在SQL Server中实现最短路径搜索的解决方法建站知
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开始这是去年的问题了,今天在整理邮件的时候才发现这个问题,感觉顶有意思的,特记录下来。
在表RelationGraph中,有三个字段(ID,Node,RelatedNode),其中Node和RelatedNode两个字段描述两个节点的连接关系;现在要求,找出从节点"p"至节点"j",最短路径(即经过的节点最少)。
图1.解析:了能够更好的描述表RelationGraph中字段Node和 RelatedNode的关系,我在这里特意使用一个图形来描述,如图2.
图2.
在图2,可清晰的看出各个节点直接如何相连,也可以清楚的看出节点"p"至节点"j"的的几种可能路径。
从上面可以看出第2种可能路径,经过的节点最少。
为了解决开始的问题,我参考了两种方法,
第1网seo优化趋势方法是,
参考单源最短路径算法:
图3.第2方法是,
针对第1种方法的改进,就是采用多源点方法,这里就是以节点"p"和节点"j"为中心向外层扩展,直到两圆外切点,如图4. :
图4.实现:在接下来,我就描述在SQL Server中,如何实现。当然我这里采用的前面说的第2种方法,以"P"和"J"为始点像中心外层层扩展。
这里提供有表RelactionGraph的create& Insert数据的脚本:
复制代码 代码如下:
use TestDB
go
if object_id('RelactionGraph') Is not null drop table RelactionGraph
create table RelactionGraph(ID int identity,Item nvarchar(50),RelactionItem nvarchar(20),constraint PK_RelactionGraph primary key(ID))
go
create nonclustered index IX_RelactionGraph_Item on RelactionGraph(Item) include(RelactionItem)
create nonclustered index IX_RelactionGraph_RelactionItem on RelactionGraph(RelactionItem) include(Item)
go
insert into RelactionGraph (Item, RelactionItem ) values
('a','b'),('a','c'),('a','d'),('a','e'),
('b','f'),('b','g'),('b','h'),
('c','i'),('c','j'),
('f','k'),('f','l'),
('k','o'),('k','p'),
('o','i'),('o','l')
go
编写一个存储过程up_GetPath复制代码 代码如下:
use TestDB go --Procedure: if object_id('up_GetPath') Is not null Drop proc up_GetPath go create proc up_GetPath ( @Node nvarchar(50), @RelatedNode nvarchar(50) ) As set nocount on declare @level smallint =1, --当前搜索的深度 @MaxLevel smallint=100, --最大可搜索深度 @Node_WhileFlag bit=1, --以@Node作为中心进行搜索时候,作为能否循环搜索的标记 @RelatedNode_WhileFlag bit=1 --以@RelatedNode作为中心进行搜索时候,作为能否循环搜索的标记 --如果直接找到两个Node存在直接关系就直接返回 if Exists(select 1 from RelationGraph where (Node=@Node And RelatedNode=@RelatedNode) or (Node=@RelatedNode And RelatedNode=@Node) ) or @Node=@RelatedNode begin select convert(nvarchar(2000),@Node + ' --> '+ @RelatedNode) As RelationGraphPath,convert(smallint,0) As StopCount return end -- if object_id('tempdb..#1') Is not null Drop Table #1 --临时表#1,存储的是以@Node作为中心向外扩展的各节点数据 if object_id('tempdb..#2') Is not null Drop Table #2 --临时表#2,存储的是以@RelatedNode作为中心向外扩展的各节点数据 create table #1( Node nvarchar(50),--相对源点 RelatedNode nvarchar(50), --相对目标 Level smallint --深度 ) create table #2(Node nvarchar(50),RelatedNode nvarchar(50),Level smallint) insert into #1 ( Node, RelatedNode, Level ) select Node, RelatedNode, @level from RelationGraph a where a.Node =@Node union --正向:以@Node作为源查询 select RelatedNode, Node, @level from RelationGraph a where a.RelatedNode = @Node --反向:以@Node作为目标进行查询 set @Node_WhileFlag=sign(@@rowcount) insert into #2 ( Node, RelatedNode, Level ) select Node, RelatedNode, @level from RelationGraph a where a.Node =@RelatedNode union --正向:以@RelatedNode作为源查询 select RelatedNode, Node, @level from RelationGraph a where a.RelatedNode = @RelatedNode --反向:以@RelatedNode作为目标进行查询 set @RelatedNode_WhileFlag=sign(@@rowcount) --如果在表RelationGraph中找不到@Node 