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<html lang="zh"><head><meta charset="utf-8"><meta name="viewport" content="width=device-width, initial-scale=1, maximum-scale=1"><meta><title>raincity's blog</title><link rel="manifest" href="/manifest.json"><meta name="application-name" content="raincity's blog"><meta name="msapplication-TileImage" content="/img/favicon.svg"><meta name="apple-mobile-web-app-capable" content="yes"><meta name="apple-mobile-web-app-title" content="raincity's blog"><meta name="apple-mobile-web-app-status-bar-style" content="default"><meta property="og:type" content="blog"><meta property="og:title" content="raincity's blog"><meta property="og:url" content="https://raincity343.github.io/"><meta property="og:site_name" content="raincity's blog"><meta property="og:locale" content="zh_CN"><meta property="og:image" content="https://raincity343.github.io/img/og_image.png"><meta property="article:author" content="raincity"><meta property="twitter:card" content="summary"><meta property="twitter:image:src" content="https://raincity343.github.io/img/og_image.png"><script type="application/ld+json">{"@context":"https://schema.org","@type":"BlogPosting","mainEntityOfPage":{"@type":"WebPage","@id":"https://raincity343.github.io"},"headline":"raincity's blog","image":["https://raincity343.github.io/img/og_image.png"],"author":{"@type":"Person","name":"raincity"},"publisher":{"@type":"Organization","name":"raincity's blog","logo":{"@type":"ImageObject","url":"https://raincity343.github.io/img/logo.svg"}},"description":""}</script><link rel="icon" href="/img/favicon.svg"><link rel="stylesheet" href="https://use.fontawesome.com/releases/v6.0.0/css/all.css"><link data-pjax rel="stylesheet" href="https://cdn.jsdelivr.net/npm/[email protected]/styles/atom-one-light.css"><link rel="stylesheet" href="https://fonts.googleapis.com/css2?family=Ubuntu:wght@400;600&family=Source+Code+Pro"><link data-pjax rel="stylesheet" href="/css/default.css"><style>body>.footer,body>.navbar,body>.section{opacity:0}</style><!--!--><!--!--><!--!--><!--!--><link rel="stylesheet" href="https://cdn.jsdelivr.net/npm/[email protected]/dist/css/lightgallery.min.css"><link rel="stylesheet" href="https://cdn.jsdelivr.net/npm/[email protected]/dist/css/justifiedGallery.min.css"><!--!--><!--!--><!--!--><!--!--><style>.pace{-webkit-pointer-events:none;pointer-events:none;-webkit-user-select:none;-moz-user-select:none;user-select:none}.pace-inactive{display:none}.pace .pace-progress{background:#3273dc;position:fixed;z-index:2000;top:0;right:100%;width:100%;height:2px}</style><script src="https://cdn.jsdelivr.net/npm/[email protected]/pace.min.js"></script><!--!--><!--!--><!-- hexo injector head_end start --><script>
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</script><!-- hexo injector head_end end --><meta name="generator" content="Hexo 7.3.0"></head><body class="is-2-column"><nav class="navbar navbar-main"><div class="container navbar-container"><div class="navbar-brand justify-content-center"><a class="navbar-item navbar-logo" href="/"><img src="/img/logo.svg" alt="raincity's blog" height="28"></a></div><div class="navbar-menu"><div class="navbar-start"><a class="navbar-item is-active" href="/">Home</a><a class="navbar-item" href="/archives">Archives</a><a class="navbar-item" href="/categories">Categories</a><a class="navbar-item" href="/tags">Tags</a><a class="navbar-item" href="/about">About</a></div><div class="navbar-end"><a class="navbar-item search" title="搜索" href="javascript:;"><i class="fas fa-search"></i></a></div></div></div></nav><section class="section"><div class="container"><div class="columns"><div class="column order-2 column-main is-8-tablet is-8-desktop is-8-widescreen"><div class="card"><article class="card-content article" role="article"><div class="article-meta is-size-7 is-uppercase level is-mobile"><div class="level-left"><span class="level-item"><time dateTime="2024-11-06T02:09:17.000Z" title="2024/11/6 10:09:17">2024-11-06</time>发表</span></div></div><p class="title is-3 is-size-4-mobile"><a class="link-muted" href="/2024/11/06/eulerian-circuits/">欧拉回路</a></p><div class="content"><p>网络上关于求欧拉回路的线性算法的资料普遍缺少证明。本文将通过分析欧拉回路的性质正向推导出这一算法。</p>
