poj 1077 Eight 八数码
http://acm.pku.edu.cn/JudgeOnline/problem?id=1077
Description
The 15-puzzle has been around for over 100 years; even if you don't know it by that name, you've seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one tile missing. Let's call the missing tile 'x'; the object of the puzzle is to arrange the tiles so that they are ordered as:
where the only legal operation is to exchange 'x' with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:
The letters in the previous row indicate which neighbor of the 'x' tile is swapped with the 'x' tile at each step; legal values are 'r','l','u' and 'd', for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing 'x' tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 x
where the only legal operation is to exchange 'x' with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
5 6 7 8 5 6 7 8 5 6 7 8 5 6 7 8
9 x 10 12 9 10 x 12 9 10 11 12 9 10 11 12
13 14 11 15 13 14 11 15 13 14 x 15 13 14 15 x
r-> d-> r->
The letters in the previous row indicate which neighbor of the 'x' tile is swapped with the 'x' tile at each step; legal values are 'r','l','u' and 'd', for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing 'x' tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
Input
You will receive a description of a configuration of the 8 puzzle. The description is just a list of the tiles in their initial positions, with the rows listed from top to bottom, and the tiles listed from left to right within a row, where the tiles are represented by numbers 1 to 8, plus 'x'. For example, this puzzle
is described by this list:
1 2 3
x 4 6
7 5 8
is described by this list:
1 2 3 x 4 6 7 5 8
Output
You will print to standard output either the word ``unsolvable'', if the puzzle has no solution, or a string consisting entirely of the letters 'r', 'l', 'u' and 'd' that describes a series of moves that produce a solution. The string should include no spaces and start at the beginning of the line.
Sample Input
2 3 4 1 5 x 7 6 8
Sample Output
ullddrurdllurdruldr
Source
South Central USA 1998
八数码 经典的BFS啊， 我就直接用的单向的BFS，也过了，双向的应该会快很多。
判重的hash函数看的discuss里的，用逆序数，9!种情况，一一对应，不会有重复的。
八数码 经典的BFS啊， 我就直接用的单向的BFS，也过了，双向的应该会快很多。
判重的hash函数看的discuss里的，用逆序数，9!种情况，一一对应，不会有重复的。
Source Code
| Problem: 1077 | User: lovecanon | |
| Memory: 9572K | Time: 125MS | |
| Language: GCC | Result: Accepted |
- Source Code
#include
<
stdio.h
>
#include
<
string
.h
>
#include
<
stdlib.h
>
struct
node
{
int
state[
3
][
3
];
int
pre;
int
dir;
}queue[
362881
];
int
hash[
362881
];
int
step[
362881
];
int
a[
3
][
3
];
int
fac(
int
i)
{
switch
(i)
{
case
0
:
return
1
;
case
1
:
return
1
;
case
2
:
return
2
;
case
3
:
return
6
;
case
4
:
return
24
;
case
5
:
return
120
;
case
6
:
return
720
;
case
7
:
return
5040
;
case
8
:
return
40320
;
}
return
0
;
}
int
HASH()
{
int
i,j,k
=
0
,b[
9
],ret
=
0
,num
=
0
;
for
(i
=
0
;i
<
3
;i
++
)
for
(j
=
0
;j
<
3
;j
++
)
b[k
++
]
=
a[i][j];
for
(i
=
0
;i
<
9
;i
++
)
{
num
=
0
;
for
(j
=
0
;j
<
i;j
++
)
if
(b[j]
>
b[i]) num
++
;
ret
+=
fac(i)
*
num;
}
return
ret;
}
void
output(
int
len)
{
int
i;
for
(i
=
len;i
>=
0
;i
--
)
{
if
(step[i]
==
1
) printf(
"
l
"
);
if
(step[i]
==
2
) printf(
"
r
"
);
if
(step[i]
==
3
) printf(
"
u
"
);
if
(step[i]
==
4
) printf(
"
d
"
);
}
printf(
"
\n
"
);
}
int
main()
{
char
s[
10
];
int
i,j,rear
=
0
,front
=
0
,tag
=
0
;
rear
++
;
for
(i
=
0
;i
<
3
;i
++
)
for
(j
=
0
;j
<
3
;j
++
)
{
scanf(
"
%s
"
,s);
if
(s[
0
]
==
'
x
'
) s[
0
]
=
'
9
'
;
queue[rear].state[i][j]
=
s[
0
]
-
'
0
'
;
}
queue[rear].pre
=
0
;queue[rear].dir
=
0
;
for
(i
=
0
;i
<
3
;i
++
)
for
(j
=
0
;j
<
3
;j
++
)
a[i][j]
=
queue[rear].state[i][j];
hash[HASH()]
=
1
;
while
(front
<
rear)
{
int
e,f,tmp,cntdir,len;
front
++
;
for
(i
=
0
;i
<
3
;i
++
)
for
(j
=
0
;j
<
3
;j
++
)
{
a[i][j]
=
queue[front].state[i][j];
if
(a[i][j]
==
9
) {e
=
i;f
=
j;}
}
if
(f
-
1
>=
0
)
{
cntdir
=
1
;
tmp
=
a[e][f];
a[e][f]
=
a[e][f
-
1
];
a[e][f
-
1
]
=
tmp;
tmp
=
HASH();
if
(tmp
==
0
)
{
int
t
=
front;
len
=
0
;
step[len
++
]
=
cntdir;
while
(queue[t].pre)
{
step[len
++
]
=
queue[t].dir;
t
=
queue[t].pre;
}
output(len);
return
0
;
}
if
(
!
