POJ Lake Counting

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

这一题算是非常基础的深度优先搜索题，就是找到‘W’，进行dfs(),运行是将'W' 改成‘.’

AC代码：

# include <cstdio>
using namespace std;
char g[110][110];
int m, n;
void dfs(int x, int y){
    g[x][y]='.';
    for(int i=-1; i<=1; i++){
    	for(int j=-1; j<=1; j++){
    		if(x+i>=0&&x+i<=m-1&&y+j<=n-1&&y+j>=0&&g[x+i][y+j]=='W'){
    			dfs(x+i, y+j);
			}
		}
	}
}
int main(){
	int i, j, k, ans;
	while(scanf("%d%d", &m, &n)!=EOF){
		getchar();
		for(i=0; i<=m-1; i++){
			scanf("%s", g[i]);
		}
		ans=0;
		for(i=0; i<=m-1; i++){
			for(j=0; j<=n-1; j++){
				if(g[i][j]=='W'){
					ans++;
					dfs(i,j);
				}
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}