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?
这个问题可以通过将数据视为图中的边并遍历所有边来解决递归地。
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;
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