Question

我们有一个由边表表示的有向图。如何在纯SQL中检测周期？

CREATE TABLE edges(id integer primary key identity, from_node int, to_node int);
CREATE NONCLUSTERED INDEX index_edges_of2 ON edges(from_node);

INSERT INTO edges(from_node,to_node) VALUES(1,2),(2,3),(3,1);

Answer 1

WITH cte AS (
    SELECT CAST('.1.' AS varchar(1000)) AS gpath, 1 AS current_node, 0 AS Cycle

    UNION ALL

    SELECT CAST(CONCAT(gpath, '.', e.to_node, '.') AS varchar(1000)) AS gpath, e.to_node AS current_node, 0 AS cycle
    FROM cte
    JOIN edges AS e ON e.from_node = cte.current_node
    WHERE CHARINDEX(CONCAT('.',e.to_node,'.'),cte.gpath)=0 AND cte.Cycle=0

    UNION ALL

    SELECT CAST(CONCAT(gpath, '.', e.to_node, '.') AS varchar(1000)) AS gpath, e.to_node AS current_node, 1 AS cycle
    FROM cte
    JOIN edges AS e ON e.from_node = cte.current_node
    WHERE CHARINDEX(CONCAT('.',e.to_node,'.'),cte.gpath)>0 AND cte.Cycle=0
)
SELECT * FROM cte;

gpath current_node周期

.1。 1 0

.1..2。 2 0

.1..2..3。 3 0

.1..2..3..1。 1 1

Answer 2

解决方案是递归CTE。但是，要使其正常工作，您需要保留已访问节点的列表。 SQL Server对此没有很好的解决方案（例如数组），因此您需要使用字符串操作。

以下将在图中列出循环：

with cte as (
      select from_node, to_node, 
             convert(varchar(max), concat(',', from_node, ',', to_node, ',')) as nodes, 1 as lev, 
             (case when from_node = to_node then 1 else 0 end) as has_cycle
      from edges e
      union all
      select cte.from_node, e.to_node,
             convert(varchar(max), concat(cte.nodes, e.to_node, ',')), lev + 1,
             (case when cte.nodes like concat('%,', e.to_node, ',%') then 1 else 0 end) as has_cycle
      from cte join
           edges e
           on e.from_node = cte.to_node
      where cte.has_cycle = 0 
     )
select *
from cte
where has_cycle = 1;

Here是db <>小提琴。

sql在有向图中检测周期

2 个答案: