{"id":54389,"date":"2019-12-25T00:00:00","date_gmt":"2019-12-24T21:00:00","guid":{"rendered":"https:\/\/prohoster.info\/blog\/blog_prohoster\/indeksiruemoe-binarnoe-derevo"},"modified":"2020-02-18T14:02:23","modified_gmt":"2020-02-18T11:02:23","slug":"indeksiruemoe-binarnoe-derevo","status":"publish","type":"post","link":"https:\/\/prohoster.info\/cs\/blog\/administrirovanie\/indeksiruemoe-binarnoe-derevo","title":{"rendered":"Indexovan\u00e9 bin\u00e1rn\u00ed strom","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"<p><img decoding=\"async\" alt=\"Indexovan\u00e9 bin\u00e1rn\u00ed strom\" src=\"\/wp-content\/uploads\/2019\/12\/8ceb987e007db02de04d29f33185e8ec.jpg\" style=\"display:block;margin: 0 auto;\" \/><\/p>\n<p>Narazil jsem na \u00fakol n\u00e1sleduj\u00edc\u00edho typu. Je t\u0159eba implementovat kontejner pro ukl\u00e1d\u00e1n\u00ed dat, kter\u00fd zajist\u00ed n\u00e1sleduj\u00edc\u00ed funk\u010dnost: <\/p>\n<p><\/p>\n<ul>\n<li>vlo\u017eit nov\u00fd prvek<\/li>\n<li>odstranit prvek podle po\u0159adov\u00e9ho \u010d\u00edsla<\/li>\n<li>z\u00edskat prvek podle po\u0159adov\u00e9ho \u010d\u00edsla<\/li>\n<li>data jsou uchov\u00e1v\u00e1na ve se\u0159azen\u00e9m stavu<\/li>\n<\/ul>\n<p><noindex><a rel=\"nofollow\" name=\"habracut\"><\/a><\/noindex><\/p>\n<p>Data se neust\u00e1le p\u0159id\u00e1vaj\u00ed a odstra\u0148uj\u00ed, struktura by m\u011bla zaji\u0161\u0165ovat rychlou reak\u010dn\u00ed dobu. Zpo\u010d\u00e1tku jsem se pokusil implementovat takovou v\u011bc pomoc\u00ed standardn\u00edch kontejner\u016f z <strong>std<\/strong>. Tato cesta se neuk\u00e1zala jako \u00fasp\u011b\u0161n\u00e1 a do\u0161el jsem k z\u00e1v\u011bru, \u017ee je pot\u0159eba n\u011bco realizovat s\u00e1m. Jako prvn\u00ed m\u011b napadlo vyu\u017e\u00edt bin\u00e1rn\u00ed vyhled\u00e1vac\u00ed strom. Proto\u017ee spl\u0148uje po\u017eadavek na rychl\u00e9 vkl\u00e1d\u00e1n\u00ed, odstra\u0148ov\u00e1n\u00ed a uchov\u00e1v\u00e1n\u00ed dat v se\u0159azen\u00e9m stavu. Zb\u00fdvalo jen vymyslet, jak indexovat v\u0161echny prvky a p\u0159epo\u010d\u00edt\u00e1vat indexy, kdy\u017e se strom m\u011bn\u00ed.<\/p>\n<p><\/p>\n<pre><code class=\"cpp\">struct node_s {    \n    data_t data;\n\n    uint64_t weight; \/\/ v\u00e1ha uzlu\n\n    node_t *left;\n    node_t *right;\n\n    node_t *parent;\n};<\/code><\/pre>\n<p><\/p>\n<p>V \u010dl\u00e1nku bude v\u00edce obr\u00e1zk\u016f a teorie ne\u017e k\u00f3du. K\u00f3d si bude mo\u017en\u00e9 prohl\u00e9dnout na odkazu n\u00ed\u017ee.<\/p>\n<p><\/p>\n<h2 id=\"ves\">Hmotnost<\/h2>\n<p><\/p>\n<p>Abychom toho dos\u00e1hli, strom pro\u0161el drobnou modifikac\u00ed, p\u0159idala se dodate\u010dn\u00e1 informace o <strong>v\u00e1ze<\/strong> uzlu. V\u00e1ha uzlu je <strong>po\u010det potomk\u016f dan\u00e9ho uzlu<\/strong> + <strong>1<\/strong> (v\u00e1ha jednotkov\u00e9ho prvku).