Lake Counting
Time Limit:1000MS Memory Limit:65536K
Total Submit:1360 Accepted:629
Description
Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.
Given a diagram of Farmer John's field, determine how many ponds he has.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.
Output
* Line 1: The number of ponds in Farmer John's field.
Sample Input
10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.
Sample Output
3
Hint
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
Source
USACO 2004 November
#include
<
iostream
>
using
namespace
std;
const
int
MAX
=
102
;
int
g[MAX][MAX];
void
travel(
int
i,
int
j)
{
int
incr[
8
][
2
]
=
{
{
0
,
-
1
}
,
{
1
,
-
1
}
,
{
1
,
0
}
,
{
1
,
1
}
,
{
0
,
1
}
,
{
-
1
,
1
}
,
{
-
1
,
0
}
,
{
-
1
,
-
1
}
}
;
int
k;
int
tmpI, tmpJ;
for
(k
=
0
; k
<
8
; k
++
)
{
tmpI
=
i
+
incr[k][
0
];
tmpJ
=
j
+
incr[k][
1
];
if
(g[tmpI][tmpJ]
==
1
)
{
g[tmpI][tmpJ]
=
-
1
;
travel(tmpI, tmpJ);
}
}
}
int
main()
{
int
n, m;
int
i, j;
int
ans
=
0
;
char
t;
cin
>>
n
>>
m;
for
(i
=
0
; i
<=
m
+
1
; i
++
)
{
g[
0
][i]
=
-
1
;
g[n
+
1
][i]
=
-
1
;
}
for
(i
=
0
; i
<=
n
+
1
; i
++
)
{
g[i][
0
]
=
-
1
;
g[i][m
+
1
]
=
-
1
;
}
for
(i
=
1
; i
<=
n; i
++
)
{
for
(j
=
1
; j
<=
m; j
++
)
{
cin
>>
t;
if
(t
==
'
W
'
)
{
g[i][j]
=
1
;
}
else
{
g[i][j]
=
0
;
}
}
}
for
(i
=
1
; i
<=
n; i
++
)
{
for
(j
=
1
; j
<=
m; j
++
)
{
if
(g[i][j]
==
1
)
{
ans
++
;
g[i][j]
=
-
1
;
travel(i, j);
}
}
}
cout
<<
ans
<<
endl;
return
0
;
}
感受递归的神奇!
posted on 2006-05-01 18:48
豪 阅读(697)
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