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图论，动态规划，二进制基础知识。
引入
我们都知道欧拉回路，也就是经过每条边正好一遍的路径。
哈密尔顿回路和欧拉回路很像，但是需要经过每个点正好一遍。
欧拉回路有稳定的解法，然而哈密尔顿回路是 NPC 的，没有多项式时间解法。
我们使用动态规划来解决哈密尔顿回路问题。"><meta property="og:type" content="article"><meta property="og:title" content="哈密尔顿回路"><meta property="og:url" content="https://schtonn.github.io/blog/posts/hamilton/index.html"><meta property="og:site_name" content="schtonn"><meta property="og:description" content="前置知识
图论，动态规划，二进制基础知识。
引入
我们都知道欧拉回路，也就是经过每条边正好一遍的路径。
哈密尔顿回路和欧拉回路很像，但是需要经过每个点正好一遍。
欧拉回路有稳定的解法，然而哈密尔顿回路是 NPC 的，没有多项式时间解法。
我们使用动态规划来解决哈密尔顿回路问题。"><meta property="og:locale" content="en_US"><meta property="article:published_time" content="2020-06-14T14:03:13.000Z"><meta property="article:modified_time" content="2022-10-19T15:02:06.667Z"><meta property="article:author" content="Alex"><meta property="article:tag" content="graph"><meta property="article:tag" content="dp"><meta name="twitter:card" content="summary"><link rel="canonical" href="https://schtonn.github.io/blog/posts/hamilton/"><script id="page-configurations">CONFIG.page={sidebar:"",isHome:!1,isPost:!0,lang:"en"}</script><title>哈密尔顿回路 | schtonn</title><noscript><style>.sidebar-inner,.use-motion .brand,.use-motion .collection-header,.use-motion .comments,.use-motion .menu-item,.use-motion .pagination,.use-motion .post-block,.use-motion .post-body,.use-motion .post-header{opacity:initial}.use-motion .site-subtitle,.use-motion .site-title{opacity:initial;top:initial}.use-motion .logo-line-before i{left:initial}.use-motion .logo-line-after i{right:initial}</style></noscript></head><body itemscope itemtype="http://schema.org/WebPage"><div class="container use-motion"><div class="headband"></div><header class="header" itemscope itemtype="http://schema.org/WPHeader"><div class="header-inner"><div class="site-brand-container"><div class="site-nav-toggle"><div class="toggle" aria-label="Toggle navigation bar"><span class="toggle-line toggle-line-first"></span><span class="toggle-line toggle-line-middle"></span><span class="toggle-line toggle-line-last"></span></div></div><div class="site-meta"><a href="/blog/" class="brand" rel="start"><span class="logo-line-before"><i></i></span><h1 class="site-title">schtonn</h1><span class="logo-line-after"><i></i></span></a><p class="site-subtitle" itemprop="description">schtonn</p></div><div class="site-nav-right"><div class="toggle popup-trigger"></div></div></div><nav class="site-nav"><ul id="menu" class="menu"><li class="menu-item menu-item-home"><a href="/blog/" rel="section"><i 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itemprop="description" content="blog"></span><script type="text/javascript" src="/blog/js/md5.js"></script><script></script><script>document.oncopy=function(e){window.event&&(e=window.event);try{var t=e.srcElement;return"INPUT"==t.tagName&&"text"==t.type.toLowerCase()||"TEXTAREA"==t.tagName}catch(e){return!1}}</script><span hidden itemprop="publisher" itemscope itemtype="http://schema.org/Organization"><meta itemprop="name" content="schtonn"></span><header class="post-header"><h1 class="post-title" itemprop="name headline"> 哈密尔顿回路</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 22:03:13" itemprop="dateCreated datePublished" datetime="2020-06-14T22:03:13+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/hamilton/#valine-comments" itemprop="discussionUrl"><span class="post-comments-count valine-comment-count" data-xid="/blog/posts/hamilton/" 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>图论，动态规划，二进制基础知识。</p><h2 id="引入">引入</h2><p>我们都知道欧拉回路，也就是经过每条边正好一遍的路径。</p><p>哈密尔顿回路和欧拉回路很像，但是需要经过每个点正好一遍。</p><p>欧拉回路有稳定的解法，然而哈密尔顿回路是 <code>NPC</code> 的，没有多项式时间解法。</p><p>我们使用动态规划来解决哈密尔顿回路问题。</p><a id="more"></a><h2 id="思路">思路</h2><p>想要解决回路的问题，首先要解决路径的问题。若有图 <span class="math inline">\(G(n,m)\)</span>，需要找出图中的一条哈密尔顿路径。</p><p>对于一条长度大于一的哈密尔顿路径，它一定是由一个更短的路径和一条连通边走过来的，所以我们定义数组 <code>F</code>，<code>F[{S}][p]</code> 表示经过了集合 <span class="math inline">\(S\)</span> 中的所有点，最后到达点 <code>t</code> 有没有路径。