POJ2386 Lake Counting (DFS)

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 35370		Accepted: 17562

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

思路：题目让求的就是水洼的个数，只有当池塘的八个方向都是陆地时该池塘才是水洼，也是说当'W'的上下左右，左上左下右上右下八个方向都是'.'时才符合题意。因此就可以从任意‘W’出发开始遍历八个方向，把遍历到的‘W’都改成‘.’，直到要遍历的都成了‘.’时遍历结束。dfs搜索的次数就是图中的水洼数目。

AC代码如下：

#include <cstdio>
using namespace std;

const int MAX_N = 105;
char map[MAX_N][MAX_N];
int n,m;

void dfs(int x, int y){
map[x][y] = '.';

for(int i = -1; i <= 1; i ++){
for(int j = -1; j <= 1; j ++){
int dx = i + x, dy = j + y;
if(dx >= 0 && dx < n && dy >= 0 && dy < m && map[dx][dy] == 'W')
dfs(dx,dy);
}
}
return ;
}

int main(){
while(~scanf("%d%d",&n,&m)){
for(int i = 0; i < n; i ++){
scanf("%s",map[i]);
}

int cnt = 0;
for(int i = 0; i < n; i ++){
for(int j = 0; j < m; j ++){
if(map[i][j] == 'W'){
dfs(i,j);
cnt ++;
}
}
}
printf("%d\n",cnt);
}
return 0;
}