<h2 id="算法流程"><a href="#算法流程" class="headerlink" title="算法流程"></a>算法流程</h2><p>基本的定义可以参考 <a target="_blank" rel="noopener" href="https://www.cnblogs.com/alex-wei/p/basic_graph_theory.html">Alex_Wei 的博客</a>，本文不再赘述。</p>
<p>算法流程部分仅推导求无向图欧拉回路的算法，求有向图欧拉回路的算法的推导过程是类似的，更改一些对应术语即可。</p>
<p>显然图中的孤立点与欧拉回路无关，本小节假设要么图中不存在孤立点，要么整张图就是一个孤立点。</p>
<p><strong>性质 1</strong>：无向图 $G$ 是欧拉图，当且仅当图 $G$ 连通，且 $\forall u \in G$，$2 \mid d(u)$。</p>
<blockquote>
<p> 证明：必要性显然，下证充分性。</p>
<p> 考虑归纳法。$|E| = 0$ 时结论显然成立。假设 $|E| < m$ 时结论成立，下证 $|E| = m$ 时结论成立。</p>
<p> 显然图中至少存在一个环（否则图将会是一棵树，树的叶子节点度数为 $1$，矛盾）。删除环上的边后图可能分裂成若干个连通块，但每个连通块中至少存在一个环上的点，并且仍然满足 $\forall u \in G$，$2 \mid d(u)$。由归纳假设，每个连通块都存在欧拉回路，用环把这些欧拉回路串起来即可得到原图的欧拉回路。</p>
</blockquote>
<p><strong>性质 2</strong>：若无向图 $G$ 是欧拉图，则 $G$ 是边双连通图。</p>
<blockquote>
<p> 证明：性质 1 的证明直接导出了这一结论。</p>
</blockquote>
<p><strong>性质 3</strong>：无向图 $G$ 是半欧拉图，当且仅当图 $G$ 连通，且 $\sum_{u \in G} [2 \nmid d(u)] = 2$。并且，设满足 $2 \nmid d(u)$ 的两个点分别为 $s, t$，则 $G$ 的欧拉迹的两个端点一定为 $s, t$。</p>
<blockquote>
<p> 证明：必要性显然，下证充分性。</p>
<p> 在 $s, t$ 之间再连一条边，则图会变成一张欧拉图。把新图的欧拉回路中 $(s, t)$ 这条边删掉即可得到原图的一条欧拉迹。</p>
</blockquote>
<p>现在考虑构造一张欧拉图的欧拉回路或一张半欧拉图的欧拉迹。考虑递归构造，定义函数 $dfs(s)$ 返回 $s$ 所在的连通块的生成子图中一条 $s \rightsquigarrow s$ 的欧拉回路或一条 $s \rightsquigarrow t$ 的欧拉迹。</p>
<p>若整张图就是一个孤立点，则可以返回一条空途径 $s$。否则任取 $s$ 的一条邻边 $(s, v)$ 并删除，有以下两种情况：</p>
<ol>
<li>图变得不连通，则 $(s, v)$ 是原图的一条割边。由性质 2，我们要求的一定是一条欧拉迹。并且，$t$ 一定在 $v$ 所在的连通块，因为假设在原图加入 $(s, t)$ 这条边则原图会变得连通。因此可以返回 $dfs(s) + dfs(v)$。</li>
<li>图仍然连通。可以返回 $s + dfs(v)$。</li>
</ol>
<p>这样我们就得到了一个 $O(m (n + m))$ 的算法，瓶颈在于判断图是否连通。</p>
<p>我们希望避免判断图是否连通，即合并两种情况。实际上这是非常简单的。可以先调用 $dfs(v)$，设结果为 $a$，再调用 $dfs(s)$，设结果为 $b$，最后返回 $b + a$。这是因为如果图不连通，则 $dfs(s)$ 和 $dfs(v)$ 的过程互不干扰，否则 $dfs(v)$ 时会把图删空，$dfs(s)$ 会直接返回 $s$。</p>
<p>这样我们就得到了一个 $O(n + m)$ 的算法，不过实现时有一些需要注意的地方：</p>
<ul>
<li>删边可以用打标记 + 当前弧优化代替。</li>
<li>$dfs$ 并不需要真的返回一条途径，只需要不断向构造的途径里 prepend 新的点。这可以通过在 $dfs$ 过程中执行 append，最后执行一次 reverse 解决。</li>
<li>不断 $dfs(s)$ 其实很蠢，这相当于枚举 $s$ 的每条出边并执行 $dfs(v)$，回溯前再 append $s$。</li>
</ul>
<p>这样就得到了我们平时实现的算法。</p>
<h2 id="参考实现"><a href="#参考实现" class="headerlink" title="参考实现"></a>参考实现</h2><p>模板题：<a target="_blank" rel="noopener" href="https://uoj.ac/problem/117">UOJ117 欧拉回路</a>。本题要求输出边的顺序，只需要把回溯前 append $s$ 这一步改成每次调用 $dfs(v)$ 之后 append $(s, v)$ 这条边即可。</p>
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class="meta">#<span class="keyword">define</span> all(x) begin(x), end(x)</span></span><br><span class="line"><span class="meta">#<span class="keyword">define</span> len(x) int((x).size())</span></span><br><span class="line"></span><br><span class="line"><span class="keyword">using</span> ll = <span class="type">long</span> <span class="type">long</span>;</span><br><span class="line"><span class="keyword">using</span> db = <span class="type">long</span> <span class="type">double</span>;</span><br><span class="line"></span><br><span class="line"><span class="keyword">template</span> <<span class="keyword">typename</span> T></span><br><span class="line"><span class="function"><span class="keyword">inline</span> <span class="type">bool</span> <span class="title">ckmin</span><span class="params">(T &x, <span class="type">const</span> T &y)</span> </span>{</span><br><span class="line"> <span class="keyword">return</span> y < x <span class="built_in">and</span> (x = y, <span class="literal">true</span>);</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="keyword">template</span> <<span class="keyword">typename</span> T></span><br><span class="line"><span class="function"><span class="keyword">inline</span> <span class="type">bool</span> <span class="title">ckmax</span><span class="params">(T &x, <span class="type">const</span> T &y)</span> </span>{</span><br><span class="line"> <span class="keyword">return</span> x < y <span class="built_in">and</span> (x = y, <span class="literal">true</span>);</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="type">int</span> n, m;</span><br><span