• SQL递归游戏你厉害吗,来过5关


    5个flash的游戏地址是
    http://www.sostart.com/article/view.php/765
    此类游戏一般都是通过穷举或者递归之类的方法来求解,对于编成语言来说都比较简单。
    这里用SQL语言的CTE递归来玩玩看。我的算法和答案如下:

    /*
        四个字段分别代表 羊,狼,草,人的位置,0表示河左边,1表示河右边。
        每次都必须有人过河,因为动物和植物们都不会划船。
    */
    declare @t table (y bit , l bit , c bit , r bit , path varchar (8000 ))
    insert into @t select 0 , 0 , 0 , 0 , ''

    ;with t
    as
    (
        select * , 0 as cc from @t union all
        select ~ y , l , c , ~ r , path + '人羊' + ltrim (r )+ '→' , cc + 1 from t where cc < 10 and y = r and y & l & c = 0 and path not like '%人羊_→' union all
        select y , ~ l , c , ~ r , path + '人狼' + ltrim (r )+ '→' , cc + 1 from t where cc < 10 and   l = r and y & l & c = 0 and y <> c and path not like '%人狼_→' union all
        select y , l , ~ c , ~ r , path + '人草' + ltrim (r )+ '→' , cc + 1 from t where   cc < 10 and c = r and y & l & c = 0 and y <> l and path not like '%人草_→' union all
        select y , l , c , ~ r , path + '人' + ltrim (r )+ '→' , cc + 1 from t where cc < 10 and   y & l & c = 0 and y <> c and y <> l and path not like '%人_→'
       
    )
    select REPLACE (REPLACE (path , '0' , '过河' ), '1' , '返回' ) as path from t where y & l & c = 1

    /*
    人羊过河→人返回→人草过河→人羊返回→人狼过河→人返回→人羊过河→
    人羊过河→人返回→人狼过河→人羊返回→人草过河→人返回→人羊过河→
    */


    /*

        前五个字段分别代表 左人数,左鬼数,右人数,右鬼数,船位置
        船位置为代表在右边,代表在左边
        当右边没有人和鬼时(gr+pr>0),不执行返回操作,递归结束
    */
    ;with t (pl , gl , pr , gr , boat , path )
    as
    (
        select 0 , 0 , 3 , 3 , cast (0 as bit ), cast ('' as varchar (8000 )) union all
        select pl + 2 as pl , gl , pr - 2 as pr , gr , ~ boat , path + '2人过河→'
            from t where boat = 0 and pr >= 2 and (pr - 2 >= gr or pr = 2 ) union all
        select pl + 1 , gl + 1 , pr - 1 , gr - 1 , ~ boat , path + '1人鬼过河→'
            from t where boat = 0 and pr >= 1 and gr >= 1 and pl >= gl   union all
        select pl , gl + 2 , pr , gr - 2 , ~ boat , path + '2鬼过河→'
            from t where boat = 0 and gr >= 2 and (pl - 2 >= gl or pl = 0 ) union all
        select pl - 1 , gl , pr + 1 , gr , ~ boat , path + '1人返回→'
            from t where boat = 1 and pl >= 1 and gr + pr > 0   and (pl - 1 >= gl or pl = 1 ) and pr + 1 >= gr union all
        select pl , gl - 1 , pr , gr + 1 , ~ boat , path + '1鬼返回→'
            from t where boat = 1 and gl >= 1 and gr + pr > 0 and (pr - 1 >= gr or pr = 0union all
        select pl - 1 , gl - 1 , pr + 1 , gr + 1 , ~ boat , path + '1人鬼返回→'
            from t where boat = 1 and pl >= 1 and gl >= 1 and gr + pr > 0 and pr >= gr and path not like '%1人鬼过河→'
    )
    select path from t where pr = 0 and gr = 0

    /*
    2鬼过河→鬼返回→鬼过河→鬼返回→人过河→人鬼返回→人过河→鬼返回→鬼过河→鬼返回→鬼过河→
    2鬼过河→鬼返回→鬼过河→鬼返回→人过河→人鬼返回→人过河→鬼返回→鬼过河→人返回→人鬼过河→
    1人鬼过河→人返回→鬼过河→鬼返回→人过河→人鬼返回→人过河→鬼返回→鬼过河→鬼返回→鬼过河→
    1人鬼过河→人返回→鬼过河→鬼返回→人过河→人鬼返回→人过河→鬼返回→鬼过河→人返回→人鬼过河→
    */

    /*
    第三关:一家人过河

    结果:过河的全部组合有万多中情况,其中满足s之内的有种方法

    关联的三个表a b c配合charindex函数分别可以表示出可以过河的两个人和可以返回的一个人
    当没有人可以过河则递归自动结束
    递归中的case语句可以表示当对岸有个人的时候就不需要返回
    */

    declare @t table (name varchar (8000 ), time int )

    insert into @t
    select '瘦人' , 1 union
    select '小胖' , 3 union
    select '姑娘' , 6 union
    select '大胖' , 8 union
    select '瘸子' , 12