<\/p>\n<p><\/p>\n<p>Funkce pro z\u00edsk\u00e1n\u00ed v\u00e1hy uzlu:<\/p>\n<p><\/p>\n<pre><code class=\"cpp\">uint64_t bntree::get_child_weight(node_t *node) {\n    if (node) {\n        return node-&gt;weight;\n    }\n\n    return 0;\n}<\/code><\/pre>\n<p><\/p>\n<p>U listu je tedy v\u00e1ha rovna <strong>0<\/strong>.<\/p>\n<p><\/p>\n<p>D\u00e1le se p\u0159esuneme k n\u00e1zorn\u00e9mu zobrazen\u00ed p\u0159\u00edkladu takov\u00e9ho stromu. <strong>\u010cernou<\/strong> barvou bude v n\u011bm zobrazen kl\u00ed\u010d uzlu (hodnota zobrazen\u00e1 nebude, proto\u017ee v tom nen\u00ed pot\u0159eba), <strong>\u010dervenou<\/strong> \u2014 v\u00e1ha uzlu, <strong>zelenou<\/strong> \u2014 index uzlu.<\/p>\n<p><\/p>\n<p>Kdy\u017e je strom pr\u00e1zdn\u00fd, jeho v\u00e1ha je 0. P\u0159id\u00e1me do n\u011bj ko\u0159enov\u00fd prvek:<\/p>\n<p>\n<img decoding=\"async\" alt=\"Indexovan\u00e9 bin\u00e1rn\u00ed strom\" src=\"\/wp-content\/uploads\/2019\/12\/2d4145039daee26582910556a40d2a5c.jpg\" style=\"display:block;margin: 0 auto;\" \/><\/p>\n<p>V\u00e1ha stromu se st\u00e1v\u00e1 1, v\u00e1ha ko\u0159enov\u00e9ho prvku 1. V\u00e1ha ko\u0159enov\u00e9ho prvku je v\u00e1hou stromu.<\/p>\n<p><\/p>\n<p>P\u0159id\u00e1me je\u0161t\u011b n\u011bkolik prvk\u016f:<\/p>\n<p>\n<img decoding=\"async\" alt=\"Indexovan\u00e9 bin\u00e1rn\u00ed strom\" src=\"\/wp-content\/uploads\/2019\/12\/803cc4a65aa3d8a2fcb20fe325351cfa.jpg\" style=\"display:block;margin: 0 auto;\" \/><br \/>\n<img decoding=\"async\" alt=\"Indexovan\u00e9 bin\u00e1rn\u00ed strom\" src=\"\/wp-content\/uploads\/2019\/12\/0aa12f41b822b4bb1e9fadb7564dc461.jpg\" style=\"display:block;margin: 0 auto;\" \/><br \/>\n<img decoding=\"async\" alt=\"Indexovan\u00e9 bin\u00e1rn\u00ed strom\" src=\"\/wp-content\/uploads\/2019\/12\/8566df9404037e92f53b315bc3304016.jpg\" style=\"display:block;margin: 0 auto;\" \/><br \/>\n<img decoding=\"async\" alt=\"Indexovan\u00e9 bin\u00e1rn\u00ed strom\" src=\"\/wp-content\/uploads\/2019\/12\/1847e6ddffdb28d0a6d9a4949ebb5f80.jpg\" style=\"display:block;margin: 0 auto;\" \/><\/p>\n<p>Poka\u017ed\u00e9, kdy\u017e se p\u0159id\u00e1v\u00e1 nov\u00fd prvek, sestupujeme skrz uzly dol\u016f a zvy\u0161ujeme po\u010d\u00edtadlo v\u00e1hy ka\u017ed\u00e9ho pro\u0161l\u00e9ho uzlu. P\u0159i vytvo\u0159en\u00ed nov\u00e9ho uzlu mu je p\u0159i\u0159azena v\u00e1ha <strong>1<\/strong>. Pokud uzel s takov\u00fdm kl\u00ed\u010dem ji\u017e existuje, p\u0159ep\u00ed\u0161eme hodnotu a p\u016fjdeme zp\u011bt k ko\u0159enu sm\u011brem nahoru, abychom zru\u0161ili zm\u011bny v\u00e1hy u v\u0161ech uzl\u016f, kter\u00e9 jsme pro\u0161li.