class="line"></span><br><span class="line"><span class="keyword">namespace</span> undirected {</span><br><span class="line"><span class="keyword">constexpr</span> <span class="type">int</span> N = <span class="number">1e5</span> + <span class="number">5</span>, M = <span class="number">4e5</span> + <span class="number">5</span>;</span><br><span class="line"></span><br><span class="line"><span class="type">int</span> tot = <span class="number">1</span>, hd[N], nxt[M], to[M], d[N];</span><br><span class="line"><span class="type">bool</span> vis[M];</span><br><span class="line">vector<<span class="type">int</span>> p;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="type">void</span> <span class="title">ae</span><span class="params">(<span class="type">int</span> u, <span class="type">int</span> v)</span> </span>{</span><br><span class="line"> ++d[u], ++d[v];</span><br><span class="line"> nxt[++tot] = hd[u], hd[u] = tot, to[tot] = v;</span><br><span class="line"> nxt[++tot] = hd[v], hd[v] = tot, to[tot] = u;</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="type">void</span> <span class="title">dfs</span><span class="params">(<span class="type">int</span> u)</span> </span>{</span><br><span class="line"> <span class="keyword">for</span> (<span class="type">int</span> &i = hd[u]; i > <span class="number">0</span>; i = nxt[i]) {</span><br><span class="line"> <span class="keyword">if</span> (vis[i / <span class="number">2</span>]) {</span><br><span class="line"> <span class="keyword">continue</span>;</span><br><span class="line"> }</span><br><span class="line"> vis[i / <span class="number">2</span>] = <span class="literal">true</span>;</span><br><span class="line"> <span class="type">int</span> tmp = i % <span class="number">2</span> == <span class="number">1</span> ? -(i / <span class="number">2</span>) : i / <span class="number">2</span>;</span><br><span class="line"> <span class="built_in">dfs</span>(to[i]);</span><br><span class="line"> p.<span class="built_in">push_back</span>(tmp);</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="type">void</span> <span class="title">proc</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="keyword">for</span> (<span class="type">int</span> i = <span class="number">1</span>; i <= m; ++i) {</span><br><span class="line"> <span class="type">int</span> u, v;</span><br><span class="line"> cin >> u >> v;</span><br><span class="line"> <span class="built_in">ae</span>(u, v);</span><br><span class="line"> }</span><br><span class="line"> <span class="type">int</span> s = <span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span> (<span class="type">int</span> i = <span class="number">1</span>; i <= n; ++i) {</span><br><span class="line"> <span class="keyword">if</span> (d[i] % <span class="number">2</span> == <span class="number">1</span>) {</span><br><span class="line"> cout << <span class="string">"NO\n"</span>;</span><br><span class="line"> <span class="keyword">return</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span> (d[i] > <span class="number">0</span>) {</span><br><span class="line"> s = i;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span> (s != <span class="number">0</span>) {</span><br><span class="line"> <span class="built_in">dfs</span>(s);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span> (<span class="built_in">len</span>(p) < m) {</span><br><span class="line"> cout << <span class="string">"NO\n"</span>;</span><br><span class="line"> <span class="keyword">return</span>;</span><br><span class="line"> }</span><br><span class="line"> cout << <span class="string">"YES\n"</span>;</span><br><span class="line"> <span class="keyword">for</span> (<span class="type">int</span> i = m; --i >= <span class="number">0</span>;) {</span><br><span class="line"> cout << p[i] << <span class="string">" \n"</span>[i == <span class="number">0</span>];</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="keyword">namespace</span> directed {</span><br><span class="line"><span class="keyword">constexpr</span> <span