【个人学习笔记11之--SQL 中的 图 树 层次结构】

在RDBMS中操作 图 树 层次结构 等特殊的数据结构时,我们通常采用2个主要方法:
1.基于迭代/递归

2.具体化描述数据结构的附加信息。
一般模型有：员工组织图(树,层次结构);料表--BOM(有向图);道路系统(无向循环图)

1.迭代/递归
迭代可以迭代图的一个节点,也可以迭代一个层次.后者比前者要快很多.
实现方法：SQL2000通过UDF(用户自定义函数),SQL2005使用CTE。

 

a.下属问题(通俗说,求子节点)
--这里我使用书上的员工表(表内容如下)
SET NOCOUNT ON;USE tempdb;GOIF OBJECT_ID('dbo.Employees') IS NOT NULL DROP TABLE dbo.Employees;GOCREATE TABLE dbo.Employees( empid INT NOT NULL PRIMARY KEY, mgrid INT NULL REFERENCES dbo.Employees, empname VARCHAR(25) NOT NULL, salary MONEY NOT NULL, CHECK (empid <> mgrid));INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(1, NULL, 'David', $10000.00);INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(2, 1, 'Eitan', $7000.00);INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(3, 1, 'Ina', $7500.00);INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(4, 2, 'Seraph', $5000.00);INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(5, 2, 'Jiru', $5500.00);INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(6, 2, 'Steve', $4500.00);INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(7, 3, 'Aaron', $5000.00);INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(8, 5, 'Lilach', $3500.00);INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(9, 7, 'Rita', $3000.00);INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(10, 5, 'Sean', $3000.00);INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(11, 7, 'Gabriel', $3000.00);INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(12, 9, 'Emilia' , $2000.00);INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(13, 9, 'Michael', $2000.00);INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(14, 9, 'Didi', $1500.00);--创建索引CREATE UNIQUE INDEX idx_unc_mgrid_empid ON dbo.Employees(mgrid, empid);go

--SQL2000 udf方法：

 

IF OBJECT_ID('dbo.fn_subordinates1') IS NOT NULL DROP FUNCTION dbo.fn_subordinates1;GOCREATE FUNCTION dbo.fn_subordinates1(@root AS INT) RETURNS @Subs TABLE( empid INT NOT NULL PRIMARY KEY NONCLUSTERED, lvl INT NOT NULL, UNIQUE CLUSTERED(lvl, empid) )ASbegin declare @lv int set @lv=0insert @Subs values(@root,@lv)while @@rowcount>0begin set @lv=@Lv+1; insert @subs select b.empid ,@Lv from @subs a join dbo.Employees b on a.empid=b.mgrid and lvl=@lv-1endreturn;end goSELECT empid, lvl FROM dbo.fn_subordinates1(3) AS S;

 

--SQL2005 CTE

 

DECLARE @root AS INT;SET @root = 3;WITH SubsCTEAS( -- Anchor member returns root node SELECT empid, empname, 0 AS lvl FROM dbo.Employees WHERE empid = @root UNION ALL -- Recursive member returns next level of children SELECT C.empid, C.empname, P.lvl + 1 FROM SubsCTE AS P JOIN dbo.Employees AS C ON C.mgrid = P.empid)SELECT * FROM SubsCTE;

 

 

--查询结果
/*empid empname lvl----------- ------------------------- -----------3 Ina 07 Aaron 19 Rita 211 Gabriel 212 Emilia 313 Michael 314 Didi 3*/

---------------如果需要限制递归的层数------------

--SQL2000

 