    ;with t (forword_name , time , path ) as (

        select   replace (a . name + b. name , c . name , '' ), b. time + c . time , a . name + b. name + '过河→' + c . name + '返回→'
            from @t a , @t b, @t c   where a . time < b. time and charindex (c . name , a . name + b. name )> 0
           
        union all
       
        select  
            case when len (forword_name )< 6 then replace (forword_name + a . name + b. name , c . name , '' ) else forword_name + a . name + b. name end ,
            case when len (forword_name )< 6 then t . time + b. time + c . time else t . time + b. time end ,
            case when len (forword_name )< 6 then path + a . name + b. name + '过河→' + c . name + '返回→' else path + a . name + b. name + '过河→' end
            from @t a , @t b, @t c , t   where a . time < b. time and charindex (c . name , forword_name + a . name + b. name )> 0
            and charindex (a . name , t . forword_name )= 0 and charindex (b. name , t . forword_name )= 0
       )
      select path , time from t where len (forword_name )= 10   and time <= 30
    /*
    瘦人小胖过河→小胖返回→大胖瘸子过河→瘦人返回→瘦人小胖过河→瘦人返回→瘦人姑娘过河→
    瘦人小胖过河→小胖返回→大胖瘸子过河→瘦人返回→瘦人小胖过河→瘦人返回→瘦人姑娘过河→
    瘦人小胖过河→小胖返回→大胖瘸子过河→瘦人返回→瘦人小胖过河→瘦人返回→瘦人姑娘过河→
    瘦人小胖过河→小胖返回→大胖瘸子过河→瘦人返回→瘦人小胖过河→瘦人返回→瘦人姑娘过河→
    瘦人小胖过河→小胖返回→大胖瘸子过河→瘦人返回→瘦人小胖过河→瘦人返回→瘦人姑娘过河→
    瘦人小胖过河→小胖返回→大胖瘸子过河→瘦人返回→瘦人姑娘过河→瘦人返回→瘦人小胖过河→
    ...共40行
    */


    /*第四关:跳马
    */

    declare @t table (p varchar (max ), x int , y int )
    insert into @t
    select 'p11' , 1 , 1 union
    select 'p12' , 1 , 2 union
    select 'p13' , 1 , 3 union
    select 'p14' , 1 , 4 union
    select 'p21' , 2 , 1 union
    select 'p22' , 2 , 2 union
    select 'p23' , 2 , 3 union
    select 'p24' , 2 , 4 union
    select 'p31' , 3 , 1 union
    select 'p32' , 3 , 2 union
    select 'p33' , 3 , 3 union
    select 'p34' , 3 , 4 union
    select 'p42' , 4 , 2 union
    select 'p43' , 4 , 3

    ;with t (pstart , p, c , x , y ) as
    (
        select   p, p, 1 , x , y from @t
            union all
        select t . pstart , t . p+ '→' + t_next . p, c + 1 , t_next . x , t_next . y from @t t_next , t where
            (
                (abs (t_next . x - t . x )= 1 and abs (t_next . y - t . y )= 2 ) or
                (abs (t_next . x - t . x )= 2 and abs (t_next . y - t . y )= 1 ) )
                and (CHARINDEX (t_next . p, t . p)= 0 or (c = 14 and t_next . p= t . pstart ))
    )
    select p from t   where c = 15

    /*
    p43→p31→p23→p11→p32→p13→p21→p42→p34→p22→p14→p33→p12→p24→p43
    p43→p31→p12→p33→p14→p22→p34→p13→p21→p42→p23→p11→p32→p24→p43
    p43→p31→p12→p24→p32→p11→p23→p42→p34→p13→p21→p33→p14→p22→p43
    p43→p24→p32→p11→p23→p42→p21→p13→p34→p22→p14→p33→p12→p31→p43
    ...共112行
    */

