給定一個由兩列「Identifier1」和「Identifier2」表示的無向圖,我們如何將彼此相關的識別碼進行分組並為它們分配唯一的群組ID?
可以透過將資料視為圖中的邊並遍歷所有邊來解決此問題遞歸地。
<code class="sql">WITH CTE_Idents AS ( SELECT Ident1 AS Ident FROM @T UNION SELECT Ident2 AS Ident FROM @T ), CTE_Pairs AS ( SELECT Ident1, Ident2 FROM @T WHERE Ident1 <> Ident2 UNION SELECT Ident2 AS Ident1, Ident1 AS Ident2 FROM @T WHERE Ident1 <> Ident2 ), CTE_Recursive AS ( SELECT CAST(CTE_Idents.Ident AS varchar(8000)) AS AnchorIdent , Ident1 , Ident2 , CAST(',' + Ident1 + ',' + Ident2 + ',' AS varchar(8000)) AS IdentPath , 1 AS Lvl FROM CTE_Pairs INNER JOIN CTE_Idents ON CTE_Idents.Ident = CTE_Pairs.Ident1 UNION ALL SELECT CTE_Recursive.AnchorIdent , CTE_Pairs.Ident1 , CTE_Pairs.Ident2 , CAST(CTE_Recursive.IdentPath + CTE_Pairs.Ident2 + ',' AS varchar(8000)) AS IdentPath , CTE_Recursive.Lvl + 1 AS Lvl FROM CTE_Pairs INNER JOIN CTE_Recursive ON CTE_Recursive.Ident2 = CTE_Pairs.Ident1 WHERE CTE_Recursive.IdentPath NOT LIKE CAST('%,' + CTE_Pairs.Ident2 + ',%' AS varchar(8000)) ), CTE_RecursionResult AS ( SELECT AnchorIdent, Ident1, Ident2 FROM CTE_Recursive ), CTE_CleanResult AS ( SELECT AnchorIdent, Ident1 AS Ident FROM CTE_RecursionResult UNION SELECT AnchorIdent, Ident2 AS Ident FROM CTE_RecursionResult ) SELECT CTE_Idents.Ident ,CASE WHEN CA_Data.XML_Value IS NULL THEN CTE_Idents.Ident ELSE CA_Data.XML_Value END AS GroupMembers ,DENSE_RANK() OVER(ORDER BY CASE WHEN CA_Data.XML_Value IS NULL THEN CTE_Idents.Ident ELSE CA_Data.XML_Value END ) AS GroupID FROM CTE_Idents CROSS APPLY ( SELECT CTE_CleanResult.Ident+',' FROM CTE_CleanResult WHERE CTE_CleanResult.AnchorIdent = CTE_Idents.Ident ORDER BY CTE_CleanResult.Ident FOR XML PATH(''), TYPE ) AS CA_XML(XML_Value) CROSS APPLY ( SELECT CA_XML.XML_Value.value('.', 'NVARCHAR(MAX)') ) AS CA_Data(XML_Value) WHERE CTE_Idents.Ident IS NOT NULL ORDER BY Ident;</code>
以上是我們如何在由兩列「Identifier1」和「Identifier2」所表示的無向圖中將相關識別碼分組,並為它們指派唯一的群組ID?的詳細內容。更多資訊請關注PHP中文網其他相關文章!