或 @RelatedNode 数据,就直接跳过后面的While过程 if not exists(select 1 from #1) or not exists(select 1 from #2) begin goto While_Out end while not exists(select 1 from #1 a inner join #2 b on b.RelatedNode=a.RelatedNode) --判断是否出现切点 and (@Node_WhileFlag|@RelatedNode_WhileFlag)>0 --判断是否能搜索 And @level<@MaxLevel --控制深度 begin if @Node_WhileFlag >0 begin insert into #1 ( Node, RelatedNode, Level ) --正向 select a.Node,a.RelatedNode,@level+1 From RelationGraph a where exists(select 1 from #1 where RelatedNode=a.Node And L企业网站建设evel=@level) And Not exists(select 1 from #1 where Node=a.Node) union --反向 select a.RelatedNode,a.Node,@level+1 From RelationGraph a where exists(select 1 from #1 where RelatedNode=a.RelatedNode And Level=@level) And Not exists(select 1 from #1 where Node=a.RelatedNode) set @Node_WhileFlag=sign(@@rowcount) end if @RelatedNode_WhileFlag >0 begin insert into #2 ( Node, RelatedNode, Level ) --正向 select a.Node,a.RelatedNode,@level+1 From RelationGraph a where exists(select 1 from #2 where RelatedNode=a.Node And Level=@level) And Not exists(select 1 from #2 where Node=a.Node) union --反向 select a.RelatedNode,a.Node,@level+1 From RelationGraph a where exists(select 1 from #2 where RelatedNode=a.RelatedNode And Level=@level) And Not exists(select 1 from #2 where Node=a.RelatedNode) set @RelatedNode_WhileFlag=sign(@@rowcount) end select @level+=1 end While_Out: --下面是构造返回的结果路径 if object_id('tempdb..#Path1') Is not null Drop Table #Path1 if object_id('tempdb..#Path2') Is not null Drop Table #Path2 ;with cte_path1 As ( select a.Node,a.RelatedNode,Level,convert(nvarchar(2000),a.Node+' -> '+a.RelatedNode) As RelationGraphPath,Convert(smallint,1) As PathLevel From #1 a where exists(select 1 from #2 where RelatedNode=a.RelatedNode) union all select b.Node,a.RelatedNode,b.Level,convert(nvarchar(2000),b.Node+' -> '+a.RelationGraphPath) As RelationGraphPath ,Convert(smallint,a.PathLevel+1) As PathLevel from cte_path1 a inner join #1 b on b.RelatedNode=a.Node and b.Level=a.Level-1 ) select * Into #Path1 from cte_path1 ;with cte_path2 As ( select a.Node,a.RelatedNode,Level,convert(nvarchar(2000),a.Node) As RelationGraphPath,Convert(smallint,1) As PathLevel From #2 a where exists(select 1 from #1 where RelatedNode=a.RelatedNode) union all select b.Node,a.RelatedNode,b.Level,convert(nvarchar(2000),a.RelationGraphPath+' -> '+b.Node) As RelationGraphPath ,Convert(smallint,a.PathLevel+1) from cte_path2 a inner join #2 b on b.RelatedNode=a.Node and b.Level=a.Level-1 ) select * Into #Path2 from cte_path2 ;with cte_result As ( select a.RelationGraphPath+' -> '+b.RelationGraphPath As RelationGraphPath,a.PathLevel+b.PathLevel -1 As StopCount,rank() over(order by a.PathLevel+b.PathLevel) As Result_row From #Path1 a inner join #Path2 b on b.RelatedNode=a.RelatedNode and b.Level=1 where a.Level=1 ) select distinct RelationGraphPath,StopCount From cte_result where Result_row=1 go
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