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name="description" content="前置知识 字符串基础操作。 引入 回文串,即 \(s&#x3D;s_{rev}\),也就是本身与反转相等。 想要找到字符串 \(s\) 中最大的回文子串,很容易想到的是一种复杂度为 \(O(n^2)\) 的朴素算法,也就是对于每一个点都向外扩展。 这里介绍一种复杂度为 \(O(n)\) 的最大回文子串算法,叫做 Manacher。"><meta property="og:type" content="article"><meta property="og:title" content="Manacher"><meta property="og:url" content="https://schtonn.github.io/blog/posts/manacher/index.html"><meta property="og:site_name" content="schtonn"><meta property="og:description" content="前置知识 字符串基础操作。 引入 回文串,即 \(s&#x3D;s_{rev}\),也就是本身与反转相等。 想要找到字符串 \(s\) 中最大的回文子串,很容易想到的是一种复杂度为 \(O(n^2)\) 的朴素算法,也就是对于每一个点都向外扩展。 这里介绍一种复杂度为 \(O(n)\) 的最大回文子串算法,叫做 Manacher。"><meta property="og:locale" content="en_US"><meta property="article:published_time" content="2020-06-14T14:47:39.000Z"><meta property="article:modified_time" content="2022-10-19T15:02:06.667Z"><meta property="article:author" content="Alex"><meta property="article:tag" content="string"><meta name="twitter:card" content="summary"><link rel="canonical" 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itemtype="http://schema.org/Organization"><meta itemprop="name" content="schtonn"></span><header class="post-header"><h1 class="post-title" itemprop="name headline"> Manacher</h1><div class="post-meta"><span class="post-meta-item"><span class="post-meta-item-icon"><i class="fa fa-calendar-o"></i></span> <span class="post-meta-item-text">Posted on</span> <time title="Created: 2020-Jun-14 22:47:39" itemprop="dateCreated datePublished" datetime="2020-06-14T22:47:39+08:00">2020-Jun-14</time></span><span class="post-meta-item"><span class="post-meta-item-icon"><i class="fa fa-calendar-check-o"></i></span> <span class="post-meta-item-text">Edited on</span> <time title="Modified: 2022-Oct-19 23:02:06" itemprop="dateModified" datetime="2022-10-19T23:02:06+08:00">2022-Oct-19</time></span><span class="post-meta-item"><span class="post-meta-item-icon"><i class="fa fa-comment-o"></i></span> <span class="post-meta-item-text">Valine:</span><a title="valine" href="/blog/posts/manacher/#valine-comments" itemprop="discussionUrl"><span class="post-comments-count valine-comment-count" data-xid="/blog/posts/manacher/" itemprop="commentCount"></span></a></span></div></header><div class="post-body" itemprop="articleBody"><h2 id="前置知识">前置知识</h2><p>字符串基础操作。</p><h2 id="引入">引入</h2><p>回文串,即 <span class="math inline">\(s=s_{rev}\)</span>,也就是本身与反转相等。</p><p>想要找到字符串 <span class="math inline">\(s\)</span> 中最大的回文子串,很容易想到的是一种复杂度为 <span class="math inline">\(O(n^2)\)</span> 的朴素算法,也就是对于每一个点都向外扩展。</p><p>这里介绍一种复杂度为 <span class="math inline">\(O(n)\)</span> 的最大回文子串算法,叫做 Manacher。</p><a id="more"></a><h2 id="最大回文子串">最大回文子串</h2><p>求最大回文子串,可以转化为求对于每一个位置 <code>i</code>,求以 <code>i</code> 为中心的回文串半径 <code>rad[i]</code>。</p><p>然而由于偶数长度的回文串没有实际的“中心”,所以我们需要把偶数长度全部转化为奇数长度。</p><p>方法是:在每一位字符之间添加一个填充符号,比如 <span class="math inline">\(\mathtt{\$ }\)</span>。如果原字符串是 <span class="math inline">\(\mathtt{schtonn}\)</span>,那么就转化为 <span class="math inline">\(\mathtt{\$s\$c\$h\$t\$o\$n\$n\$ }\)</span>。这样偶数长度的回文串 <span class="math inline">\(\mathtt{nn}\)</span>,就变成了奇数长度的 <span class="math inline">\(\mathtt{\$n\$n\$ }\)</span>。</p><h2 id="manacher">Manacher</h2><p>Manacher 算法维护了 <code>l</code> 和 <code>r</code>,记录<strong>右边界最靠右</strong>的回文子串的边界。</p><p>现在我们迭代 <code>i</code>,计算 <code>rad[i]</code>: - 如果 <code>i&gt;r</code>: 直接运行朴素算法,尝试拓展 <code>rad[i]</code>,直到不符合回文串性质或者碰到边界。 - 如果 <code>i&lt;=r</code> 这时可以尝试利用之前已计算出的结果。对于之前计算出的回文串,我们在<strong>大多数情况下</strong>认为 <code>rad[i]</code> 至少为之前计算的回文串的对应位置的值。通俗的说就是在一个大回文串内部的小回文串大小是相等的。</p><p>比如字符串 <span class="math inline">\(\mathtt{abacabaab}\)</span>,我们先将其处理成 <span class="math inline">\(\mathtt{a\$b\$a\$c\$a\$b\$a\$a\$b}\)</span>。</p><p>在第三位 <span class="math inline">\(\mathtt{b}\)</span> 前都是朴素算法:</p><p><span class="math display">\[\begin{aligned}\mathtt{\underline{\underline a\space\underline\$\space b\space\$\space a}\space\$\space c\space\$\space a\space\$\space b\space\$\space a\space\$\space a\space\$\space b}\\\mathtt{1\space1\space3\space0\space0\space0\space0\space0\space0\space0\space0\space0\space0\space0\space0\space0\space0}\end{aligned}\]</span></p><p>到了第四位 <span class="math inline">\(\mathtt{\$}\)</span>,算法认定这一位的 <code>rad</code> 至少为第二位的 <code>rad</code> 值,所以从 1 开始计算朴素算法。