POJ 3083 Children of the Candy Corn
解题思路:
1）靠左走，方向优先级为左->前->右->后
2）靠右走，方向优先级为右->前->左->后
2）最短路，广度优先
NULL
#include
<
iostream
>
using
namespace
std;
char
map[
40
][
41
];
bool
visit[
40
][
40
];
const
int
dir[
4
][
2
]
=
{
0
,
1
,
1
,
0
,
0
,
-
1
,
-
1
,
0
};
int
queue[
1600
][
2
];
int
w,h,sx,sy,ex,ey;
inline
bool
IsOk(
int
x,
int
y,
int
d)
{
x
+=
dir[d][
0
],y
+=
dir[d][
1
];
if
(x
>=
0
&&
x
<
h
&&
y
>=
0
&&
y
<
w
&&
(map[x][y]
==
'
.
'
||
map[x][y]
==
'
E
'
)
&&!
visit[x][y])
return
true
;
return
false
;
}
int
CalStep(
bool
IsLeft)
{
int
d,tx,ty,td,curd,step;
d
=
IsLeft
?
1
:
3
;
step
=
curd
=
1
,tx
=
sx,ty
=
sy;
while
(tx
!=
ex
||
ty
!=
ey)
{
td
=
(curd
+
(
4
-
d))
%
4
;
if
(IsOk(tx,ty,td))tx
+=
dir[td][
0
],ty
+=
dir[td][
1
],curd
=
td,step
++
;
else
if
(IsOk(tx,ty,curd))tx
+=
dir[curd][
0
],ty
+=
dir[curd][
1
],step
++
;
else
if
(IsOk(tx,ty,td
=
(curd
+
d)
%
4
))tx
+=
dir[td][
0
],ty
+=
dir[td][
1
],curd
=
td,step
++
;
else
td
=
(curd
+
2
)
%
4
,tx
+=
dir[td][
0
],ty
+=
dir[td][
1
],curd
=
td,step
++
;
}
return
step;
}
int
BFS(
int
s,
int
e)
{
int
i,j,tx,ty,p;
for
(i
=
s,p
=
e;i
<=
e;i
++
)
for
(j
=
0
;j
<
4
;j
++
)
if
(IsOk(queue[i][
0
],queue[i][
1
],j))
{
queue[
++
p][
0
]
=
queue[i][
0
]
+
dir[j][
0
];
queue[p][
1
]
=
queue[i][
1
]
+
dir[j][
1
];
visit[queue[p][
0
]][queue[p][
1
]]
=
true
;
if
(queue[p][
0
]
==
ex
&&
queue[p][
1
]
==
ey)
return
1
;
}
return
1
+
BFS(e
+
1
,p);
}
int
main()
{
int
i,j,n,a,b,c;
scanf(
"
%d
"
,
&
n);
while
(n
--
)
{
memset(visit,
0
,
sizeof
(visit));
scanf(
"
%d %d
"
,
&
w,
&
h);
for
(i
=
0
;i
<
h;i
++
)
{
scanf(
"
%s
"
,map[i]);
for
(j
=
0
;j
<
w;j
++
)
{
if
(map[i][j]
==
'
S
'
)sx
=
i,sy
=
j;
else
if
(map[i][j]
==
'
E
'
)ex
=
i,ey
=
j;
}
}
queue[
0
][
0
]
=
sx,queue[
0
][
1
]
=
sy;
a
=
CalStep(
true
),b
=
CalStep(
false
);
c
=
BFS(
0
,
0
)
+
1
;
printf(
"
%d %d %d\n
"
, a, b, c);
}
return
0
;
}