span><span class="p">[</span><span class="n">nr</span><span class="p">][</span><span class="n">nc</span><span class="p">]</span> <span class="o">=</span> <span class="mi">2</span>
<a id="__codelineno-3-27" name="__codelineno-3-27" href="#__codelineno-3-27"></a> <span class="n">fresh</span> <span class="o">-=</span> <span class="mi">1</span>
<a id="__codelineno-3-28" name="__codelineno-3-28" href="#__codelineno-3-28"></a> <span class="n">queue</span><span class="o">.</span><span class="n">append</span><span class="p">((</span><span class="n">nr</span><span class="p">,</span> <span class="n">nc</span><span class="p">))</span>
<a id="__codelineno-3-29" name="__codelineno-3-29" href="#__codelineno-3-29"></a>
<a id="__codelineno-3-30" name="__codelineno-3-30" href="#__codelineno-3-30"></a> <span class="k">return</span> <span class="n">time</span> <span class="k">if</span> <span class="n">fresh</span> <span class="o">==</span> <span class="mi">0</span> <span class="k">else</span> <span class="o">-</span><span class="mi">1</span>
</code></pre></div>
<ul>
<li><strong>关键洞察</strong>:多源 BFS 同时处理所有源。这给出了从<em>任何</em>源的最短距离,这正是"最后一个新鲜橘子腐烂需要多长时间"。</li>
</ul>
<h2 id="dfs">模式:DFS(深度优先搜索)<a class="headerlink" href="#dfs" title="Permanent link">&para;</a></h2>
<ul>
<li>DFS 尽可能深地探索,然后回溯。它使用栈(显式栈或通过递归使用调用栈)。DFS 是以下问题的首选:<ul>
<li>环检测</li>
<li>拓扑排序</li>
<li>连通分量</li>
<li>回溯 / 穷举搜索</li>
<li>带约束的寻路</li>
</ul>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-4-1" name="__codelineno-4-1" href="#__codelineno-4-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">dfs</span><span class="p">(</span><span class="n">graph</span><span class="p">,</span> <span class="n">node</span><span class="p">,</span> <span class="n">visited</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
<a id="__codelineno-4-2" name="__codelineno-4-2" href="#__codelineno-4-2"></a> <span class="k">if</span> <span class="n">visited</span> <span class="ow">is</span> <span class="kc">None</span><span class="p">:</span>
<a id="__codelineno-4-3" name="__codelineno-4-3" href="#__codelineno-4-3"></a> <span class="n">visited</span> <span class="o">=</span> <span class="nb">set</span><span class="p">()</span>
<a id="__codelineno-4-4" name="__codelineno-4-4" href="#__codelineno-4-4"></a> <span class="n">visited</span><span class="o">.</span><span class="n">add</span><span class="p">(</span><span class="n">node</span><span class="p">)</span>
<a id="__codelineno-4-5" name="__codelineno-4-5" href="#__codelineno-4-5"></a> <span class="k">for</span> <span class="n">neighbour</span> <span class="ow">in</span> <span class="n">graph</span><span class="p">[</span><span class="n">node</span><span class="p">]:</span>
<a id="__codelineno-4-6" name="__codelineno-4-6" href="#__codelineno-4-6"></a> <span class="k">if</span> <span class="n">neighbour</span> <span class="ow">not</span> <span class="ow">in</span> <span class="n">visited</span><span class="p">:</span>
<a id="__codelineno-4-7" name="__codelineno-4-7" href="#__codelineno-4-7"></a> <span class="n">dfs</span><span class="p">(</span><span class="n">graph</span><span class="p">,</span> <span class="n">neighbour</span><span class="p">,</span> <span class="n">visited</span><span class="p">)</span>
</code></pre></div>
<h3 id="_5">中等:课程表(环检测)<a class="headerlink" href="#_5" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p><strong>问题</strong>:给定 <span class="arithmatex">\(n\)</span> 门课程和先修条件,判断是否能完成所有课程(即,没有循环依赖)。</p>
</li>
<li>