IF OBJECT_ID('dbo.fn_subordinates2') IS NOT NULL DROP FUNCTION dbo.fn_subordinates2;GOCREATE FUNCTION dbo.fn_subordinates2 (@root AS INT, @maxlevels AS INT = NULL) RETURNS @Subs TABLE( empid INT NOT NULL PRIMARY KEY NONCLUSTERED, lvl INT NOT NULL, UNIQUE CLUSTERED(lvl, empid) )ASBEGIN DECLARE @lvl AS INT; SET @lvl = 0; -- 如果@maxlevels输入的是NULL,把他变为最大的INT类型 SET @maxlevels = COALESCE(@maxlevels, 2147483647); INSERT INTO @Subs(empid, lvl) SELECT empid, @lvl FROM dbo.Employees WHERE empid = @root; WHILE @@rowcount > 0 AND @lvl < @maxlevels -- 这里注意加一个小于最大的递归数的条件,是小于 不是小于等于 BEGIN SET @lvl = @lvl + 1; INSERT INTO @Subs(empid, lvl) SELECT C.empid, @lvl --这里跟上面处理都一样 FROM @Subs AS P JOIN dbo.Employees AS C ON P.lvl = @lvl - 1 AND C.mgrid = P.empid; END RETURN;ENDGOSELECT empid, lvlFROM dbo.fn_subordinates2(3, NULL) AS S;

 

--查询结果

/*empid lvl----------- -----------3 07 19 211 212 313 314 3*/-----

SELECT empid, lvl
FROM dbo.fn_subordinates2(2, NULL) AS S;
/*empid lvl----------- -----------2 04 15 16 18 210 2*/

PS：这里控制返回的层数 你当然可以用最开始的方法,然后再筛选语句里控制层数
 如SELECT empid, lvl FROM dbo.fn_subordinates1(3) AS S where lvl<3;  但是控制本身的递归只能在UDF里面了
 
 --sql2005方法类似

WITH SubsCTEAS( SELECT empid, empname, 0 AS lvl FROM dbo.Employees WHERE empid = @root UNION ALL SELECT C.empid, C.empname, P.lvl + 1 FROM SubsCTE AS P JOIN dbo.Employees AS C ON C.mgrid = P.empid AND P.lvl < @maxlevels -- 这里控制递归数)SELECT * FROM SubsCTE;

 

--还有一种偏方：但是不推荐 虽然结果正确 但是会报错
WITH SubsCTEAS( SELECT empid, empname, 0 AS lvl FROM dbo.Employees WHERE empid = @root UNION ALL SELECT C.empid, C.empname, P.lvl + 1 FROM SubsCTE AS P JOIN dbo.Employees AS C ON C.mgrid = P.empid)SELECT * FROM SubsCTEOPTION (MAXRECURSION 2);--就是这里的MAXRECURSION 控制递归

 

--查询结果

/* empid empname lvl----------- ------------------------- -----------3 Ina 07 Aaron 19 Rita 211 Gabriel 2消息 530，级别 16，状态 1，第 4 行语句被终止。完成执行语句前已用完最大递归 2。 */ 
 

b.祖先(通俗说：求父节点)
其实思路跟子节点差不多

--SQL2000

IF OBJECT_ID('dbo.fn_managers') IS NOT NULL DROP FUNCTION dbo.fn_managers;GOCREATE FUNCTION dbo.fn_managers(@empid AS INT, @maxlevels AS INT = NULL) RETURNS @Mgrs TABLE( empid INT NOT NULL PRIMARY KEY, lvl INT NOT NULL)ASBEGIN IF NOT EXISTS(SELECT * FROM dbo.Employees WHERE empid = @empid) --这里判断是否存在经理,不存在 马上返回 RETURN; DECLARE @lvl AS INT,@mgrid int ; SET @lvl = 0; set @mgrid=@empid --赋值给@mgrid 我这是为了更加清楚变量的意思 其实不用这个@Mgrid变量的 -- 如果@maxlevels输入的是NULL,把他变为最大的INT类型 SET @maxlevels = COALESCE(@maxlevels, 2147483647); WHILE @lvl <= @maxlevels and @mgrid is not null --注意这里是小于等于 而且新插入的经理不可以为NULL，为null说明是老大了 BEGIN --插入当前经理 INSERT INTO @Mgrs(empid, lvl) VALUES(@mgrid, @lvl); --这里第一次插入是自己,后面就是各自的经理了 SET @lvl = @lvl + 1; --获得下一个级别的经理 SET @mgrid = (SELECT mgrid FROM dbo.Employees WHERE empid = @mgrid); END RETURN;ENDGOSELECT empid, lvlFROM dbo.fn_managers(8, NULL) AS M;

--查询结果

/*empid lvl----------- -----------1 32 25 18 0*/

SELECT empid, lvl
FROM dbo.fn_managers(8, 2) AS M;
/*empid lvl----------- -----------2 25 18 0*/

 

 

--sql2005
DECLARE @empid AS INT, @maxlevels AS INT;SET @empid = 8;SET @maxlevels = 2;WITH MgrsCTEAS( SELECT empid, mgrid, empname, 0 AS lvl FROM dbo.Employees WHERE empid = @empid UNION ALL SELECT P.empid, P.mgrid, P.empname, C.lvl + 1 FROM MgrsCTE AS C JOIN dbo.Employees AS P ON C.mgrid = P.empid AND C.lvl < @maxlevels--限制递归数...)SELECT * FROM MgrsCTE;

 

C.层次显示

 

--SQL2000,思路跟下属问题一模一样 只是多了个PATH
IF OBJECT_ID('dbo.fn_subordinates3') IS NOT NULL DROP FUNCTION dbo.fn_subordinates3;GOCREATE FUNCTION dbo.fn_subordinates3 (@root AS INT, @maxlevels AS INT = NULL) RETURNS @Subs TABLE( empid INT NOT NULL PRIMARY KEY NONCLUSTERED, lvl INT NOT NULL, path VARCHAR(900) NOT NULL UNIQUE CLUSTERED(lvl, empid) )ASBEGIN DECLARE @lvl AS INT; SET @lvl = 0; SET @maxlevels = COALESCE(@maxlevels, 2147483647); INSERT INTO @Subs(empid, lvl, path) SELECT empid, @lvl, CAST(empid AS VARCHAR(100))--ZHU YI ZHE LI FROM dbo.Employees WHERE empid = @root; WHILE @@rowcount > 0 AND @lvl < @maxlevels BEGIN SET @lvl = @lvl + 1; INSERT INTO @Subs(empid, lvl, path) SELECT C.empid, @lvl, P.path +'-'+ CAST(C.empid AS VARCHAR(100)) --和上面类型保持一致 FROM @Subs AS P JOIN dbo.Employees AS C ON P.lvl = @lvl - 1 AND C.mgrid = P.empid; END RETURN;ENDGO

--这里的显示分2种
--显示1
select empid ,pos=REPLICATE('-',lvl)+rtrim(empid)
FROM dbo.fn_subordinates3(1, NULL) AS S
ORDER BY PATH
/*empid pos----------- ------------1 12 -24 --45 --510 ---108 ---86 --63 -37 --711 ---119 ---912 ----1213 ----1314 ----14*/

--还有一种
SELECT empid,  path
FROM dbo.fn_subordinates3(1, NULL) AS S
ORDER BY PATH
/*empid path----------- ----------------1 12 1-24 1-2-45 1-2-510 1-2-5-108 1-2-5-86 1-2-63 1-37 1-3-711 1-3-7-119 1-3-7-912 1-3-7-9-1213 1-3-7-9-1314 1-3-7-9-14*/

--当然上面的显示方式还很多,自己可以控制 比如不加ORDER 就可以排出不一样的

 

--SQL2005
DECLARE @root AS INT;SET @root = 1;WITH SubsCTEAS( SELECT empid, empname, 0 AS lvl, -- Path of root = '.' + empid + '.' CAST(CAST(empid AS VARCHAR(10)) AS VARCHAR(MAX)) AS path FROM dbo.Employees WHERE empid = @root UNION ALL SELECT C.empid, C.empname, P.lvl + 1, CAST(P.path+'-' + CAST(C.empid AS VARCHAR(10)) AS VARCHAR(MAX)) AS path FROM SubsCTE AS P JOIN dbo.Employees AS C ON C.mgrid = P.empid)SELECT empid, REPLICATE(' | ', lvl) + empname AS empnameFROM SubsCTEORDER BY path ;

 

--结果查询

/*empid empname----------- -------------------------1 David2 | Eitan4 | | Seraph5 | | Jiru10 | | | Sean8 | | | Lilach6 | | Steve3 | Ina7 | | Aaron11 | | | Gabriel9 | | | Rita12 | | | | Emilia13 | | | | Michael14 | | | | Didi*/

 

 

