给定一个由两列“Identifier1”和“Identifier2”表示的无向图,我们如何对彼此相关的标识符进行分组并为它们分配唯一的组ID?
可以通过将数据视为图中的边并遍历所有边来解决此问题递归地。
<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>
以上是我们如何在由两列'Identifier1”和'Identifier2”表示的无向图中对相关标识符进行分组,并为它们分配唯一的组ID?的详细内容。更多信息请关注PHP中文网其他相关文章!