<p><strong>模式</strong>:在有向图中检测环。使用带有三种状态的 DFS:未访问、正在进行(在当前 DFS 路径上)、已完成。</p>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-5-1" name="__codelineno-5-1" href="#__codelineno-5-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">can_finish</span><span class="p">(</span><span class="n">num_courses</span><span class="p">,</span> <span class="n">prerequisites</span><span class="p">):</span>
<a id="__codelineno-5-2" name="__codelineno-5-2" href="#__codelineno-5-2"></a> <span class="n">graph</span> <span class="o">=</span> <span class="p">{</span><span class="n">i</span><span class="p">:</span> <span class="p">[]</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">num_courses</span><span class="p">)}</span>
<a id="__codelineno-5-3" name="__codelineno-5-3" href="#__codelineno-5-3"></a> <span class="k">for</span> <span class="n">course</span><span class="p">,</span> <span class="n">prereq</span> <span class="ow">in</span> <span class="n">prerequisites</span><span class="p">:</span>
<a id="__codelineno-5-4" name="__codelineno-5-4" href="#__codelineno-5-4"></a> <span class="n">graph</span><span class="p">[</span><span class="n">course</span><span class="p">]</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">prereq</span><span class="p">)</span>
<a id="__codelineno-5-5" name="__codelineno-5-5" href="#__codelineno-5-5"></a>
<a id="__codelineno-5-6" name="__codelineno-5-6" href="#__codelineno-5-6"></a> <span class="c1"># 0 = 未访问, 1 = 进行中, 2 = 已完成</span>
<a id="__codelineno-5-7" name="__codelineno-5-7" href="#__codelineno-5-7"></a> <span class="n">state</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="n">num_courses</span>
<a id="__codelineno-5-8" name="__codelineno-5-8" href="#__codelineno-5-8"></a>
<a id="__codelineno-5-9" name="__codelineno-5-9" href="#__codelineno-5-9"></a> <span class="k">def</span><span class="w"> </span><span class="nf">has_cycle</span><span class="p">(</span><span class="n">node</span><span class="p">):</span>
<a id="__codelineno-5-10" name="__codelineno-5-10" href="#__codelineno-5-10"></a> <span class="k">if</span> <span class="n">state</span><span class="p">[</span><span class="n">node</span><span class="p">]</span> <span class="o">==</span> <span class="mi">1</span><span class="p">:</span>
<a id="__codelineno-5-11" name="__codelineno-5-11" href="#__codelineno-5-11"></a> <span class="k">return</span> <span class="kc">True</span> <span class="c1"># 回边 → 环</span>
<a id="__codelineno-5-12" name="__codelineno-5-12" href="#__codelineno-5-12"></a> <span class="k">if</span> <span class="n">state</span><span class="p">[</span><span class="n">node</span><span class="p">]</span> <span class="o">==</span> <span class="mi">2</span><span class="p">:</span>
<a id="__codelineno-5-13" name="__codelineno-5-13" href="#__codelineno-5-13"></a> <span class="k">return</span> <span class="kc">False</span> <span class="c1"># 已经完全探索过</span>
<a id="__codelineno-5-14" name="__codelineno-5-14" href="#__codelineno-5-14"></a>
<a id="__codelineno-5-15" name="__codelineno-5-15" href="#__codelineno-5-15"></a> <span class="n">state</span><span class="p">[</span><span class="n">node</span><span class="p">]</span> <span class="o">=</span> <span class="mi">1</span> <span class="c1"># 标记为进行中</span>
<a id="__codelineno-5-16" name="__codelineno-5-16" href="#__codelineno-5-16"></a> <span class="k">for</span> <span class="n">neighbour</span> <span class="ow">in</span> <span class="n">graph</span><span class="p">[</span><span class="n">node</span><span class="p">]:</span>
