USACO chapter 2 section 2.4 Overfencing
USER: tian tianbing [tbbd4261]
TASK: maze1
LANG: C++
Compiling...
Compile: OK
Executing...
Test 1: TEST OK [0.011 secs, 3052 KB]
Test 2: TEST OK [0.000 secs, 3052 KB]
Test 3: TEST OK [0.000 secs, 3052 KB]
Test 4: TEST OK [0.000 secs, 3052 KB]
Test 5: TEST OK [0.011 secs, 3052 KB]
Test 6: TEST OK [0.000 secs, 3088 KB]
Test 7: TEST OK [0.022 secs, 3052 KB]
Test 8: TEST OK [0.000 secs, 3052 KB]
Test 9: TEST OK [0.011 secs, 3052 KB]
Test 10: TEST OK [0.151 secs, 3052 KB]
All tests OK.
Your program ('maze1') produced all correct answers! This is your
submission #2 for this problem. Congratulations!
DFS从两个入口各搜索一次，更新即可。
先写了一个代码，太乱，花了不少时间，后来重写了，要注意的代码的整洁。
/*
ID:tbbd4261
PROG:maze1
LANG:C++
*/
#include
<
fstream
>
using
namespace
std;
ifstream fin(
"
maze1.in
"
);
ofstream fout(
"
maze1.out
"
);
char
wall[
205
][
80
]
=
{
0
};
int
step[
105
][
40
]
=
{
0
};
char
ch;
int
x1
=
0
,y1
=
0
,x2
=
0
,y2
=
0
,i,j,w,h;
int
dx[
4
]
=
{
-
1
,
1
,
0
,
0
},dy[
4
]
=
{
0
,
0
,
-
1
,
1
};
void
input()
{
fin
>>
w
>>
h;
fin.
get
();
for
(i
=
1
; i
<=
2
*
h
+
1
; i
++
)
{
for
(j
=
1
; j
<=
2
*
w
+
1
; j
++
)
{
fin.
get
(ch); wall[i][j]
=
ch;
if
((i
==
1
||
i
==
2
*
h
+
1
||
j
==
1
||
j
==
2
*
w
+
1
)
&&
ch
==
'
'
)
{
if
(
!
x1){ x1
=
i; y1
=
j; }
else
{x2
=
i; y2
=
j; }
}
}
fin.
get
();
}
}
bool
valid(
int
i,
int
j)
{
return
i
>=
2
&&
i
<=
2
*
h
&&
j
>=
2
&&
j
<=
2
*
w
&&
(i
%
2
==
0
)
&&
(j
%
2
==
0
) ; }
bool
valid2(
int
i,
int
j){
return
i
>=
1
&&
i
<=
h
&&
j
>=
1
&&
j
<=
w; }
void
deal(
int
&
x,
int
&
y)
{
for
( i
=
0
; i
<
4
; i
++
)
if
(valid(x
+
dx[i],y
+
dy[i]))
{ x
=
x
+
dx[i]; y
=
y
+
dy[i]; }
x
/=
2
; y
/=
2
;
}
void
dfs(
int
i,
int
j)
{
for
(
int
k
=
0
; k
<
4
; k
++
)
if
(valid2(i
+
dx[k],j
+
dy[k])
&&
wall[
2
*
i
+
dx[k]][j
*
2
+
dy[k]]
==
'
'
)
if
((step[i
+
dx[k]][j
+
dy[k]]
>
step[i][j]
+
1
)
||
step[i
+
dx[k]][j
+
dy[k]]
==
0
)
{ step[i
+
dx[k]][j
+
dy[k]]
=
step[i][j]
+
1
; dfs(i
+
dx[k],j
+
dy[k]); }
}
int
main()
{
input();
deal(x1,y1);
deal(x2,y2);
step[x1][y1]
=
1
;
step[x2][y2]
=
1
;
dfs(x1,y1);
dfs(x2,y2);
int
max
=
1
;
for
(i
=
1
; i
<=
h; i
++
)
for
(j
=
1
; j
<=
w; j
++
)
if
(step[i][j]
>
max)max
=
step[i][j];
fout
<<
max
<<
endl;
return
0
;
}
