ACM资料 10.27KB

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资源文件列表:

并查集.zip 大约有1个文件

并查集.ppt 44.5KB

资源介绍:

ACM 地方都是法师东方时代

<html xmlns="http://www.w3.org/1999/xhtml"><meta charset="utf-8"><meta name="generator" content="pdf2htmlEX"><meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1"><link rel="stylesheet" href="/image.php?url=https://csdnimg.cn/release/download_crawler_static/css/base.min.css"><link rel="stylesheet" href="/image.php?url=https://csdnimg.cn/release/download_crawler_static/css/fancy.min.css"><link rel="stylesheet" href="/image.php?url=https://csdnimg.cn/release/download_crawler_static/536692/raw.css"><script src="/image.php?url=https://csdnimg.cn/release/download_crawler_static/js/compatibility.min.js"></script><script src="/image.php?url=https://csdnimg.cn/release/download_crawler_static/js/pdf2htmlEX.min.js"></script><script>try{pdf2htmlEX.defaultViewer = new pdf2htmlEX.Viewer({});}catch(e){}</script><div id="sidebar" style="display: none"><div id="outline"></div></div><div id="pf1" class="pf w0 h0" data-page-no="1"><div class="pc pc1 w0 h0"><img class="bi x0 y0 w1 h1" alt="" src="/image.php?url=https://csdnimg.cn/release/download_crawler_static/536692/bg1.jpg"><div class="c x0 y1 w2 h2"><div class="t m0 x1 h3 y2 ff1 fs0 fc0 sc0 ls0 ws0"> <span class="_ _0"> </span><span class="fc1"> </span></div><div class="t m0 x2 h4 y3 ff2 fs1 fc1 sc1 ls0 ws0">并查集</div><div class="t m0 x3 h5 y4 ff1 fs2 fc2 sc0 ls0 ws0">Y<span class="_ _1"></span>ali <span class="_ _2"> </span><span class="ff2">朱全民</span></div></div></div><div class="pi" data-data='{"ctm":[1.333333,0.000000,0.000000,1.333333,0.000000,0.000000]}'></div></div></html><div id="pf2" class="pf w0 h0" data-page-no="2"><div class="pc pc2 w0 h0"><img class="bi x0 y0 w1 h1" alt="" src="/image.php?url=https://csdnimg.cn/release/download_crawler_static/536692/bg2.jpg"><div class="c x0 y5 w3 h0"><div class="t m0 x1 h3 y6 ff1 fs0 fc1 sc0 ls0 ws0"> <span class="_ _3"> </span> </div></div><div class="c x0 y1 w2 h2"><div class="t m0 x4 h5 y7 ff2 fs2 fc2 sc2 ls0 ws0">分离集合</div><div class="t m0 x4 h6 y8 ff3 fs3 fc1 sc0 ls0 ws0"></div><div class="t m0 x5 h7 y9 ff2 fs4 fc1 sc0 ls0 ws0">在有的问题中<span class="_ _4"></span>，需要对不相交<span class="_ _4"></span>的集合<span class="_ _5"> </span><span class="ff1">(disjoint </span></div><div class="t m0 x5 h7 ya ff1 fs4 fc1 sc0 ls0 ws0">set)<span class="_ _6"> </span><span class="ff2">进行这样两种<span class="_ _4"></span>操作：</span></div><div class="t m0 x6 h8 yb ff1 fs5 fc1 sc0 ls0 ws0">–</div><div class="t m0 x7 h9 yc ff2 fs6 fc1 sc0 ls0 ws0">检索某元素属于哪个集合</div><div class="t m0 x6 h8 yd ff1 fs5 fc1 sc0 ls0 ws0">–</div><div class="t m0 x7 h9 ye ff2 fs6 fc1 sc0 ls0 ws0">合并两个集合</div><div class="t m0 x4 h6 yf ff3 fs3 fc1 sc0 ls0 ws0"></div><div class="t m0 x5 h7 y10 ff2 fs4 fc1 sc0 ls0 ws0">能够维护这两<span class="_ _4"></span>个操作的数据结<span class="_ _4"></span>构，我们称之为</div><div class="t m0 x5 h7 y11 ff2 fs4 fc1 sc0 ls0 ws0">并查集。</div></div></div><div class="pi" data-data='{"ctm":[1.333333,0.000000,0.000000,1.333333,0.000000,0.000000]}'></div></div><div id="pf3" class="pf w0 h0" data-page-no="3"><div class="pc pc3 w0 h0"><img class="bi x0 y0 w1 h1" alt="" src="/image.php?url=https://csdnimg.cn/release/download_crawler_static/536692/bg3.jpg"><div class="c x0 y5 w3 h0"><div class="t m0 x1 h3 y6 ff1 fs0 fc1 sc0 ls0 ws0"> <span class="_ _3"> </span> </div></div><div class="c x0 y1 w2 h2"><div class="t m0 x4 h5 y7 ff2 fs2 fc2 sc2 ls0 ws0">并查集的森林实现</div><div class="t m0 x4 h6 y8 ff3 fs3 fc1 sc0 ls0 ws0"></div><div class="t m0 x5 h7 y9 ff2 fs4 fc1 sc0 ls0 ws0">一般来说我们<span class="_ _4"></span>用森林的结构实<span class="_ _4"></span>现并查集</div><div class="t m0 x4 h6 y12 ff3 fs3 fc1 sc0 ls0 ws0"></div><div class="t m0 x5 h7 y13 ff2 fs4 fc1 sc0 ls0 ws0">在森林中，每<span class="_ _4"></span>棵树代表一个集<span class="_ _4"></span>合。用树根来表</div><div class="t m0 x5 h7 y14 ff2 fs4 fc1 sc0 ls0 ws0">示这个集合。</div><div class="t m0 x4 h6 y15 ff3 fs3 fc1 sc0 ls0 ws0"></div><div class="t m0 x5 h7 y16 ff2 fs4 fc1 sc0 ls0 ws0">合并操作：两<span class="_ _4"></span>个集合<span class="_ _5"> </span><span class="ff1">S1<span class="_ _5"> </span></span>、<span class="_ _5"> </span><span class="ff1">S2<span class="_ _5"> </span></span>合并，将其中</div><div class="t m0 x5 h7 y17 ff2 fs4 fc1 sc0 ls0 ws0">的一个树根作<span class="_ _4"></span>为另一个树根的<span class="_ _4"></span>子树即可。</div><div class="t m0 x4 h6 y18 ff3 fs3 fc1 sc0 ls0 ws0"></div><div class="t m0 x5 h7 y19 ff2 fs4 fc1 sc0 ls0 ws0">查找操作：对<span class="_ _4"></span>于一个元素<span class="_ _5"> </span><span class="ff1">u<span class="_ _5"> </span></span>的查找，顺着<span class="_ _5"> </span><span class="ff1">u<span class="_ _5"> </span></span>往</div><div class="t m0 x5 h7 y1a ff2 fs4 fc1 sc0 ls0 ws0">上找，直到线<span class="_ _4"></span>索到根节点，也<span class="_ _4"></span>就确定了<span class="_ _5"> </span><span class="ff1">u<span class="_ _5"> </span></span>所在</div><div class="t m0 x5 h7 y1b ff2 fs4 fc1 sc0 ls0 ws0">的集合。</div></div></div><div class="pi" data-data='{"ctm":[1.333333,0.000000,0.000000,1.333333,0.000000,0.000000]}'></div></div><div id="pf4" class="pf w0 h0" data-page-no="4"><div class="pc pc4 w0 h0"><img class="bi x0 y0 w1 h1" alt="" src="/image.php?url=https://csdnimg.cn/release/download_crawler_static/536692/bg4.jpg"><div class="c x0 y5 w3 h0"><div class="t m0 x1 h3 y6 ff1 fs0 fc1 sc0 ls0 ws0"> <span class="_ _3"> </span> </div></div><div class="c x0 y1 w2 h2"><div class="t m0 x4 h5 y7 ff2 fs2 fc2 sc2 ls0 ws0">两个优化</div><div class="t m0 x4 h6 y8 ff3 fs3 fc1 sc0 ls0 ws0"></div><div class="t m0 x5 h7 y9 ff2 fs4 fc1 sc0 ls0 ws0">启发式合并：</div><div class="t m0 x4 h7 y13 ff2 fs4 fc1 sc0 ls0 ws0"> <span class="_ _6"> </span>在合并集合<span class="_ _5"> </span><span class="ff1">S1<span class="_ _5"> </span></span>、<span class="_ _6"> </span><span class="ff1">S2<span class="_ _5"> </span></span>的时候，我们让较</div><div class="t m0 x5 h7 y14 ff2 fs4 fc1 sc0 ls0 ws0">小的树成为较大的<span class="_ _4"></span>树的子树。