<a id="__codelineno-5-17" name="__codelineno-5-17" href="#__codelineno-5-17"></a> <span class="k">if</span> <span class="n">has_cycle</span><span class="p">(</span><span class="n">neighbour</span><span class="p">):</span>
<a id="__codelineno-5-18" name="__codelineno-5-18" href="#__codelineno-5-18"></a> <span class="k">return</span> <span class="kc">True</span>
<a id="__codelineno-5-19" name="__codelineno-5-19" href="#__codelineno-5-19"></a> <span class="n">state</span><span class="p">[</span><span class="n">node</span><span class="p">]</span> <span class="o">=</span> <span class="mi">2</span> <span class="c1"># 标记为已完成</span>
<a id="__codelineno-5-20" name="__codelineno-5-20" href="#__codelineno-5-20"></a> <span class="k">return</span> <span class="kc">False</span>
<a id="__codelineno-5-21" name="__codelineno-5-21" href="#__codelineno-5-21"></a>
<a id="__codelineno-5-22" name="__codelineno-5-22" href="#__codelineno-5-22"></a> <span class="k">for</span> <span class="n">course</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">num_courses</span><span class="p">):</span>
<a id="__codelineno-5-23" name="__codelineno-5-23" href="#__codelineno-5-23"></a> <span class="k">if</span> <span class="n">has_cycle</span><span class="p">(</span><span class="n">course</span><span class="p">):</span>
<a id="__codelineno-5-24" name="__codelineno-5-24" href="#__codelineno-5-24"></a> <span class="k">return</span> <span class="kc">False</span>
<a id="__codelineno-5-25" name="__codelineno-5-25" href="#__codelineno-5-25"></a> <span class="k">return</span> <span class="kc">True</span>
</code></pre></div>
<ul>
<li><strong>为什么需要三种状态</strong>:两种状态(已访问/未访问)无法区分"我正在探索这个节点"和"我已完成对这个节点的探索"。找到一个当前正在被探索的节点(状态 = 1)意味着我们发现了环。找到一个已经完全探索的节点(状态 = 2)只是交叉边,不是环。</li>
</ul>
<h3 id="ii">中等:课程表 II(拓扑排序)<a class="headerlink" href="#ii" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p><strong>问题</strong>:返回一个有效的课程顺序(拓扑排序)。</p>
</li>
<li>
<p><strong>模式(Kahn 算法——基于 BFS</strong>:从没有入边的节点(入度为 0)开始。处理它们,减少它们邻居的入度。重复。</p>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-6-1" name="__codelineno-6-1" href="#__codelineno-6-1"></a><span class="kn">from</span><span class="w"> </span><span class="nn">collections</span><span class="w"> </span><span class="kn">import</span> <span class="n">deque</span>
<a id="__codelineno-6-2" name="__codelineno-6-2" href="#__codelineno-6-2"></a>
<a id="__codelineno-6-3" name="__codelineno-6-3" href="#__codelineno-6-3"></a><span class="k">def</span><span class="w"> </span><span class="nf">find_order</span><span class="p">(</span><span class="n">num_courses</span><span class="p">,</span> <span class="n">prerequisites</span><span class="p">):</span>
<a id="__codelineno-6-4" name="__codelineno-6-4" href="#__codelineno-6-4"></a> <span class="n">graph</span> <span class="o">=</span> <span class="p">{</span><span class="n">i</span><span class="p">:</span> <span class="p">[]</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">num_courses</span><span class="p">)}</span>
<a id="__codelineno-6-5" name="__codelineno-6-5" href="#__codelineno-6-5"></a> <span class="n">indegree</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="n">num_courses</span>
<a id="__codelineno-6-6" name="__codelineno-6-6" href="#__codelineno-6-6"></a>
<a id="__codelineno-6-7" name="__codelineno-6-7" href="#__codelineno-6-7"></a> <span class="k">for</span> <span class="n">course</span><span class="p">,</span> <span class="n">prereq</span> <span class="ow">in</span> <span class="n">prerequisites</span><span class="p">:</span>