D.检测循环中异常
说白了 就是检查表里有没出现1-2-4-1这种头接尾的圈.这在现实中是不可能的.一个经理部可能是它手下的手下.-- ||
我们对表做手脚,把老大的经理改成是它的一个员工
UPDATE dbo.Employees SET mgrid = 14 WHERE empid = 1;DECLARE @root AS INT;SET @root = 1;WITH SubsCTEAS( SELECT empid, empname, 0 AS lvl, CAST( CAST(empid AS VARCHAR(10)) AS VARCHAR(MAX)) AS path, -- 第一层是不会出现圈圈的 0 AS cycle --0表示没有圈 1表示出现圈 FROM dbo.Employees WHERE empid = @root UNION ALL SELECT C.empid, C.empname, P.lvl + 1, CAST(P.path + '-'+CAST(C.empid AS VARCHAR(10)) AS VARCHAR(MAX)) AS path, -- 这里需要检查 CASE WHEN P.path LIKE '%' + CAST(C.empid AS VARCHAR(10)) + '%' THEN 1 ELSE 0 END FROM SubsCTE AS P JOIN dbo.Employees AS C ON C.mgrid = P.empid where p.cycle=0 --保证不会继续在错误的地方继续循环下去)SELECT empid, empname, cycle, pathFROM SubsCTEwhere cycle=1

 

--查询结果

/*empid empname cycle path----------- ------------------------- ----------- ------------------1 David 1 1-3-7-9-14-1*/
--这样管理员可以轻易找到那个错误的地方.
改过来：UPDATE dbo.Employees SET mgrid = NULL WHERE empid = 1;

 

 

2.具体化路径
这里就是新增加2列，一列是级别 一列是节点路径，这样可以避免每次都去计算.
2个优点：不需要递归,只需要基于集合 ；查询可以使用到路径索引

 

a.维护数据

1.添加不管理员工的员工
SET NOCOUNT ON;USE tempdb;GOIF OBJECT_ID('dbo.Employees') IS NOT NULL DROP TABLE dbo.Employees;GOCREATE TABLE dbo.Employees( empid INT NOT NULL PRIMARY KEY NONCLUSTERED, mgrid INT NULL REFERENCES dbo.Employees, empname VARCHAR(25) NOT NULL, salary MONEY NOT NULL, lvl INT NOT NULL, path VARCHAR(900) NOT NULL UNIQUE CLUSTERED);CREATE UNIQUE INDEX idx_unc_mgrid_empid ON dbo.Employees(mgrid, empid);GOIF OBJECT_ID('dbo.usp_insertemp') IS NOT NULL DROP PROC dbo.usp_insertemp;GOCREATE PROC dbo.usp_insertemp @empid INT, @mgrid INT, @empname VARCHAR(25), @salary MONEYASSET NOCOUNT ON;-- 如果插入的是一个没有经理的老大IF @mgrid IS NULL INSERT INTO dbo.Employees(empid, mgrid, empname, salary, lvl, path) VALUES(@empid, @mgrid, @empname, @salary, 0, CAST(@empid AS VARCHAR(10)));--插入有经理的小弟ELSE INSERT INTO dbo.Employees(empid, mgrid, empname, salary, lvl, path) SELECT @empid, @mgrid, @empname, @salary, lvl + 1, path +'-'+ CAST(@empid AS VARCHAR(10)) FROM dbo.Employees WHERE empid = @mgrid;GOEXEC dbo.usp_insertemp @empid = 1, @mgrid = NULL, @empname = 'David', @salary = $10000.00;EXEC dbo.usp_insertemp @empid = 2, @mgrid = 1, @empname = 'Eitan', @salary = $7000.00;EXEC dbo.usp_insertemp @empid = 3, @mgrid = 1, @empname = 'Ina', @salary = $7500.00;EXEC dbo.usp_insertemp @empid = 4, @mgrid = 2, @empname = 'Seraph', @salary = $5000.00;EXEC dbo.usp_insertemp @empid = 5, @mgrid = 2, @empname = 'Jiru', @salary = $5500.00;EXEC dbo.usp_insertemp @empid = 6, @mgrid = 2, @empname = 'Steve', @salary = $4500.00;EXEC dbo.usp_insertemp @empid = 7, @mgrid = 3, @empname = 'Aaron', @salary = $5000.00;EXEC dbo.usp_insertemp @empid = 8, @mgrid = 5, @empname = 'Lilach', @salary = $3500.00;EXEC dbo.usp_insertemp @empid = 9, @mgrid = 7, @empname = 'Rita', @salary = $3000.00;EXEC dbo.usp_insertemp @empid = 10, @mgrid = 5, @empname = 'Sean', @salary = $3000.00;EXEC dbo.usp_insertemp @empid = 11, @mgrid = 7, @empname = 'Gabriel', @salary = $3000.00;EXEC dbo.usp_insertemp @empid = 12, @mgrid = 9, @empname = 'Emilia', @salary = $2000.00;EXEC dbo.usp_insertemp @empid = 13, @mgrid = 9, @empname = 'Michael', @salary = $2000.00;EXEC dbo.usp_insertemp @empid = 14, @mgrid = 9, @empname = 'Didi', @salary = $1500.00 ;
----检测
SELECT empid, mgrid, empname, salary, lvl, path
FROM dbo.Employees
ORDER BY path;
/*empid mgrid empname salary lvl path----------- ----------- ------------------------- --------------------- ----------- ---------1 NULL David 10000.00 0 12 1 Eitan 7000.00 1 1-24 2 Seraph 5000.00 2 1-2-45 2 Jiru 5500.00 2 1-2-510 5 Sean 3000.00 3 1-2-5-108 5 Lilach 3500.00 3 1-2-5-86 2 Steve 4500.00 2 1-2-63 1 Ina 7500.00 1 1-37 3 Aaron 5000.00 2 1-3-711 7 Gabriel 3000.00 3 1-3-7-119 7 Rita 3000.00 3 1-3-7-912 9 Emilia 2000.00 4 1-3-7-9-1213 9 Michael 2000.00 4 1-3-7-9-1314 9 Didi 1500.00 4 1-3-7-9-14*/