<br \/>\nPokud doch\u00e1z\u00ed k odstran\u011bn\u00ed uzlu, sestupujeme dol\u016f a dekrementujeme v\u00e1hy pro\u0161l\u00fdch uzl\u016f. <\/p>\n<p><\/p>\n<h2 id=\"indeksy\">Indexy<\/h2>\n<p><\/p>\n<p>P\u0159ejdeme k tomu, jak indexovat uzly. Uzly neukl\u00e1daj\u00ed sv\u016fj vlastn\u00ed index, ten se vypo\u010d\u00edt\u00e1v\u00e1 na z\u00e1klad\u011b hmotnosti uzl\u016f. Kdyby si uchov\u00e1valy sv\u016fj index, bylo by pot\u0159eba <strong>O(n)<\/strong> hodn\u011b \u010dasu na aktualizaci index\u016f v\u0161ech uzl\u016f po ka\u017ed\u00e9 zm\u011bn\u011b stromu.<br \/>\nP\u0159ejdeme k n\u00e1zorn\u00e9mu zn\u00e1zorn\u011bn\u00ed. Na\u0161e stromov\u00e1 struktura je pr\u00e1zdn\u00e1, p\u0159id\u00e1me do n\u00ed prvn\u00ed uzel:<\/p>\n<p>\n<img decoding=\"async\" alt=\"Indexovan\u00e9 bin\u00e1rn\u00ed strom\" src=\"\/wp-content\/uploads\/2019\/12\/8a3b176f318cd077b1cf50ddf232e0da.jpg\" style=\"display:block;margin: 0 auto;\" \/><\/p>\n<p>Prvn\u00ed uzel m\u00e1 index <strong>0<\/strong>, a nyn\u00ed nast\u00e1vaj\u00ed dva p\u0159\u00edpady. V prvn\u00edm se zm\u011bn\u00ed index ko\u0159enov\u00e9ho prvku, ve druh\u00e9m se nezm\u011bn\u00ed.<\/p>\n<p>\n<img decoding=\"async\" alt=\"Indexovan\u00e9 bin\u00e1rn\u00ed strom\" src=\"\/wp-content\/uploads\/2019\/12\/cb863be8f42385d3bbb2f46700a8cff3.jpg\" style=\"display:block;margin: 0 auto;\" \/><\/p>\n<p>Lev\u00e9 podstrom ko\u0159ene v\u00e1\u017e\u00ed 1.<\/p>\n<p><\/p>\n<p>Druh\u00fd p\u0159\u00edpad:<\/p>\n<p>\n<img decoding=\"async\" alt=\"Indexovan\u00e9 bin\u00e1rn\u00ed strom\" src=\"\/wp-content\/uploads\/2019\/12\/35eec59a81fc8b056c7e91daa3ee508e.jpg\" style=\"display:block;margin: 0 auto;\" \/><\/p>\n<p>Index ko\u0159ene se nezm\u011bnil, proto\u017ee hmotnost jeho lev\u00e9ho podstromu z\u016fstala 0.<\/p>\n<p><\/p>\n<p>Jak se po\u010d\u00edt\u00e1 index uzlu, je to hmotnost jeho lev\u00e9ho podstromu plus \u010d\u00edslo p\u0159edan\u00e9 od rodi\u010de. Co to je za \u010d\u00edslo?, To je \u010d\u00edta\u010d index\u016f, ze za\u010d\u00e1tku je roven <strong>0<\/strong>, proto\u017ee ko\u0159en nem\u00e1 rodi\u010de. D\u00e1le v\u0161e z\u00e1vis\u00ed na tom, zda sestupujeme k lev\u00e9mu d\u00edt\u011bti nebo prav\u00e9mu. Pokud k lev\u00e9mu, pak se k \u010d\u00edta\u010di nic nep\u0159id\u00e1v\u00e1. Pokud k prav\u00e9mu, p\u0159id\u00e1v\u00e1me index aktu\u00e1ln\u00edho uzlu.<\/p>\n<p>\n<img decoding=\"async\" alt=\"Indexovan\u00e9 bin\u00e1rn\u00ed strom\" src=\"\/wp-content\/uploads\/2019\/12\/d328174370ef52c646d8689cce977302.jpg\" style=\"display:block;margin: 0 auto;\" \/><\/p>\n<p>Nap\u0159\u00edklad, jak se po\u010d\u00edt\u00e1 index prvku s kl\u00ed\u010dem 8 (prav\u00e9 d\u00edt\u011b ko\u0159ene). To je \u201eIndex ko\u0159ene\u201c + \u201ev\u00e1ha lev\u00e9ho podstromu uzlu s kl\u00ed\u010dem 8\u201c + \u201e1\u201c == 3 + 2 + 1 == <strong>6<\/strong><br \/>\nIndexem prvku s kl\u00ed\u010dem 6 bude \u201eIndex ko\u0159ene\u201c + 1 == 3 + 1 == <strong>4<\/strong><\/p>\n<p><\/p>\n<p>T\u00edm p\u00e1dem, pro z\u00edsk\u00e1n\u00ed a odstran\u011bn\u00ed prvku podle indexu je pot\u0159eba \u010das <strong>O(log n)<\/strong>, proto\u017ee abychom z\u00edskali po\u017eadovan\u00fd prvek, mus\u00edme nejprve naj\u00edt (sestoupit od ko\u0159ene k tomuto prvku).