<a id="__codelineno-6-8" name="__codelineno-6-8" href="#__codelineno-6-8"></a> <span class="n">graph</span><span class="p">[</span><span class="n">prereq</span><span class="p">]</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">course</span><span class="p">)</span>
<a id="__codelineno-6-9" name="__codelineno-6-9" href="#__codelineno-6-9"></a> <span class="n">indegree</span><span class="p">[</span><span class="n">course</span><span class="p">]</span> <span class="o">+=</span> <span class="mi">1</span>
<a id="__codelineno-6-10" name="__codelineno-6-10" href="#__codelineno-6-10"></a>
<a id="__codelineno-6-11" name="__codelineno-6-11" href="#__codelineno-6-11"></a> <span class="n">queue</span> <span class="o">=</span> <span class="n">deque</span><span class="p">([</span><span class="n">i</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">num_courses</span><span class="p">)</span> <span class="k">if</span> <span class="n">indegree</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">==</span> <span class="mi">0</span><span class="p">])</span>
<a id="__codelineno-6-12" name="__codelineno-6-12" href="#__codelineno-6-12"></a> <span class="n">order</span> <span class="o">=</span> <span class="p">[]</span>
<a id="__codelineno-6-13" name="__codelineno-6-13" href="#__codelineno-6-13"></a>
<a id="__codelineno-6-14" name="__codelineno-6-14" href="#__codelineno-6-14"></a> <span class="k">while</span> <span class="n">queue</span><span class="p">:</span>
<a id="__codelineno-6-15" name="__codelineno-6-15" href="#__codelineno-6-15"></a> <span class="n">node</span> <span class="o">=</span> <span class="n">queue</span><span class="o">.</span><span class="n">popleft</span><span class="p">()</span>
<a id="__codelineno-6-16" name="__codelineno-6-16" href="#__codelineno-6-16"></a> <span class="n">order</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">node</span><span class="p">)</span>
<a id="__codelineno-6-17" name="__codelineno-6-17" href="#__codelineno-6-17"></a> <span class="k">for</span> <span class="n">neighbour</span> <span class="ow">in</span> <span class="n">graph</span><span class="p">[</span><span class="n">node</span><span class="p">]:</span>
<a id="__codelineno-6-18" name="__codelineno-6-18" href="#__codelineno-6-18"></a> <span class="n">indegree</span><span class="p">[</span><span class="n">neighbour</span><span class="p">]</span> <span class="o">-=</span> <span class="mi">1</span>
<a id="__codelineno-6-19" name="__codelineno-6-19" href="#__codelineno-6-19"></a> <span class="k">if</span> <span class="n">indegree</span><span class="p">[</span><span class="n">neighbour</span><span class="p">]</span> <span class="o">==</span> <span class="mi">0</span><span class="p">:</span>
<a id="__codelineno-6-20" name="__codelineno-6-20" href="#__codelineno-6-20"></a> <span class="n">queue</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">neighbour</span><span class="p">)</span>
<a id="__codelineno-6-21" name="__codelineno-6-21" href="#__codelineno-6-21"></a>
<a id="__codelineno-6-22" name="__codelineno-6-22" href="#__codelineno-6-22"></a> <span class="k">return</span> <span class="n">order</span> <span class="k">if</span> <span class="nb">len</span><span class="p">(</span><span class="n">order</span><span class="p">)</span> <span class="o">==</span> <span class="n">num_courses</span> <span class="k">else</span> <span class="p">[]</span> <span class="c1"># 空 = 存在环</span>
</code></pre></div>
<ul>
<li><strong>陷阱</strong>:如果结果中的节点数少于图中的节点数,则存在环(某些节点的入度从未降到 0)。</li>
</ul>
<h2 id="_6">最短路径<a class="headerlink" href="#_6" title="Permanent link">&para;</a></h2>
