POJ - Lake Counting(DFS)

[align=center]Lake Counting[/align]
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

/*
*  假设从任意一点W开始，不停地把邻接的部分用'.'代替。1次DFS之后与初始动这个W连接的所有W都被替换成'.'因此直到图中不再存在W为止。
*  总共进行DFS的次数就是最后的结果咯！8个方向对应了8种状态转移，每个各自作为DFS的参数至多被调用一次，所以时间复杂度为：O(8*N*M)。
*
*/
#include <iostream>

using namespace std;

char pos[101][101];
int n,m;
void dfs(int x, int y)
{
pos[x][y] = '.';
for(int dx = -1; dx <= 1; dx++)
for(int dy = -1; dy <= 1; dy++)
{
int nx = x + dx, ny = y + dy;
if(0<=nx&&nx<n&&0<=ny&&ny<m&&pos[nx][ny]=='W')
dfs(nx,ny);
}
return ;
}
int main(void)
{
while(cin>>n>>m)
{
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
cin>>pos[i][j];
int res = 0;
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
{
if(pos[i][j] == 'W')
{
dfs(i,j);
res++;
}
}
cout<<res<<endl;
}
return 0;
}