POJ 2386 Lake Counting (DFS)

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.

#include<iostream>

using namespace std;

const int MAX = 110;
char map[MAX][MAX];
int vis[MAX][MAX];

void dfs(int x, int y)
{
if(vis[x][y] == 1 || map[x][y] == '.'|| map[x][y] == 0)	return;//若曾经访问过这个格子，或当前格子是空地，或者当前格子为0（即出界）则返回
vis[x][y] = 1;
//递归访问周围8个格子
dfs(x-1,y-1);	dfs(x-1,y);		dfs(x-1,y+1);
dfs(x,y-1);						dfs(x,y+1);
dfs(x+1,y-1);	dfs(x+1,y);		dfs(x+1,y+1);
}

int main()
{
memset(map,0,sizeof(map));//初始化为0，作为边界
memset(vis,0,sizeof(vis));//初始化为0，表示全未访问过
int m,n,cnt = 0;
cin >> m >> n;

for(int i = 1;i <= m;++i)
for(int j = 1;j <= n;++j)
cin >> map[i][j];//将地图放在矩阵中间，周围有一圈0，方便对出界的判断

for(int i = 1;i <= m;++i)
for(int j = 1;j <= n;++j)
{
if(!vis[i][j] && map[i][j] == 'W')//若找到未被访问过的格子且当前格子是水
{
++cnt;
dfs(i,j);
}
}
cout << cnt <<endl;

return 0;
}