 

 

2.移动子树
比如说某个部门来了个新老大,原来部门老大和手下的人都要跟着他.这个时候表里的路径和级别都要更新
Select Empid, Replicate(' | ', Lvl) + Empname As Empname, Lvl, Path
From Dbo.Employees
Order By Path;

--这个是移动之前的层次分布

/*empid empname lvl path ----------- ----------------- ----------- ------------1 David 0 1 2 | Eitan 1 1-24 | | Seraph 2 1-2-4 5 | | Jiru 2 1-2-5 10 | | | Sean 3 1-2-5-10 6 | | Steve 2 1-2-6 3 | Ina 1 1-3 7 | | Aaron 2 1-3-7 11 | | | Gabriel 3 1-3-7-11 9 | | | Rita 3 1-3-7-9 13 | | | | Michael 4 1-3-7-9-13 14 | | | | Didi 4 1-3-7-9-14 */

 

==接下来我们移动
IF OBJECT_ID('dbo.usp_movesubtree') IS NOT NULL DROP PROC dbo.usp_movesubtree;GOCREATE PROC dbo.usp_movesubtree @root INT,--旧经理 @mgrid INT--新经理ASSET NOCOUNT ON;BEGIN TRAN; UPDATE E SET lvl = E.lvl + NM.lvl - OM.lvl,-- 级别=当前的级别+(新经理级别-原经理级别) path = STUFF(E.path, 1, LEN(OM.path), NM.path) -- 路径=自己的路径移除原来经理那部分再加上新的经理的那部分 --这里的 OM.path 是被替换经理的经理的路径 比如旧经理是1-3-7 那么OM.PATH 就是1-3 --NM.path 是新经理的路径了 比如要换新经理的EMPID是10 所以NM.path 就是 1-2-5-10 --所以如果本来旧经理一个手下比如说是1-3-7-10 现在整个替换过来就是1-2-5-10-7-10 FROM dbo.Employees AS E -- E = 员工 JOIN dbo.Employees AS R -- R = 根 ON R.empid = @root AND E.path LIKE R.path + '%' JOIN dbo.Employees AS OM -- OM = 旧经理 ON OM.empid = R.mgrid JOIN dbo.Employees AS NM -- NM = 新经理 ON NM.empid = @mgrid; -- 更新旧的经理的经理为新来的经理 UPDATE dbo.Employees SET mgrid = @mgrid WHERE empid = @root; COMMIT TRAN;GOBEGIN TRAN; EXEC dbo.usp_movesubtree @root = 7, @mgrid = 10; -- After moving subtree SELECT empid, REPLICATE(' | ', lvl) + empname AS empname, lvl, path FROM dbo.Employees ORDER BY path;ROLLBACK TRAN;