<h3 id="dijkstra">Dijkstra 算法<a class="headerlink" href="#dijkstra" title="Permanent link">&para;</a></h3>
<ul>
<li><strong>非负</strong>加权图中从源点找到到所有其他节点的最短路径。使用优先队列(最小堆)。</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-7-1" name="__codelineno-7-1" href="#__codelineno-7-1"></a><span class="kn">import</span><span class="w"> </span><span class="nn">heapq</span>
<a id="__codelineno-7-2" name="__codelineno-7-2" href="#__codelineno-7-2"></a>
<a id="__codelineno-7-3" name="__codelineno-7-3" href="#__codelineno-7-3"></a><span class="k">def</span><span class="w"> </span><span class="nf">dijkstra</span><span class="p">(</span><span class="n">graph</span><span class="p">,</span> <span class="n">start</span><span class="p">):</span>
<a id="__codelineno-7-4" name="__codelineno-7-4" href="#__codelineno-7-4"></a> <span class="c1"># graph: {node: [(neighbour, weight), ...]}</span>
<a id="__codelineno-7-5" name="__codelineno-7-5" href="#__codelineno-7-5"></a> <span class="n">dist</span> <span class="o">=</span> <span class="p">{</span><span class="n">node</span><span class="p">:</span> <span class="nb">float</span><span class="p">(</span><span class="s1">&#39;inf&#39;</span><span class="p">)</span> <span class="k">for</span> <span class="n">node</span> <span class="ow">in</span> <span class="n">graph</span><span class="p">}</span>
<a id="__codelineno-7-6" name="__codelineno-7-6" href="#__codelineno-7-6"></a> <span class="n">dist</span><span class="p">[</span><span class="n">start</span><span class="p">]</span> <span class="o">=</span> <span class="mi">0</span>
<a id="__codelineno-7-7" name="__codelineno-7-7" href="#__codelineno-7-7"></a> <span class="n">heap</span> <span class="o">=</span> <span class="p">[(</span><span class="mi">0</span><span class="p">,</span> <span class="n">start</span><span class="p">)]</span>
<a id="__codelineno-7-8" name="__codelineno-7-8" href="#__codelineno-7-8"></a>
<a id="__codelineno-7-9" name="__codelineno-7-9" href="#__codelineno-7-9"></a> <span class="k">while</span> <span class="n">heap</span><span class="p">:</span>
<a id="__codelineno-7-10" name="__codelineno-7-10" href="#__codelineno-7-10"></a> <span class="n">d</span><span class="p">,</span> <span class="n">node</span> <span class="o">=</span> <span class="n">heapq</span><span class="o">.</span><span class="n">heappop</span><span class="p">(</span><span class="n">heap</span><span class="p">)</span>
<a id="__codelineno-7-11" name="__codelineno-7-11" href="#__codelineno-7-11"></a> <span class="k">if</span> <span class="n">d</span> <span class="o">&gt;</span> <span class="n">dist</span><span class="p">[</span><span class="n">node</span><span class="p">]:</span>
<a id="__codelineno-7-12" name="__codelineno-7-12" href="#__codelineno-7-12"></a> <span class="k">continue</span> <span class="c1"># 过期条目</span>
<a id="__codelineno-7-13" name="__codelineno-7-13" href="#__codelineno-7-13"></a>
<a id="__codelineno-7-14" name="__codelineno-7-14" href="#__codelineno-7-14"></a> <span class="k">for</span> <span class="n">neighbour</span><span class="p">,</span> <span class="n">weight</span> <span class="ow">in</span> <span class="n">graph</span><span class="p">[</span><span class="n">node</span><span class="p">]:</span>
<a id="__codelineno-7-15" name="__codelineno-7-15" href="#__codelineno-7-15"></a> <span class="n">new_dist</span> <span class="o">=</span> <span class="n">d</span> <span class="o">+</span> <span class="n">weight</span>
<a id="__codelineno-7-16" name="__codelineno-7-16" href="#__codelineno-7-16"></a> <span class="k">if</span> <span class="n">new_dist</span> <span class="o">&lt;</span> <span class="n">dist</span><span class="p">[</span><span class="n">neighbour</span><span class="p">]:</span>
<a id="__codelineno-7-17" name="__codelineno-7-17" href="#__codelineno-7-17"></a> <span class="n">dist</span><span class="p">[</span><span class="n">neighbour</span><span class="p">]</span> <span class="o">=</span> <span class="n">new_dist</span>
<a id="__codelineno-7-18" name="__codelineno-7-18" href="#__codelineno-7-18"></a> <span class="n">heapq</span><span class="o">.</span><span class="n">heappush</span><span class="p">(</span><span class="n">heap</span><span class="p">,</span> <span class="p">(</span><span class="n">new_dist</span><span class="p">,</span> <span class="n">neighbour</span><span class="p">))</span>
<a id="__codelineno-7-19" name="__codelineno-7-19" href="#__codelineno-7-19"></a>
<a id="__codelineno-7-20" name="__codelineno-7-20" href="#__codelineno-7-20"></a> <span class="k">return</span> <span class="n">dist</span>
</code></pre></div>
<ul>
<li>
<p>时间:使用二叉堆为 <span class="arithmatex">\(O((|V| + |E|) \log |V|)\)</span></p>
</li>
<li>
<p><strong>陷阱</strong><code>if d &gt; dist[node]: continue</code> 这行是必须的。没有它,你会处理过期的堆条目,可能退化到 <span class="arithmatex">\(O(|V|^2)\)</span></p>
</li>
<li>
<p><strong>陷阱</strong>:Dijkstra 不适用于负权重。如果一条边有负权重,贪心假设(一旦节点被确定,其距离就是最优的)就不成立了。应改用 Bellman-Ford。</p>
</li>
</ul>
<h3 id="_7">困难:网络延迟时间<a class="headerlink" href="#_7" title="Permanent link">&para;</a></h3>
<ul>
<li><strong>问题</strong>:给定 <span class="arithmatex">\(n\)</span> 个节点和加权有向边,找出信号从源点到达所有节点所需的时间。如果并非所有节点都可到达,返回 -1。</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-8-1" name="__codelineno-8-1" href="#__codelineno-8-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">network_delay</span><span class="p">(</span><span class="n">times</span><span class="p">,</span> <span class="n">n</span><span class="p">,</span> <span class="n">k</span><span class="p">):</span>
<a id="__codelineno-8-2" name="__codelineno-8-2" href="#__codelineno-8-2"></a> <span class="n">graph</span> <span class="o">=</span> <span class="p">{</span><span class="n">i</span><span class="p">:</span> <span class="p">[]</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">n</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)}</span>
<a id="__codelineno-8-3" name="__codelineno-8-3" href="#__codelineno-8-3"></a> <span class="k">for</span> <span class="n">u</span><span class="p">,</span> <span class="n">v</span><span class="p">,</span> <span class="n">w</span> <span class="ow">in</span> <span class="n">times</span><span class="p">:</span>
<a id="__codelineno-8-4" name="__codelineno-8-4" href="#__codelineno-8-4"></a> <span class="n">graph</span><span class="p">[</span><span class="n">u</span><span class="p">]</span><span class="o">.</span><span class="n">append</span><span class="p">((</span><span class="n">v</span><span class="p">,</span> <span class="n">w</span><span class="p">))</span>
<a id="__codelineno-8-5" name="__codelineno-8-5" href="#__codelineno-8-5"></a>
<a id="__codelineno-8-6" name="__codelineno-8-6" href="#__codelineno-8-6"></a> <span class="n">dist</span> <span class="o">=</span> <span class="n">dijkstra</span><span class="p">(</span><span class="n">graph</span><span class="p">,</span> <span class="n">k</span><span class="p">)</span>
<a id="__codelineno-8-7" name="__codelineno-8-7" href="#__codelineno-8-7"></a> <span class="n">max_time</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="n">dist</span><span class="o">.</span><span class="n">values</span><span class="p">())</span>