hash[tmp])
{
rear
++
;
for
(i
=
0
;i
<
3
;i
++
)
for
(j
=
0
;j
<
3
;j
++
)
queue[rear].state[i][j]
=
a[i][j];
queue[rear].dir
=
cntdir;queue[rear].pre
=
front;
hash[tmp]
=
1
;
}
tmp
=
a[e][f];
a[e][f]
=
a[e][f
-
1
];
a[e][f
-
1
]
=
tmp;
}
if
(f
+
1
<
3
)
{
cntdir
=
2
;
tmp
=
a[e][f];
a[e][f]
=
a[e][f
+
1
];
a[e][f
+
1
]
=
tmp;
tmp
=
HASH();
if
(tmp
==
0
)
{
int
t
=
front;
len
=
0
;
step[len
++
]
=
cntdir;
while
(queue[t].pre)
{
step[len
++
]
=
queue[t].dir;
t
=
queue[t].pre;
}
output(len);
return
0
;
}
if
(
!
hash[tmp])
{
rear
++
;
for
(i
=
0
;i
<
3
;i
++
)
for
(j
=
0
;j
<
3
;j
++
)
queue[rear].state[i][j]
=
a[i][j];
queue[rear].dir
=
cntdir;queue[rear].pre
=
front;
hash[tmp]
=
1
;
}
tmp
=
a[e][f];
a[e][f]
=
a[e][f
+
1
];
a[e][f
+
1
]
=
tmp;
}
if
(e
-
1
>=
0
)
{
cntdir
=
3
;
tmp
=
a[e][f];
a[e][f]
=
a[e
-
1
][f];
a[e
-
1
][f]
=
tmp;
tmp
=
HASH();
if
(tmp
==
0
)
{
int
t
=
front;
len
=
0
;
step[len
++
]
=
cntdir;
while
(queue[t].pre)
{
step[len
++
]
=
queue[t].dir;
t
=
queue[t].pre;
}
output(len);
return
0
;
}
if
(
!
hash[tmp])
{
rear
++
;
for
(i
=
0
;i
<
3
;i
++
)
for
(j
=
0
;j
<
3
;j
++
)
queue[rear].state[i][j]
=
a[i][j];
queue[rear].dir
=
cntdir;queue[rear].pre
=
front;
hash[tmp]
=
1
;
}
tmp
=
a[e][f];
a[e][f]
=
a[e
-
1
][f];
a[e
-
1
][f]
=
tmp;
}
if
(e
+
1
<
3
)
{
cntdir
=
4
;
tmp
=
a[e
+
1
][f];
a[e
+
1
][f]
=
a[e][f];
a[e][f]
=
tmp;
tmp
=
HASH();
if
(tmp
==
0
)
{
int
t
=
front;
len
=
0
;
step[len
++
]
=
cntdir;
while
(queue[t].pre)
{
step[len
++
]
=
queue[t].dir;
t
=
queue[t].pre;
}
output(len);
return
0
;
}
if
(
!
hash[tmp])
{
rear
++
;
for
(i
=
0
;i
<
3
;i
++
)
for
(j
=
0
;j
<
3
;j
++
)
queue[rear].state[i][j]
=
a[i][j];
queue[rear].dir
=
cntdir;queue[rear].pre
=
front;
hash[tmp]
=
1
;
}
tmp
=
a[e
+
1
][f];
a[e
+
1
][f]
=
a[e][f];
a[e][f]
=
tmp;
}
}
printf(
"
unsolvable\n
"
);
return
0
;
}