</p><p>如果 <code>F[{s}][p]</code> 为真，那么一定有 <code>F[{s}-p][q]</code> 为真且 <span class="math inline">\(q\rightarrow p\)</span> 有边，也就是有一个没有 <code>p</code> 的成立集合，并且最终能加入 <code>p</code>。</p><p>初始值设为，对于图中每一个点 <code>i</code>，<code>F[{i}][i]=true</code>。最终寻找：<span class="math inline">\(\vee^i_{i\in n} F[\{n\}][i]\)</span>（经过所有点，最终到达任意一点）。</p><p>此时我们就得到了状态转移方程：</p><p><span class="math display">\[F[\{S\}][p](p\in S)=\sum^q_{q\in S}F[\{S\}-p][q]\cdot Edge[q][p]\]</span></p><p>通过枚举 <code>S</code>，<code>q</code> 和 <code>p</code>，就可以求出哈密尔顿路径了。</p><h2 id="状态压缩">状态压缩</h2><p>此时，聪明的小朋友就会问了：你一个数组的下标怎么表示集合？</p><p>这时就要请出我们的大法师：状态压缩了。我们把本应是集合的数据压缩成一个下标。</p><p>运用二进制，把在集合中的点标记为一，不在集合中的标记为零，这样集合就转成下标了。</p><p>比如一共有三个点，其中点一，点二在集合中，那么我们就用 <span class="math inline">\((011)_2\)</span> 来表示。简直聪明至极！</p><p>作为一个n个点的图，<code>S</code> 可以以十进制顺序枚举，从 <span class="math inline">\((00\cdots 0)_2\)</span> 到 <span class="math inline">\((11\cdots 1)_2\)</span>，这样确保了前置条件是完整的。内里嵌套两层循环枚举 <code>q</code> 和 <code>p</code>。</p><p>作为二进制，当然要有专门用在二进制上的操作。</p><h3 id="按位操作">按位操作</h3><p>有几种不同的按位二进制操作：</p><ol type="1"><li><p>按位与</p><p>符号：<code>&</code></p><p>两个都是一就得一，否则得零。</p> <span class="math inline">\((01011101)_2\&(10101011)_2=(00001001)_2\)</span></li><li><p>按位或</p><p>符号：<code>|</code></p><p>只要有一就得一，否则得零。</p> <span class="math inline">\((01000101)_2|(10001011)_2=(11001111)_2\)</span></li><li><p>按位异或</p><p>符号：<code>^</code></p><p>两个不同就得一，否则得零。</p><p>注意和幂区分。</p><p>（<span class="math inline">\(\LaTeX\)</span>中打不出来这个符号，用另一个符号代替。）</p> <span class="math inline">\((11011010)_2\oplus(10101010)_2=(01110000)_2\)</span></li><li><p>左移，右移</p><p>符号：<code><<</code>和<code>>></code></p><p>顾名思义，就是把二进制位移动。边界用0代替。</p><p><span class="math inline">\((10101010)_2<<3=(01010000)_2\)</span></p></li></ol><h3 id="高级操作">高级操作</h3><ol type="1"><li><p>如果我们要判断集合S中第i位是不是1，那就用这个：<code>S&(1<<(i-1))</code>。</p></li><li><p>将集合S中已知为1的第i位变为0：<code>S^(1<<(i-1))</code>。</p></li></ol><h3 id="核心代码">核心代码</h3><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></pre></td><td class="code"><pre><span class="line"><span class="keyword">for</span>(<span class="keyword">int</span> s=<span class="number">1</span>;s<(<span class="number">1</span><<n);++s){</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> p=<span class="number">1</span>;p<=n;++p){</span><br><span class="line"> <span class="keyword">if</span>(s&(<span class="number">1</span><<(p<span class="number">-1</span>))==<span class="number">0</span>)<span class="keyword">continue</span>;<span class="comment">//判断p是否为S中元素</span></span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> q=<span class="number">1</span>;q<=n;++q){</span><br><span class="line"> <span class="keyword">if</span>(s&(<span class="number">1</span><<(q<span class="number">-1</span>))==<span class="number">0</span>)<span class="keyword">continue</span>;<span class="comment">//判断q是否为S中元素</span></span><br><span class="line"> <span class="keyword">if</span>(!e[q][p])<span class="keyword">continue</span>;<span class="comment">//判断有没有q到p的边</span></span><br><span class="line"> <span class="keyword">if</span>(dp[s^(<span class="number">1</span><<(p<span class="number">-1</span>))][q]){<span class="comment">//判断dp[{s}-p][q]是否为真</span></span><br><span class="line"> dp[s][p]=<span class="literal">true</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure><h2 id="扩展">扩展</h2><h3 id="记录">记录</h3><p>有些题目要求输出哈密尔顿路径。其实并不难，根本上只需要加上这一句：<code>g[s][p]=q;</code>，通过g数组记录路径。</p><h3 id="回路">回路</h3><p>对于哈密尔顿回路，有一个方法是让路径只允许从点1开始，最后看能不能走回点1，也就是初始值设为 <code>F[{1}][1]=true</code>，最终判定 <span class="math inline">\(F[\{n\}][i]\)</span>。</p><h3 id="代码">代码</h3><p>输出回路的完整代码</p><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><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string">"bits/stdc++.h"</span></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">const</span> <span class="keyword">int</span> N=<span class="number">1</span><<<span class="number">20</span>;</span><br><span class="line"><span class="keyword">int</span> n,m,e[<span class="number">25</span>][<span class="number">25</span>];</span><br><span class="line"><span class="keyword">bool</span> dp[N][<span class="number">25</span>];<span class="comment">//dp[i][j]表示状态为i，最后到达j点有没有路径</span></span><br><span class="line"><span class="keyword">int</span> g[N][<span class="number">25</span>];</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span></span>{</span><br><span class="line"> <span class="built_in">cin</span>>>n>>m;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=m;i++){</span><br><span class="line"> <span class="keyword">int</span> u,v;</span><br><span class="line"> <span class="built_in">cin</span>>>u>>v;</span><br><span class="line"> e[u][v]=<span class="literal">true</span>;</span><br><span class="line"> }</span><br><span class="line"> dp[<span class="number">1</span>][<span class="number">1</span>]=<span class="literal">true</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> s=<span class="number">1</span>;s<(<span class="number">1</span><<n);++s){</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> p=<span class="number">1</span>;p<=n;++p){</span><br><span class="line"> <span class="keyword">if</span>(s&(<span class="number">1</span><<(p<span class="number">-1</span>))==<span class="number">0</span>)<span class="keyword">continue</span>;<span class="comment">//判断p是否为S中元素</span></span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> q=<span class="number">1</span>;q<=n;++q){</span><br><span class="line"> <span class="keyword">if</span>(s&(<span class="number">1</span><<(q<span class="number">-1</span>))==<span class="number">0</span>)<span class="keyword">continue</span>;<span class="comment">//判断q是否为S中元素</span></span><br><span class="line"> <span class="keyword">if</span>(!e[q][p])<span class="keyword">continue</span>;<span class="comment">//判断有没有q到p的边</span></span><br><span class="line"> <span class="keyword">if</span>(dp[s^(<span class="number">1</span><<(p<span class="number">-1</span>))][q]){<span class="comment">//判断dp[{s}-p][q]是否为真</span></span><br><span class="line"> dp[s][p]=<span class="literal">true</span>;</span><br><span class="line"> g[s][p]=q;</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">int</span> s=(<span class="number">1</span><<n)<span class="number">-1</span>;</span><br><span class="line"> <span class="keyword">bool</span> flag=<span class="number">0</span>;</span><br><span class="line"> <span class="built_in">stack</span><<span class="keyword">int</span>>output;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;i++){</span><br><span class="line"> <span class="keyword">if</span>(dp[s][i]&&e[i][<span class="number">1</span>]){</span><br><span class="line"> <span class="keyword">int</span> p=i;</span><br><span class="line"> output.push(p);</span><br><span class="line"> <span class="keyword">do</span>{</span><br><span class="line"> output.push(g[s][p]);</span><br><span class="line"> <span class="keyword">int</span> t=p;</span><br><span class="line"> p=g[s][p];</span><br><span class="line"> s-=(<span class="number">1</span><<(t<span class="number">-1</span>));</span><br><span class="line"> }<span class="keyword">while</span>(s!=<span class="number">1</span>);</span><br><span class="line"> flag=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">break</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span>(!flag){</span><br><span class="line"> <span class="built_in">cout</span><<<span class="string">"No Answer"</span><<<span class="built_in">endl</span>;</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">while</span>(!output.empty()){</span><br><span class="line"> <span class="built_in">cout</span><<output.top()<<<span class="string">' '</span>;</span><br><span class="line"> output.pop();</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p></p></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/hamilton/" title="哈密尔顿回路">https://schtonn.github.io/blog/posts/hamilton/</a></li><li class="post-copyright-license"> <strong>Copyright Notice:</strong> All articles in this 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