    ;with t as (
        select 17 as f1 , 26 as f2 , 20 as f3 , 19 as f4 , 31 as f5 , cast ('' as   varchar (8000 )) as path
            union all select f1 + 8 , f2 + 8 , f3 , f4 , f5 , path + '12上→' from t where f1 < 21 and f2 < 21
            union all select f1 + 8 , f2 , f3 + 8 , f4 , f5 , path + '13上→' from t where f1 < 21 and f3 < 21
            union all select f1 + 8 , f2 , f3 , f4 + 8 , f5 , path + '14上→' from t where f1 < 21 and f4 < 21
            union all select f1 + 8 , f2 , f3 , f4 , f5 + 8 , path + '15上→' from t where f1 < 21 and f5 < 21
            union all select f1 , f2 + 8 , f3 + 8 , f4 , f5 , path + '23上→' from t where f2 < 21 and f2 < 21
            union all select f1 , f2 + 8 , f3 , f4 + 8 , f5 , path + '24上→' from t where f2 < 21 and f4 < 21
            union all select f1 , f2 + 8 , f3 , f4 , f5 + 8 , path + '25上→' from t where f2 < 21 and f5 < 21
            union all select f1 , f2 , f3 + 8 , f4 + 8 , f5 , path + '34上→' from t where f3 < 21 and f4 < 21
            union all select f1 , f2 , f3 + 8 , f4 , f5 + 8 , path + '35上→' from t where f3 < 21 and f5 < 21
            union all select f1 , f2 , f3 , f4 + 8 , f5 + 8 , path + '45上→' from t where f4 < 21 and f5 < 21
            union all select f1 - 13 , f2 - 13 , f3 , f4 , f5 , path + '12下→' from t where f1 > 25 and f2 > 25
            union all select f1 - 13 , f2 , f3 - 13 , f4 , f5 , path + '13下→' from t where f1 > 25 and f3 > 25
            union all select f1 - 13 , f2 , f3 , f4 - 13 , f5 , path + '14下→' from t where f1 > 25 and f4 > 25
            union all select f1 - 13 , f2 , f3 , f4 , f5 - 13 , path + '15下→' from t where f1 > 25 and f5 > 25
            union all select f1 , f2 - 13 , f3 - 13 , f4 , f5 , path + '23下→' from t where f2 > 25 and f2 > 25
            union all select f1 , f2 - 13 , f3 , f4 - 13 , f5 , path + '24下→' from t where f2 > 25 and f4 > 25
            union all select f1 , f2 - 13 , f3 , f4 , f5 - 13 , path + '25下→' from t where f2 > 25 and f5 > 25
            union all select f1 , f2 , f3 - 13 , f4 - 13 , f5 , path + '34下→' from t where f3 > 25 and f4 > 25
            union all select f1 , f2 , f3 - 13 , f4 , f5 - 13 , path + '35下→' from t where f3 > 25 and f5 > 25
            union all select f1 , f2 , f3 , f4 - 13 , f5 - 13 , path + '45下→' from t where f4 > 25 and f5 > 25
           
            union all select f1 + 8 , f2 + 8 , f3 , f4 , f5 , path + '12上→' from t where f1 < 21 and f2 < 23 and f3 between 21 and 25 and f4 between 21 and 25 and f5 between 21 and 25
            union all select f1 + 8 , f2 , f3 + 8 , f4 , f5 , path + '13上→' from t where f1 < 21 and f3 < 23 and f3 between 21 and 25 and f4 between 21 and 25 and f5 between 21 and 25
            union all select f1 + 8 , f2 , f3 , f4 + 8 , f5 , path + '14上→' from t where f1 < 21 and f4 < 23 and f3 between 21 and 25 and f4 between 21 and 25 and f5 between 21 and 25
            union all select f1 + 8 , f2 , f3 , f4 , f5 + 8 , path + '15上→' from t where f1 < 21 and f5 < 23 and f3 between 21 and 25 and f4 between 21 and 25 and f5 between 21 and 25
            union all select f1 , f2 + 8 , f3 + 8 , f4 , f5 , path + '23上→' from t where f2 < 21 and f2 < 23 and f3 between 21 and 25 and f4 between 21 and 25 and f5 between 21 and 25
            union all select f1 , f2 + 8 , f3 , f4 + 8 , f5 , path + '24上→' from t where f2 < 21 and f4 < 23 and f3 between 21 and 25 and f4 between 21 and 25 and f5 between 21 and 25
            union all select f1 , f2 + 8 , f3 , f4 , f5 + 8 , path + '25上→' from t where f2 < 21 and f5 < 23 and f3 between 21 and 25 and f4 between 21 and 25 and f5 between 21 and 25
            union all select f1 , f2 , f3 + 8 , f4 + 8 , f5 , path + '34上→' from t where f3 < 21 and f4 < 23 and f3 between 21 and 25 and f4 between 21 and 25 and f5 between 21 and 25
            union all select f1 , f2 , f3 + 8 , f4 , f5 + 8 , path + '35上→' from t where f3 < 21 and f5 < 23 and f3 between 21 and 25 and f4 between 21 and 25 and f5 between 21 and 25
            union all select f1 , f2 , f3 , f4 + 8 , f5 + 8 , path + '45上→' from t where f4 < 21 and f5 < 23 and f3 between 21 and 25 and f4 between 21 and 25 and f5 between 21 and 25
    )

    select path from t
    where f1 between 21 and 25
    and f2 between 21 and 25
    and f3 between 21 and 25
    and f4 between 21 and 25
    and f5 between 21 and 25

    /*
    34上→45下→45上→35下→35上→15上→25下→25上→
    34上→45下→45上→35下→35上→14上→24下→24上→
    34上→35下→35上→45下→45上→15上→25下→25上→
    34上→35下→35上→45下→45上→14上→24下→24上→
    */
  • 相关阅读:
    .htaccess 文件不起作用?| (Apache Rewrite)
    Putty 如何解决中文乱码(笔记)
    如何删除鼠标右键“发送到”的多余项
    开发环境搭建笔记
    Javascript String 包
    第十五章:使用canvas绘图
    慕课css3 2章边框和3章颜色相关
    第5章引用类型
    第一章 JavaScript简介
    第五章引用类型(Object类型、Array类型)
  • 原文地址:https://www.cnblogs.com/cl1024cl/p/6204804.html
Copyright © 2020-2023  润新知