</p><p>经过一段时间后:</p><p><span class="math display">\[\begin{aligned}\mathtt{a\space\$\space b\space\$\space a\space\$\space c\space\$\space a\space\$\space \underline{b\space\$\space a\space\$\space a\space\$\space b}}\\\mathtt{1\space1\space3\space1\space2\space1\space7\space1\space2\space1\space4\space1\space2\space4\space2\space1\space0}\end{aligned}\]</span></p><p>这时应该计算最后一位的 <span class="math inline">\(\mathtt{b}\)</span> 了,我们根据算法看出 <code>rad</code> 至少是对应位置的 4。然而正确结果却小于四,这是因为这部分已经超出了回文串的边界。所以 <code>rad[i]</code> 应该至少为<em>到 <code>r</code> 的距离</em>和<em>对应位置 <code>rad</code> 值</em>中的最小值。</p><h2 id="代码">代码</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br></pre></td><td class="code"><pre><span class="line">$include<span class="string">"bits/stdc++.h"</span></span><br><span class="line"><span class="keyword">using</span> <span class="keyword">namespace</span> <span class="built_in">std</span>;</span><br><span class="line"><span class="keyword">int</span> len,rad[<span class="number">2000020</span>];</span><br><span class="line"><span class="keyword">int</span> maxid,maxr;</span><br><span class="line"><span class="keyword">char</span> fakes[<span class="number">1000010</span>],s[<span class="number">2000020</span>];</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span></span>&#123;</span><br><span class="line"> <span class="built_in">cin</span>&gt;&gt;fakes;</span><br><span class="line"> len=<span class="built_in">strlen</span>(fakes);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i&lt;=len*<span class="number">2</span>;i+=<span class="number">2</span>)&#123;</span><br><span class="line"> s[i]=<span class="string">'$'</span>;</span><br><span class="line"> s[i+<span class="number">1</span>]=fakes[(i+<span class="number">1</span>)/<span class="number">2</span>];</span><br><span class="line"> &#125;</span><br><span class="line"> len=<span class="built_in">strlen</span>(s);</span><br><span class="line"> <span class="keyword">int</span> l=<span class="number">0</span>,r=<span class="number">-1</span>;<span class="comment">//区间边界</span></span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i&lt;len;i++)&#123;</span><br><span class="line"> <span class="keyword">int</span> k;</span><br><span class="line"> <span class="keyword">if</span>(i&gt;r)k=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">else</span> k=min(rad[l+r-i],r-i);</span><br><span class="line"> <span class="keyword">while</span>(<span class="number">0</span>&lt;=i-k &amp;&amp; i+k&lt;len &amp;&amp; s[i-k]==s[i+k])&#123;</span><br><span class="line"> k++;</span><br><span class="line"> &#125;</span><br><span class="line"> rad[i]=k--;</span><br><span class="line"> <span class="keyword">if</span>(i+k&gt;r)&#123;</span><br><span class="line"> l=i-k;</span><br><span class="line"> r=i+k;</span><br><span class="line"> &#125;</span><br><span class="line"> <span class="keyword">if</span>(rad[i]&gt;maxr)&#123;</span><br><span class="line"> maxid=i;</span><br><span class="line"> maxr=rad[i];</span><br><span class="line"> &#125;</span><br><span class="line"> &#125;</span><br><span class="line"> <span class="keyword">int</span> ans=rad[maxid]<span class="number">-1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=maxid-ans;i&lt;=maxid+ans;i++)&#123;</span><br><span class="line"> <span class="keyword">if</span>(s[i]!=<span class="string">'$'</span>)<span class="built_in">cout</span>&lt;&lt;s[i];</span><br><span class="line"> &#125;</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure><h2 id="我踩的坑">我踩的坑</h2><ul><li>加上了 <span class="math inline">\(\mathtt{\$ }\)</span> 之后,数组空间需要开大两倍。</li></ul></div><div><ul class="post-copyright"><li class="post-copyright-author"> <strong>Post author:</strong> Alex</li><li class="post-copyright-link"> <strong>Post link:</strong> <a href="https://schtonn.github.io/blog/posts/manacher/" title="Manacher">https://schtonn.github.io/blog/posts/manacher/</a></li><li class="post-copyright-license"> 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