<a id="__codelineno-8-8" name="__codelineno-8-8" href="#__codelineno-8-8"></a> <span class="k">return</span> <span class="n">max_time</span> <span class="k">if</span> <span class="n">max_time</span> <span class="o">&lt;</span> <span class="nb">float</span><span class="p">(</span><span class="s1">&#39;inf&#39;</span><span class="p">)</span> <span class="k">else</span> <span class="o">-</span><span class="mi">1</span>
</code></pre></div>
<h2 id="_8">强连通分量<a class="headerlink" href="#_8" title="Permanent link">&para;</a></h2>
<ul>
<li>
<p>在有向图中,<strong>强连通分量(SCC</strong>是一个最大节点集合,其中每个节点都能到达其他所有节点。</p>
</li>
<li>
<p><strong>Kosaraju 算法</strong>:(1) 在原始图上进行 DFS,记录完成顺序。(2) 转置图(反转所有边)。(3) 按完成顺序的逆序在转置图上进行 DFS。第3步中的每个 DFS 树就是一个 SCC。</p>
</li>
<li>
<p><strong>何时使用</strong>:寻找循环依赖、2-SAT、将有向图压缩为 SCC 的 DAG。</p>
</li>
</ul>
<hr />
<h2 id="_9">常见陷阱总结<a class="headerlink" href="#_9" title="Permanent link">&para;</a></h2>
<table>
<thead>
<tr>
<th>陷阱</th>
<th>示例</th>
<th>修复</th>
</tr>
</thead>
<tbody>
<tr>
<td>在出队时标记已访问</td>
<td>同一节点被多次入队</td>
<td>在入队时标记已访问</td>
</tr>
<tr>
<td>有向图中只有两种状态</td>
<td>无法区分回边和交叉边</td>
<td>使用三种状态:未访问/进行中/已完成</td>
</tr>
<tr>
<td>Dijkstra 用于负权重</td>
<td>错误的最短路径</td>
<td>使用 Bellman-Ford</td>
</tr>
<tr>
<td>忘记 <code>if d &gt; dist[node]: continue</code></td>
<td>处理过期堆条目</td>
<td>总是跳过当前距离更差的情况</td>
</tr>
<tr>
<td>网格边界检查</td>
<td>索引越界</td>
<td><code>0 &lt;= nr &lt; rows and 0 &lt;= nc &lt; cols</code></td>
</tr>
<tr>
<td>没有考虑 time=0 的边界情况</td>
<td>腐烂橘子:没有新鲜橘子</td>
<td>在 BFS 之前检查 <code>fresh == 0</code></td>
</tr>
<tr>
<td>将有向图构建为无向图</td>
<td>先修条件是单向的</td>
<td>只在一个方向添加边</td>
</tr>
</tbody>
</table>
<hr />
<h2 id="neetcode">课后练习题(NeetCode<a class="headerlink" href="#neetcode" title="Permanent link">&para;</a></h2>
<h3 id="bfs_1">BFS 模式<a class="headerlink" href="#bfs_1" title="Permanent link">&para;</a></h3>
<ul>
<li><a href="https://neetcode.io/problems/count-number-of-islands">岛屿数量</a> — 网格 BFS/DFS</li>
<li><a href="https://neetcode.io/problems/rotting-fruit">腐烂的橘子</a> — 多源 BFS</li>
<li><a href="https://neetcode.io/problems/clone-graph">克隆图</a> — BFS + 哈希表克隆</li>
<li><a href="https://neetcode.io/problems/pacific-atlantic-water-flow">太平洋大西洋水流</a> — 从两个海洋开始的 BFS</li>
<li><a href="https://neetcode.io/problems/word-ladder">单词接龙</a> — 隐式图上的 BFS</li>
</ul>
<h3 id="dfs_1">DFS 模式<a class="headerlink" href="#dfs_1" title="Permanent link">&para;</a></h3>
<ul>
<li><a href="https://neetcode.io/problems/max-area-of-island">岛屿的最大面积</a> — 带面积计数的 DFS</li>
<li><a href="https://neetcode.io/problems/course-schedule">课程表</a> — 有向图中的环检测</li>
<li><a href="https://neetcode.io/problems/course-schedule-ii">课程表 II</a> — 拓扑排序</li>
<li><a href="https://neetcode.io/problems/count-connected-components">连通分量数量</a> — DFS 或并查集</li>
<li><a href="https://neetcode.io/problems/valid-tree">图是否是树</a> — 连通 + 无环</li>
</ul>
<h3 id="_10">最短路径<a class="headerlink" href="#_10" title="Permanent link">&para;</a></h3>
<ul>
<li><a href="https://neetcode.io/problems/network-delay-time">网络延迟时间</a> — Dijkstra</li>
<li><a href="https://neetcode.io/problems/cheapest-flight-path">K 站中转内最便宜的航班</a> — 带约束的修改版 BFS/Bellman-Ford</li>
<li><a href="https://neetcode.io/problems/swim-in-rising-water">上升水温游泳</a> — 二分查找 + BFS 或网格上的 Dijkstra</li>
</ul>
<h3 id="_11">进阶<a class="headerlink" href="#_11" title="Permanent link">&para;</a></h3>
<ul>
<li><a href="https://neetcode.io/problems/foreign-dictionary">外星文字典</a> — 从字符顺序进行拓扑排序</li>
</ul>
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