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 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 <