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 引入
 并查集是一种快到爆炸的集合算法，可以进行两项基本操作：合并两个集合（并）、查询两个参数是否在一个集合内（查）。这也是它名字的由来。">
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          并查集
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        <h3 id="前置知识">前置知识</h3>
<p>哈哈，简单到爆，没有。</p>
<h3 id="引入">引入</h3>
<p>并查集是一种快到爆炸的集合算法，可以进行两项基本操作：合并两个集合（并）、查询两个参数是否在一个集合内（查）。这也是它名字的由来。 <a id="more"></a></p>
<h3 id="速度">速度</h3>
<p>他有多快呢？ <span class="math inline">\(O(*log n)\)</span> <span class="math inline">\(*log\)</span>有多可怕： |<span class="math inline">\(n\)</span>|<span class="math inline">\(*log n\)</span>| |-|-| |<span class="math inline">\((1,2]\)</span>|<span class="math inline">\(1\)</span>| |<span class="math inline">\((2,4]\)</span>|<span class="math inline">\(2\)</span>| |<span class="math inline">\((4,16]\)</span>|<span class="math inline">\(3\)</span>| |<span class="math inline">\((16,65536]\)</span>|<span class="math inline">\(4\)</span>| |<span class="math inline">\((65536,2^{65536}]\)</span>|<span class="math inline">\(5\)</span>| 所以n是<span class="math inline">\(2^{65536}\)</span>的时候复杂度还只有5。 <a href="https://sites.google.com/site/largenumbers/home/appendix/a/numbers/265536" target="_blank" rel="noopener" title="哈哈！不跨越GREATWALL打不开"><span class="math inline">\(2^{65536}\)</span></a>有多大，我把他拷进来的时候整个电脑卡死了。我不得不强制重启，然后重新写一遍这段。他有19729位。想通了吧？</p>
<h3 id="如何实现">如何实现</h3>
<p>这么高端的算法，是怎么实现的呢？ 其实</p>
<p>它的本质就是一个数组<span class="math inline">\(f[]\)</span>，和一个函数<span class="math inline">\(getf()\)</span> <span class="math inline">\(f[]\)</span>存的实际上就是几棵树。 <span class="math inline">\(f[i]\)</span>就是<span class="math inline">\(i\)</span>的父亲。 <span class="math inline">\(getf(i)\)</span>做的操作就是递归顺着<span class="math inline">\(f[i]\)</span>找<span class="math inline">\(i\)</span>所在的树的根。 <span class="math inline">\(getf()\)</span>代码： <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></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">getf</span><span class="params">(<span class="keyword">int</span> x)</span></span>&#123;</span><br><span class="line">	<span class="keyword">if</span>(f[x]==x)<span class="keyword">return</span> x;</span><br><span class="line">	<span class="keyword">else</span> <span class="keyword">return</span> getf(f[x]);</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure> <span class="math inline">\(...\)</span></p>
<p>那这个算法就很低端了<span class="math inline">\(...\)</span></p>
<p>那还讲个鬼啊<span class="math inline">\(...\)</span></p>
<p>所以</p>
<h3 id="超级优化">超级优化</h3>
<p>我们首先随机造出一些操作： <figure class="highlight plain"><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">10</span><br><span class="line">merge 1 5</span><br><span class="line">merge 5 2</span><br><span class="line">merge 1 3</span><br><span class="line">check 2 3</span><br><span class="line">merge 3 4</span><br><span class="line">merge 6 7</span><br><span class="line">check 1 7</span><br><span class="line">merge 7 8</span><br><span class="line">merge 8 9</span><br><span class="line">merge 1 9</span><br><span class="line">check 7 4</span><br></pre></td></tr></table></figure> 其中，merge代表合并，check代表查询。 如果按照刚才所的算法，那么在第一次查询之前，就会出来这样的森林：<img src="https://img-blog.csdnimg.cn/20200105142700396.png" alt="Insert mother fucker" /> 到最后，就形成了这样一个繁杂的森林，要找到一个点的根，就需要走很长一段路。这就拉长了时间。 <img src="https://img-blog.csdnimg.cn/20200105142531778.png" alt="Insert mother fucker" /> 为了缩减时间，超级优化就出现了：路径压缩。 路径压缩其实也很简单：在<span class="math inline">\(getf()\)</span>查找根节点的同时，把自己也链接到根节点上，使得树的深度不超过2。 代码： <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></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">getf</span><span class="params">(<span class="keyword">int</span> x)</span></span>&#123;</span><br><span class="line">	<span class="keyword">if</span>(f[x]==x)<span class="keyword">return</span> x;</span><br><span class="line">	<span class="keyword">else</span> <span class="keyword">return</span> f[x]=getf(f[x]);</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure> 发现没有，和之前的代码相比，只改了一个地方，就让时间大大压缩。 这时我们在模拟一下。 第一次合并：<img src="https://img-blog.csdnimg.cn/20200105143342729.png" alt="Insert mother fucker" /> 第二次合并时，首先寻找2,5两个节点的根节点，2的根就是2，5的根是1，于是直接把2链接到1上。 <img src="https://img-blog.csdnimg.cn/20200105144059847.png" alt="Insert mother fucker" /> 第三次，第四次合并把3链接到了1上，然后把4顺着3也链接到了1上，第五次连接了6和7。 <img src="https://img-blog.csdnimg.cn/20200105143532966.png" alt="Insert mother fucker" /> 第七次第八次链接成了<span class="math inline">\(9\rightarrow8\rightarrow7\rightarrow6\)</span>一长串，然后经过路径压缩都链接到6上了。 <img src="https://img-blog.csdnimg.cn/20200105143644622.png" alt="Insert mother fucker" /> 最后一次，把9和1链接起来了，这时深度又超过了2，一下还压缩不下去，不过没关系，查询的时候就会把它压缩的。<img src="https://img-blog.csdnimg.cn/20200105143847598.png" alt="Insert mother fucker" /> 比如查询7和4的时候就会分别寻找7和4的根节点，一路递归找上去的时候就直接把路径压缩好了，除了8还链接在6上，其他全部链接到1上了。 <img src="https://img-blog.csdnimg.cn/20200105144933595.png" alt="Insert mother fucker" /> 多么有趣啊！</p>
<h3 id="代码">代码</h3>
<p>自己想去吧，核心代码和思路都给出来了。 有一个巨大的坑，就是<span class="math inline">\(f[i]\)</span>要预设成<span class="math inline">\(i\)</span>，不然会爆炸。 加油！<del>克服恐惧的最好办法就是面对恐忄</del>快去写吧！</p>

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