Question

前几天我遇到了一个有趣的问题。它涉及安排一系列“锦标赛”。以下是约束：

有7支球队名为A，B，......，G
锦标赛包含3支球队，锦标赛中的所有球队与其他2支球队进行1场比赛。即比赛由3场比赛组成
每个团队举办两场比赛
每支队伍在其举办的锦标赛旁边参加四场比赛
当系列赛完成后，所有球队都应该两次打完所有其他球队

每次旅行都与距离有关。目标是找到一个时间表，使最短和最长的旅行团队之间的差异很小。

不完全是旅行商问题，但有点相关。如果有强烈意见，我会很乐意删除该标签。

一些示例数据：

create table d
(   source      varchar(20) not null
,   destination      varchar(20) not null
,   distance    int not null
,       primary key (source, destination) );

insert into d (source, destination, distance)
select 'A', destination, distance
from (values  ('B',204)
          ,   ('C',31)
          ,   ('D',185)
          ,   ('E',213)
          ,   ('F',142)
          ,   ('G',165)
     ) T(destination, distance);

insert into d (source, destination, distance)
select 'B', destination, distance
from (values ('C',230)
        ,    ('D',102)
        ,    ('E',169)
        ,    ('F',152)
        ,    ('G',224)
     ) T(destination, distance);

insert into d (source, destination, distance)
select 'C', destination, distance
from (values ('D',200)
            ,('E',224)
            ,('F',157)
            ,('G',135)
     ) T(destination, distance);

insert into d (source, destination, distance)
select 'D', destination, distance
from (values ('E',68)
            ,('F',51)
            ,('G',123)
     ) T(destination, distance);

insert into d (source, destination, distance)
select 'E', destination, distance
from (values ('F',52)
        ,    ('G',88)
     ) T(destination, distance);

insert into d (source, destination, distance)
select 'F', destination, distance
from (values   ('G',80)
     ) T(destination, distance);

create view dv as
select source, destination, distance from d
union all
select destination, source, distance from d;

我开始假设每个团队选择4次旅行并从那里过滤是最容易的。我们称这个关系为team_travel。通过将A的team_travels与B的team_travel等交叉连接，我们可以获得team_travels的所有可能组合。从那里我们可以排除那些不包含4'A'，4'B'等的那些。通过对最大和最小行程距离的差异进行排序，我们可以得到结果。

这很丑陋，当然可以改进：

create or replace function cnt_occ(s varchar(28), c varchar(1))
returns int
    return values length(s) - length(replace(s,c,null));