这里<span class="_ _4"></span>可以是深度、</div><div class="t m0 x5 h7 y1c ff2 fs4 fc1 sc0 ls0 ws0">节点个数等启发函<span class="_ _4"></span>数来比较树的大<span class="_ _4"></span>小。</div><div class="t m0 x4 h6 y18 ff3 fs3 fc1 sc0 ls0 ws0"></div><div class="t m0 x5 h7 y19 ff2 fs4 fc1 sc0 ls0 ws0">路径压缩：</div><div class="t m0 x4 h7 y1d ff2 fs4 fc1 sc0 ls0 ws0"> <span class="_ _6"> </span>我们在查找完<span class="_ _5"> </span><span class="ff1">u<span class="_ _5"> </span></span>至根节点的路径之后<span class="_ _4"></span>，</div><div class="t m0 x5 h7 y1e ff2 fs4 fc1 sc0 ls0 ws0">一般将这条路径上<span class="_ _4"></span>的所有节点的父<span class="_ _4"></span>节点都设为</div><div class="t m0 x5 h7 y1f ff2 fs4 fc1 sc0 ls0 ws0">根节点，这样可以<span class="_ _4"></span>大大减少之后的<span class="_ _4"></span>查找次数。</div></div></div><div class="pi" data-data='{"ctm":[1.333333,0.000000,0.000000,1.333333,0.000000,0.000000]}'></div></div><div id="pf5" class="pf w0 h0" data-page-no="5"><div class="pc pc5 w0 h0"><img class="bi x0 y0 w1 h1" alt="" src="/image.php?url=https://csdnimg.cn/release/download_crawler_static/536692/bg5.jpg"><div class="c x0 y5 w3 h0"><div class="t m0 x1 h3 y6 ff1 fs0 fc1 sc0 ls0 ws0"> <span class="_ _3"> </span> </div></div><div class="c x0 y1 w2 h2"><div class="t m0 x4 h5 y7 ff2 fs2 fc2 sc2 ls0 ws0">并查集的时间复杂度</div><div class="t m0 x4 h6 y8 ff3 fs3 fc1 sc0 ls0 ws0"></div><div class="t m0 x5 h7 y9 ff2 fs4 fc1 sc0 ls0 ws0">可以证明，经<span class="_ _4"></span>过启发式合并和<span class="_ _4"></span>路径压缩之后的</div><div class="t m0 x5 h7 ya ff2 fs4 fc1 sc0 ls0 ws0">并查集，执行<span class="_ _5"> </span><span class="ff1">m<span class="_ _5"> </span></span>次查找<span class="_ _4"></span>的复杂度为<span class="_ _5"> </span><span class="ff1">O(m<span class="ff4">α</span>(m))</span></div><div class="t m0 x4 h6 y15 ff3 fs3 fc1 sc0 ls0 ws0"></div><div class="t m0 x5 h7 y16 ff2 fs4 fc1 sc0 ls0 ws0">其中<span class="_ _5"> </span><span class="ff4">α<span class="ff1">(m)<span class="_ _5"> </span></span></span>是<span class="_ _5"> </span><span class="ff1">Ack<span class="_ _7"></span>ermann<span class="_ _5"> </span></span>函数的某个反函数，</div><div class="t m0 x5 h7 y17 ff2 fs4 fc1 sc0 ls0 ws0">你可以近似的<span class="_ _4"></span>认为它是小于<span class="_ _5"> </span><span class="ff1">5<span class="_ _5"> </span></span>的。所以并<span class="_ _4"></span>查集</div><div class="t m0 x5 h7 y20 ff2 fs4 fc1 sc0 ls0 ws0">的单次查找操<span class="_ _4"></span>作的时间复杂度<span class="_ _4"></span>也几乎是常数级</div><div class="t m0 x5 h7 y21 ff2 fs4 fc1 sc0 ls0 ws0">的。</div></div></div><div class="pi" data-data='{"ctm":[1.333333,0.000000,0.000000,1.333333,0.000000,0.000000]}'></div></div>

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