--移动后结果

 

/*empid empname lvl path ----------- ------------------------------ ----------- ----------------------1 David 0 1 2 | Eitan 1 1-2 4 | | Seraph 2 1-2-4 5 | | Jiru 2 1-2-5 10 | | | Sean 3 1-2-5-10 7 | | | | Aaron 4 1-2-5-10-7 11 | | | | | Gabriel 5 1-2-5-10-7-11 9 | | | | | Rita 5 1-2-5-10-7-9 12 | | | | | | Emilia 6 1-2-5-10-7-9-12 13 | | | | | | Michael 6 1-2-5-10-7-9-13 14 | | | | | | Didi 6 1-2-5-10-7-9-14 8 | | | Lilach 3 1-2-5-8 6 | | Steve 2 1-2-6 3 | Ina 1 1-3 */

--这个是移动之后 大家注意观察7 和 10 两位同志

 

3.移除子树
就是把某个部门从公司取消掉
--首先查看公司部门当前分布
SELECT empid, REPLICATE(' | ', lvl) + empname AS empname, lvl, pathFROM dbo.EmployeesORDER BY path;

 

--查询结果
/*empid empname lvl path ----------- ------------- ------------- -----------1 David 0 1 2 | Eitan 1 1-2 4 | | Seraph 2 1-2-4 5 | | Jiru 2 1-2-5 8 | | | Lilach 3 1-2-5-8 6 | | Steve 2 1-2-6 3 | Ina 1 1-3 7 | | Aaron 2 1-3-7 11 | | | Gabriel 3 1-3-7-11 9 | | | Rita 3 1-3-7-9 12 | | | | Emilia 4 1-3-7-9-12 13 | | | | Michael 4 1-3-7-9-13 14 | | | | Didi 4 1-3-7-9-14 */
--接着我们来开始一个部门 比如移除Aaron手下的人

BEGIN TRAN; DELETE FROM dbo.Employees WHERE path LIKE (SELECT M.path + '%' FROM dbo.Employees as M WHERE M.empid = 7); --删除之后显示 SELECT empid, REPLICATE(' | ', lvl) + empname AS empname, lvl, path FROM dbo.Employees ORDER BY path;ROLLBACK TRAN;

 

--查询结果

/*empid empname lvl path ----------- ------------------ ----------- ------------1 David 0 1 2 | Eitan 1 1-2 4 | | Seraph 2 1-2-4 5 | | Jiru 2 1-2-5 10 | | | Sean 3 1-2-5-10 8 | | | Lilach 3 1-2-5-8 6 | | Steve 2 1-2-6 3 | Ina 1 1-3*/

 

4.查询

这里的查询可就轻松多啦 ~因为路径都有啦~
--查询EMPID为3的手下一批人

SELECT REPLICATE(' | ', E.lvl - M.lvl) + E.empnameFROM dbo.Employees AS E JOIN dbo.Employees AS M ON M.empid = 3 -- root AND E.path LIKE M.path + '%' --and E.path LIKE M.path + '_%'--这里那个3号就没了 --and E.lvl - M.lvl <= 2 --限制级数 --WHERE NOT EXISTS --自己本身不是经理的,返回的都是员工 (SELECT * FROM dbo.Employees AS E2 WHERE E2.mgrid = E.empid);ORDER BY E.path;

 

--查询结果
/*------------------Ina | Aaron | | Gabriel | | Rita | | | Emilia | | | Michael | | | Didi*/

 

b.嵌套集合
是书的作者认为用于树建模最完美最高明的解决方案.
惭愧..太难..准备翻第二次书的时候再研究...