https://algorist.com/algowiki_v2/index.php?action=history&feed=atom&title=TADM2E_6.19 2026-04-30T20:39:44Z Revision history for this page on the wiki MediaWiki 1.28.0 https://algorist.com/algowiki_v2/index.php?title=TADM2E_6.19&diff=899&oldid=prev 2020-07-23T09:37:09Z

<p>6.19 alternative solution</p> <table class="diff diff-contentalign-left" data-mw="interface"> <col class='diff-marker' /> <col class='diff-content' /> <col class='diff-marker' /> <col class='diff-content' /> <tr style='vertical-align: top;' lang='en'> <td colspan='2' style="background-color: white; color:black; text-align: center;">← Older revision</td> <td colspan='2' style="background-color: white; color:black; text-align: center;">Revision as of 09:37, 23 July 2020</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l2" >Line 2:</td> <td colspan="2" class="diff-lineno">Line 2:</td></tr> <tr><td class='diff-marker'>&#160;</td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'>&#160;</td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"></td></tr> <tr><td class='diff-marker'>&#160;</td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>After running the algorithm we need to search main diagonal for minimum amount, which is additional O(n).</div></td><td class='diff-marker'>&#160;</td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>After running the algorithm we need to search main diagonal for minimum amount, which is additional O(n).</div></td></tr> <tr><td colspan="2">&#160;</td><td class='diff-marker'>+</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr> <tr><td colspan="2">&#160;</td><td class='diff-marker'>+</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">----</ins></div></td></tr> <tr><td colspan="2">&#160;</td><td class='diff-marker'>+</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">=== Alternative solution ===</ins></div></td></tr> <tr><td colspan="2">&#160;</td><td class='diff-marker'>+</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">&lt;pre&gt;</ins></div></td></tr> <tr><td colspan="2">&#160;</td><td class='diff-marker'>+</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">1. Traverse all the vertices of the graph and check if they have edges pointing to themselves, if so, return the weight of the minimal loop&#160; # O(n*m)</ins></div></td></tr> <tr><td colspan="2">&#160;</td><td class='diff-marker'>+</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">2. Use the Floyd-Warshall algorithm to find the shortest distances between any two vertices of the graph&#160; # O(n^3) </ins></div></td></tr> <tr><td colspan="2">&#160;</td><td class='diff-marker'>+</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">3. Traversing the distance table obtained in the previous step, return the minimum value of D[i][j] + D[j][i]&#160; # O(n^2)</ins></div></td></tr> <tr><td colspan="2">&#160;</td><td class='diff-marker'>+</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">&lt;/pre&gt;</ins></div></td></tr> <tr><td colspan="2">&#160;</td><td class='diff-marker'>+</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">--[[User:Bkarpov96|Bkarpov96]] ([[User talk:Bkarpov96|talk]]) 09:37, 23 July 2020 (UTC)</ins></div></td></tr> </table>

Bkarpov96 https://algorist.com/algowiki_v2/index.php?title=TADM2E_6.19&diff=382&oldid=prev 2015-12-21T19:24:58Z

<p>Created page with &quot;Simple application of Floyd-Warshall algorithm for directed graphs O(n^3). After running the algorithm we need to search main diagonal for minimum amount, which is additional...&quot;</p> <p><b>New page</b></p><div>Simple application of Floyd-Warshall algorithm for directed graphs O(n^3).<br /> <br /> After running the algorithm we need to search main diagonal for minimum amount, which is additional O(n).</div>

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