with t(s,d,n) as (
    select source, destination, distance
    from dv
), team_travel(s, d1,d2,d3,d4,n1,n2,n3,n4) as (
    select t1.s, t1.d, t2.d, t3.d, t4.d, t1.n,t2.n,t3.n,t4.n
    from t as t1
    join t as t2
        on t2.d > t1.d
        and t1.s = t2.s
    join t as t3
        on t3.d > t2.d
        and t1.s = t3.s
    join t as t4
        on t4.d > t3.d
        and t1.s = t4.s
    where t1.s not in (t1.d,t2.d,t3.d,t4.d)
)
select
       a1.s||'->'||a1.d1||a1.d2||a1.d3||a1.d4,a1.n1+a1.n2+a1.n3+a1.n4
    ,  a2.s||'->'||a2.d1||a2.d2||a2.d3||a2.d4,a2.n1+a2.n2+a2.n3+a2.n4
    ,  a3.s||'->'||a3.d1||a3.d2||a3.d3||a3.d4,a3.n1+a3.n2+a3.n3+a3.n4
    ,  a4.s||'->'||a4.d1||a4.d2||a4.d3||a4.d4,a4.n1+a4.n2+a4.n3+a4.n4
    ,  a5.s||'->'||a5.d1||a5.d2||a5.d3||a5.d4,a5.n1+a5.n2+a5.n3+a5.n4
    ,  a6.s||'->'||a6.d1||a6.d2||a6.d3||a6.d4,a6.n1+a6.n2+a6.n3+a6.n4
    ,  a7.s||'->'||a7.d1||a7.d2||a7.d3||a7.d4,a7.n1+a7.n2+a7.n3+a7.n4
    ,  greatest(a1.n1+a1.n2+a1.n3+a1.n4
            ,  a2.n1+a2.n2+a2.n3+a2.n4
            ,  a3.n1+a3.n2+a3.n3+a3.n4
            ,  a4.n1+a4.n2+a4.n3+a4.n4
            ,  a5.n1+a5.n2+a5.n3+a5.n4
            ,  a6.n1+a6.n2+a6.n3+a6.n4
            ,  a7.n1+a7.n2+a7.n3+a7.n4)
    -  least(a1.n1+a1.n2+a1.n3+a1.n4
        ,  a2.n1+a2.n2+a2.n3+a2.n4
        ,  a3.n1+a3.n2+a3.n3+a3.n4
        ,  a4.n1+a4.n2+a4.n3+a4.n4
        ,  a5.n1+a5.n2+a5.n3+a5.n4
        ,  a6.n1+a6.n2+a6.n3+a6.n4
        ,  a7.n1+a7.n2+a7.n3+a7.n4) as min_diff
from       ( select * from team_travel where s = 'A' ) as a1
cross join ( select * from team_travel where s = 'B' ) as a2
cross join ( select * from team_travel where s = 'C' ) as a3
cross join ( select * from team_travel where s = 'D' ) as a4
cross join ( select * from team_travel where s = 'E' ) as a5
cross join ( select * from team_travel where s = 'F' ) as a6
cross join ( select * from team_travel where s = 'G' ) as a7
where cnt_occ(a1.d1||a1.d2||a1.d3||a1.d4
            ||a2.d1||a2.d2||a2.d3||a2.d4
            ||a3.d1||a3.d2||a3.d3||a3.d4
            ||a4.d1||a4.d2||a4.d3||a4.d4
            ||a5.d1||a5.d2||a5.d3||a5.d4
            ||a6.d1||a6.d2||a6.d3||a6.d4
            ||a7.d1||a7.d2||a7.d3||a7.d4, 'A') = 4
  and cnt_occ(a1.d1||a1.d2||a1.d3||a1.d4
            ||a2.d1||a2.d2||a2.d3||a2.d4
            ||a3.d1||a3.d2||a3.d3||a3.d4
            ||a4.d1||a4.d2||a4.d3||a4.d4
            ||a5.d1||a5.d2||a5.d3||a5.d4
            ||a6.d1||a6.d2||a6.d3||a6.d4
            ||a7.d1||a7.d2||a7.d3||a7.d4, 'B') = 4
  and cnt_occ(a1.d1||a1.d2||a1.d3||a1.d4
            ||a2.d1||a2.d2||a2.d3||a2.d4
            ||a3.d1||a3.d2||a3.d3||a3.d4
            ||a4.d1||a4.d2||a4.d3||a4.d4
            ||a5.d1||a5.d2||a5.d3||a5.d4
            ||a6.d1||a6.d2||a6.d3||a6.d4
            ||a7.d1||a7.d2||a7.d3||a7.d4, 'C') = 4
  and cnt_occ(a1.d1||a1.d2||a1.d3||a1.d4
            ||a2.d1||a2.d2||a2.d3||a2.d4
            ||a3.d1||a3.d2||a3.d3||a3.d4
            ||a4.d1||a4.d2||a4.d3||a4.d4
            ||a5.d1||a5.d2||a5.d3||a5.d4
            ||a6.d1||a6.d2||a6.d3||a6.d4
            ||a7.d1||a7.d2||a7.d3||a7.d4, 'D') = 4
  and cnt_occ(a1.d1||a1.d2||a1.d3||a1.d4
            ||a2.d1||a2.d2||a2.d3||a2.d4
            ||a3.d1||a3.d2||a3.d3||a3.d4
            ||a4.d1||a4.d2||a4.d3||a4.d4
            ||a5.d1||a5.d2||a5.d3||a5.d4
            ||a6.d1||a6.d2||a6.d3||a6.d4
            ||a7.d1||a7.d2||a7.d3||a7.d4, 'E') = 4
  and cnt_occ(a1.d1||a1.d2||a1.d3||a1.d4
            ||a2.d1||a2.d2||a2.d3||a2.d4
            ||a3.d1||a3.d2||a3.d3||a3.d4
            ||a4.d1||a4.d2||a4.d3||a4.d4
            ||a5.d1||a5.d2||a5.d3||a5.d4
            ||a6.d1||a6.d2||a6.d3||a6.d4
            ||a7.d1||a7.d2||a7.d3||a7.d4, 'F') = 4
  and cnt_occ(a1.d1||a1.d2||a1.d3||a1.d4
            ||a2.d1||a2.d2||a2.d3||a2.d4
            ||a3.d1||a3.d2||a3.d3||a3.d4
            ||a4.d1||a4.d2||a4.d3||a4.d4
            ||a5.d1||a5.d2||a5.d3||a5.d4
            ||a6.d1||a6.d2||a6.d3||a6.d4
            ||a7.d1||a7.d2||a7.d3||a7.d4, 'G') = 4
order by min_diff
fetch first 100 rows only

必须有比这更优雅的东西，思想？

最小距离调度方案

0 个答案: