其实我也在研究同样的问题。不幸的是，我不认为可以找到一个非常有效的解决方案-至少不容易使用SQL。只需删除“distinct”和自消除查询，以观察工作集变得有多大。也就是说，下面的解决方案将起作用。

drop table if exists foobar;
drop function if exists addset(v int[], a int);
/* our vertices table */
create table foobar (
   src int,
   dst int
);

/* Create a small function to treat an array like a set, 
   not strictly necessary but convenient */
create function addset(v int[], a int) returns int[]
as $$
begin
    return (select array_agg(e order by e) 
                   from (select unnest(v || a) as e) f);
end
$$ language plpgsql;

/* fill our table with vertices, note the ordering of each value */
insert into foobar (src, dst) 
     values (1,2), (1,3), (2,3),  (3,4), (4,5), (6,7), (6,8);
/* use a recursive query to extend the sets */
with recursive foo_union (v) as (
    select array[src, dst] as v from foobar /* starter sets */
    union all
    /* join self to original array; i can use a CTE as a 'starter',
       but that is not necessary here */
    select addset(v, dst) as v from foo_union u, foobar f
        where src = any(v) and not dst = any(v)
 ) select distinct v from foo_union a where not exists (
     /* eliminate the many overlapping results */
     select * from foo_union b where b.v @> a.v and b.v != a.v
 );

但同样，这在更大的数据集上是完全不切实际的;任何其它解决方案都需要迭代。

这段纯sql代码在5分钟内将大约35000条记录分组（8个内核/32 gb内存）。好好享受

--table with RELATIONS, idea is to place every related item in a bucket
create table RELATIONS
(
    bucket int,        -- initially 0
    bucketsub int,    -- initially 0
    relnr1 float,    
    relnr2 float    -- relnr1 = a, relnr2 = b means a and b are related
)

create table ##BucketRelnrs ( relnr float ); --table functions as temp list
declare @bucket int; 
declare @bucketsub int;
declare @nrOfUpdates int;
declare @relnr float;

set @bucket=0;
set @relnr=0;
WHILE @relnr>=0 and @bucket<50000 --to prevent the while loop from hanging.
BEGIN
    set @bucket = @bucket+1
    set @bucketsub=1;

    set @relnr = (select isnull(min(relnr1),-1) from RELATIONS where bucket=0); --pick the smallest relnr that has not been assigned a bucket yet
    set @nrOfUpdates = (select count(*) from RELATIONS where bucket=0 and (relnr1=@relnr or relnr2=@relnr));
    update RELATIONS set bucket=@bucket, bucketsub=@bucketsub where bucket=0 and (relnr1=@relnr or relnr2=@relnr);
    set @bucketsub = @bucketsub+1;

    WHILE @nrOfUpdates>0 and @bucketsub<=10    --to prevent the inner while loop from hanging, actually determines the number of iterations
    BEGIN
        --refill temp list with newly found related relnrs
        truncate table ##BucketRelnrs;
        insert into ##BucketRelnrs
        select distinct relnr1 from RELATIONS where bucket=@bucket
        union select distinct relnr2 from RELATIONS where bucket=@bucket;

        --calculate the number of relations that will be updated next, if zero then stop iteration
        set @nrOfUpdates =
        (
            select count(*) from RELATIONS where bucket=0
            and (relnr1 in (select relnr from ##BucketRelnrs) or relnr2 in (select relnr from ##BucketRelnrs))
        );

        --update the RELATIONS table
        update RELATIONS set bucket=@bucket, bucketsub=@bucketsub where bucket=0
        and (relnr1 in (select relnr from ##BucketRelnrs) or relnr2 in (select relnr from ##BucketRelnrs));

        set @bucketsub = @bucketsub+1;
    END
END

drop table ##BucketRelnrs; --clean temp table

如果你需要增量地维护它，我会在你的顶点表中添加第三个int列height，将set_id作为父指针而不是连接组件，并在关系表中添加外键约束和触发器来执行父指针插入。
然后，要查看连接的组件，可以使用递归视图。
如果你已经有一个大表，你需要一个高效的批处理过程，那么主要的问题是，标准的方法不是缓存遗忘的，并且具有非常差的缓存行为，而SQL基本上是为了阻止这种情况而设计的。