POJ 3322 Bloxorz I (bfs)
题意：一款著名的智力游戏：滚木块。输入地图，输出要完成目标最少的步数。
思路：求最少的步数，那么一般就要用BFS，思路很清晰，就是状态保存的时候要注意了，题目数据时500*500的矩阵，木块是3类状态，竖向的、横向的、站立的，因为存储空间的限制，就只能分开保存这三类状态...这类题一定要细心细心在细心.......
#include
<
iostream
>
#include
<
cstdio
>
#include
<
algorithm
>
#include
<
memory.h
>
#include
<
cmath
>
#include
<
bitset
>
#include
<
queue
>
#include
<
vector
>
using
namespace
std;
const
int
BORDER
=
(
1
<<
20
)
-
1
;
const
int
MAXSIZE
=
37
;
const
int
MAXN
=
1010
;
const
int
INF
=
0x4ffffff
;
#define
CLR(x,y) memset(x,y,sizeof(x))
#define
ADD(x) x=((x+1)&BORDER)
#define
IN(x) scanf("%d",&x)
#define
OUT(x) printf("%d\n",x)
#define
MIN(m,v) (m)<(v)?(m):(v)
#define
MAX(m,v) (m)>(v)?(m):(v)
#define
ABS(x) ((x)>0?(x):-(x))
#define
SET_NODE(no,a,b,c,d) {no.state=a,no.x=b,no.y=c,no.val=d;}
int
n,m,ex,ey;
int
ans;
bool
visit[
3
][MAXN][MAXN];
//
0:oo 1:horizon 2:vertical
char
g[MAXN][MAXN];
int
dir[
5
][
4
][
3
]
=
{
{{
0
,
0
,
0
},{
0
,
0
,
0
},{
0
,
0
,
0
},{
0
,
0
,
0
}},
{{
-
2
,
0
,
4
},{
1
,
0
,
4
},{
0
,
-
2
,
2
},{
0
,
1
,
2
}},
{{
-
1
,
0
,
2
},{
1
,
0
,
2
},{
0
,
-
1
,
1
},{
0
,
2
,
1
}},
{{
0
,
0
,
0
},{
0
,
0
,
0
},{
0
,
0
,
0
},{
0
,
0
,
0
}},
{{
-
1
,
0
,
1
},{
2
,
0
,
1
},{
0
,
-
1
,
4
},{
0
,
1
,
4
}}};
typedef
struct
{
int
state;
int
x,y;
int
val;
}NODE;
NODE start_node;
int
init()
{
CLR(visit,
0
);
CLR(g,
0
);
ans
=
0
;
return
0
;
}
int
input()
{
int
i,j,tmp,k;
int
pos[
2
][
2
];
int
tag
=
0
;
for
(i
=
0
; i
<
n;
++
i)
scanf(
"
%s
"
,g[i]);
for
(i
=
0
; i
<
n;
++
i)
for
(j
=
0
; j
<
m;
++
j)
{
if
(g[i][j]
!=
'
#
'
)
{
if
(g[i][j]
==
'
X
'
)
{
pos[tag][
0
]
=
i;
pos[tag][
1
]
=
j;
++
tag;
}
else
if
(g[i][j]
==
'
O
'
)
{
g[i][j]
=
'
.
'
;
ex
=
i;
ey
=
j;
}
}
}
if
(tag
==
1
)
{
SET_NODE(start_node,
1
,pos[
0
][
0
],pos[
0
][
1
],
0
);
visit[
0
][start_node.x][start_node.y]
=
true
;
return
0
;
}
if
(pos[
0
][
0
]
==
pos[
1
][
0
])
{
tmp
=
MIN(pos[
0
][
1
],pos[
1
][
1
]);
SET_NODE(start_node,
2
,pos[
0
][
0
],tmp,
0
);
visit[
1
][start_node.x][start_node.y]
=
true
;
return
0
;
}
if
(pos[
0
][
1
]
==
pos[
1
][
1
])
{
tmp
=
MIN(pos[
0
][
0
], pos[
1
][
0
]);
SET_NODE(start_node,
4
,tmp,pos[
0
][
1
],
0
);
visit[
2
][start_node.x][start_node.y]
=
true
;
}
return
0
;
}
bool
_in(
const
int
&
x,
const
int
&
y)
{
if
(x
<
0
||
x
>=
n
||
y
<
0
||
y
>=
m)
return
false
;
return
true
;
}
bool
_check(
const
NODE
&
node)
{
int
x,y,a,b;
if
(node.x
<
0
||
node.y
<
0
)
return
false
;
if
(node.state
&
1
)
{
if
(
!
visit[
0
][node.x][node.y]
&&
g[node.x][node.y]
==
'
.
'
)
{
visit[
0
][node.x][node.y]
=
true
;
return
true
;
}
else
return
false
;
}
if
(node.state
&
2
)
{
x
=
a
=
node.x;
y
=
node.y;
b
=
y
+
1
;
if
(visit[
1
][x][y])
return
false
;
if
(g[x][y]
==
'
#
'
||
g[a][b]
==
'
#
'
)
return
false
;
visit[
1
][x][y]
=
true
;
}
else
{
x
=
node.x;
y
=
b
=
node.y;
a
=
x
+
1
;
if
(visit[
2
][x][y])
return
false
;
if
(g[x][y]
==
'
#
'
||
g[a][b]
==
'
#
'
)
return
false
;
visit[
2
][x][y]
=
true
;
}
return
true
;
}
int
bfs()
{
int
a,b,x,y,i,j,tmp;
NODE t_node,node;
queue
<
NODE
>
que;
while
(
!
que.empty())
que.pop();
que.push(start_node);
while
(
!
que.empty())
{
node
=
que.front();
que.pop();
if
((node.state
&
1
)
&&
node.x
==
ex
&&
node.y
==
ey)
return
node.val;
for
(
int
i
=
0
; i
<
4
;
++
i)
{
x
=
node.x
+
dir[node.state][i][
0
];
y
=
node.y
+
dir[node.state][i][
1
];
SET_NODE(t_node,dir[node.state][i][
2
],x,y,node.val
+
1
);
if
(_check(t_node))
que.push(t_node);
}
}
return
-
1
;
}
int
work()
{
ans
=
bfs();
if
(ans
==
-
1
)
printf(
"
Impossible\n
"
);
else
printf(
"
%d\n
"
,ans);
return
0
;
}
int
main()
{
while
(scanf(
"
%d%d
"
,
&
n,
&
m)
!=
EOF)
{
if
(
!
n
&&
!
m)
break
;
init();
input();
work();
}
return
0
;
}