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
=
0)
union
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→
*/