class="type">int</span> N = <span class="number">1e5</span> + <span class="number">5</span>, M = <span class="number">2e5</span> + <span class="number">5</span>;</span><br><span class="line"></span><br><span class="line"><span class="type">int</span> tot, hd[N], nxt[M], to[M], inD[N], outD[N];</span><br><span class="line">vector<<span class="type">int</span>> p;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="type">void</span> <span class="title">ae</span><span class="params">(<span class="type">int</span> u, <span class="type">int</span> v)</span> </span>{</span><br><span class="line"> ++inD[v], ++outD[u];</span><br><span class="line"> nxt[++tot] = hd[u], hd[u] = tot, to[tot] = v;</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="type">void</span> <span class="title">dfs</span><span class="params">(<span class="type">int</span> u)</span> </span>{</span><br><span class="line"> <span class="type">int</span> &i = hd[u];</span><br><span class="line"> <span class="keyword">while</span> (i > <span class="number">0</span>) {</span><br><span class="line"> <span class="type">int</span> tmp = i;</span><br><span class="line"> i = nxt[i];</span><br><span class="line"> <span class="built_in">dfs</span>(to[tmp]);</span><br><span class="line"> p.<span class="built_in">push_back</span>(tmp);</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="type">void</span> <span class="title">proc</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="keyword">for</span> (<span class="type">int</span> i = <span class="number">1</span>; i <= m; ++i) {</span><br><span class="line"> <span class="type">int</span> u, v;</span><br><span class="line"> cin >> u >> v;</span><br><span class="line"> <span class="built_in">ae</span>(u, v);</span><br><span class="line"> }</span><br><span class="line"> <span class="type">int</span> s = <span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span> (<span class="type">int</span> i = <span class="number">1</span>; i <= n; ++i) {</span><br><span class="line"> <span class="keyword">if</span> (inD[i] != outD[i]) {</span><br><span class="line"> cout << <span class="string">"NO\n"</span>;</span><br><span class="line"> <span class="keyword">return</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span> (inD[i] > <span class="number">0</span>) {</span><br><span class="line"> s = i;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span> (s != <span class="number">0</span>) {</span><br><span class="line"> <span class="built_in">dfs</span>(s);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span> (<span class="built_in">len</span>(p) < m) {</span><br><span class="line"> cout << <span class="string">"NO\n"</span>;</span><br><span class="line"> <span class="keyword">return</span>;</span><br><span class="line"> }</span><br><span class="line"> cout << <span class="string">"YES\n"</span>;</span><br><span class="line"> <span class="keyword">for</span> (<span class="type">int</span> i = m; --i >= <span class="number">0</span>;) {</span><br><span class="line"> cout << p[i] << <span class="string">" \n"</span>[i == <span class="number">0</span>];</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="type">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> cin.<span class="built_in">tie</span>(<span class="literal">nullptr</span>)-><span class="built_in">sync_with_stdio</span>(<span class="literal">false</span>);</span><br><span class="line"> cin.<span class="built_in">exceptions</span>(cin.failbit);</span><br><span class="line"></span><br><span class="line"> <span class="type">int</span> tc;</span><br><span class="line"> cin >> tc >> n >> m;</span><br><span class="line"> <span class="keyword">if</span> (tc == <span class="number">1</span>) {</span><br><span class="line"> undirected::<span class="built_in">proc</span>();</span><br><span class="line"> } <span class="keyword">else</span> {</span><br><span class="line"> directed::<span class="built_in">proc</span>();</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
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