<\/p>\n<p><\/p>\n<h2 id=\"glubina\">Hloubka<\/h2>\n<p><\/p>\n<p>Na z\u00e1klad\u011b hmotnosti lze tak\u00e9 vypo\u010d\u00edtat hloubku stromu. Ta je nezbytn\u00e1 pro vyv\u00e1\u017een\u00ed.<br \/>\nK tomu je t\u0159eba hmotnost aktu\u00e1ln\u00edho uzlu zaokrouhlit na prvn\u00ed \u010d\u00edslo ve stejn\u00e9 mocnin\u011b 2, kter\u00e9 je v\u011bt\u0161\u00ed nebo rovno dan\u00e9 hmotnosti, a vz\u00edt od n\u011bj bin\u00e1rn\u00ed logaritmus. Tak z\u00edsk\u00e1me hloubku stromu za p\u0159edpokladu, \u017ee je vyv\u00e1\u017een\u00e9. Strom se vyva\u017euje po vlo\u017een\u00ed nov\u00e9ho prvku. Teorii o tom, jak vyva\u017eovat stromy, popisovat nebudu. V zdrojov\u00fdch k\u00f3dech je uvedena funkce vyva\u017eov\u00e1n\u00ed.<\/p>\n<p><\/p>\n<p>K\u00f3d pro p\u0159evod hmotnosti na hloubku.<\/p>\n<p><\/p>\n<pre><code class=\"cpp\">\/*\n * \u0412\u043e\u0437\u0432\u0440\u0430\u0449\u0430\u0435\u0442 \u043f\u0435\u0440\u0432\u043e\u0435 \u0447\u0438\u0441\u043b\u043e \u0432 \u0441\u0442\u0435\u043f\u0435\u043d\u0438 2, \u043a\u043e\u0442\u043e\u0440\u043e\u0435 \u0431\u043e\u043b\u044c\u0448\u0435 \u0438\u043b\u0438 \u0440\u043e\u0432\u043d\u043e x\n *\/\nuint64_t bntree::cpl2(uint64_t x) {\n    x = x - 1;\n    x = x | (x &gt;&gt; 1);\n    x = x | (x &gt;&gt; 2);\n    x = x | (x &gt;&gt; 4);\n    x = x | (x &gt;&gt; 8);\n    x = x | (x &gt;&gt; 16);\n    x = x | (x &gt;&gt; 32);\n\n    return x + 1;\n}\n\n\/*\n * \u0414\u0432\u043e\u0438\u0447\u043d\u044b\u0439 \u043b\u043e\u0433\u0430\u0440\u0438\u0444\u043c \u043e\u0442 \u0447\u0438\u0441\u043b\u0430\n *\/\nlong bntree::ilog2(long d) {\n    int result;\n    std::frexp(d, &amp;result);\n    return result - 1;\n}\n\n\/*\n * \u0412\u0435\u0441 \u043a \u0433\u043b\u0443\u0431\u0438\u043d\u0435\n *\/\nuint64_t bntree::weight_to_depth(node_t *p) {\n    if (p == NULL) {\n        return 0;\n    }\n\n    if (p-&gt;weight == 1) {\n        return 1;\n    } else if (p-&gt;weight == 2) {\n        return 2;\n    }\n\n    return this-&gt;ilog2(this-&gt;cpl2(p-&gt;weight));\n}<\/code><\/pre>\n<p><\/p>\n<h2 id=\"itogi\">Z\u00e1v\u011bry<\/h2>\n<p><\/p>\n<ul>\n<li>Vlo\u017een\u00ed nov\u00e9ho prvku prob\u00edh\u00e1 za <strong>O(log n)<\/strong><\/li>\n<li>Odstran\u011bn\u00ed prvku podle po\u0159adov\u00e9ho \u010d\u00edsla prob\u00edh\u00e1 za <strong>O(log n)<\/strong><\/li>\n<li>Z\u00edsk\u00e1n\u00ed prvku podle po\u0159adov\u00e9ho \u010d\u00edsla prob\u00edh\u00e1 za <strong>O(log n)<\/strong><\/li>\n<\/ul>\n<p><\/p>\n<p>Rychlost\u00ed <strong>O(log n)<\/strong> plat\u00edme za to, \u017ee v\u0161echna data jsou uchov\u00e1v\u00e1na ve se\u0159azen\u00e9 podob\u011b. <\/p>\n<p><\/p>\n<p>Kde m\u016f\u017ee takov\u00e1 struktura b\u00fdt u\u017eite\u010dn\u00e1 \u2014 nev\u00edm. Jen \u00fakol, abych se je\u0161t\u011b jednou pod\u00edval, jak stromy funguj\u00ed. D\u011bkuji za pozornost.<\/p>\n<p><\/p>\n<h2 id=\"ssylki\">Odkazy<\/h2>\n<p><\/p>\n<ul>\n<li><noindex><a rel=\"nofollow\" href=\"https:\/\/github.com\/dvjdjvu\/bntree\">Zdrojov\u00fd k\u00f3d stromu<\/a><\/noindex><\/li>\n<\/ul>\n<p><\/p>\n<p>V projektu se nach\u00e1zej\u00ed testovac\u00ed \u00fadaje pro kontrolu rychlosti pr\u00e1ce. Strom se zapl\u0148uje <strong>1000000<\/strong> prvky. A doch\u00e1z\u00ed k sekven\u010dn\u00edmu odstra\u0148ov\u00e1n\u00ed, vkl\u00e1d\u00e1n\u00ed a z\u00edsk\u00e1v\u00e1n\u00ed prvk\u016f <strong>1000000<\/strong> kr\u00e1t. To znamen\u00e1 <strong>3000000<\/strong> operac\u00ed. V\u00fdsledek se uk\u00e1zal jako docela dobr\u00fd ~ 8 sekund.<\/p>\n<p>Zdroj: <a content=\"nofollow\" rel=\"nofollow\" href=\"https:\/\/habr.com\/ru\/post\/481372\/\">habr.com<\/a><\/p>","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"<p>\u041f\u043e\u043f\u0430\u043b\u0430\u0441\u044c \u043c\u043d\u0435 \u0437\u0430\u0434\u0430\u0447\u0430 \u0441\u043b\u0435\u0434\u0443\u044e\u0449\u0435\u0433\u043e \u0432\u0438\u0434\u0430. \u041d\u0435\u043e\u0431\u0445\u043e\u0434\u0438\u043c\u043e \u0440\u0435\u0430\u043b\u0438\u0437\u043e\u0432\u0430\u0442\u044c \u043a\u043e\u043d\u0442\u0435\u0439\u043d\u0435\u0440 \u0445\u0440\u0430\u043d\u0435\u043d\u0438\u044f \u0434\u0430\u043d\u043d\u044b\u0445 \u043e\u0431\u0435\u0441\u043f\u0435\u0447\u0438\u0432\u0430\u044e\u0449\u0438\u0439 \u0441\u043b\u0435\u0434\u0443\u044e\u0449\u0438\u0439 \u0444\u0443\u043d\u043a\u0446\u0438\u043e\u043d\u0430\u043b: 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\u0441\u0442\u0440\u0443\u043a\u0442\u0443\u0440\u0430 \u0434\u043e\u043b\u0436\u043d\u0430 \u043e\u0431\u0435\u0441\u043f\u0435\u0447\u0438\u0432\u0430\u0442\u044c \u0431\u044b\u0441\u0442\u0440\u0443\u044e \u0441\u043a\u043e\u0440\u043e\u0441\u0442\u044c \u0440\u0430\u0431\u043e\u0442\u044b. \u0421\u043d\u0430\u0447\u0430\u043b\u0430 \u043f\u044b\u0442\u0430\u043b\u0441\u044f \u0440\u0435\u0430\u043b\u0438\u0437\u043e\u0432\u0430\u0442\u044c \u0442\u0430\u043a\u0443\u044e \u0432\u0435\u0449\u044c \u0438\u0441\u043f\u043e\u043b\u044c\u0437\u0443\u044f \u0441\u0442\u0430\u043d\u0434\u0430\u0440\u0442\u043d\u044b\u0435 \u043a\u043e\u043d\u0442\u0435\u0439\u043d\u0435\u0440\u044b \u0438\u0437 std. \u042d\u0442\u043e\u0442 \u043f\u0443\u0442\u044c \u043d\u0435 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