# POJ - 2386：Lake Counting

xiaoxiao2021-03-01  10

### Lake Counting

#### 题目

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.

#### 输入

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.

#### 输出

Line 1: The number of ponds in Farmer John’s field.

#### 输入样例

10 12 W…WW. .WWW…WWW …WW…WW. …WW. …W… …W…W… .W.W…WW. W.W.W…W. .W.W…W. …W…W.

3

#### 提示

OUTPUT DETAILS: There are three ponds: one in the upper left, one in the lower left,and one along the right side.

#### 参考代码

#include<stdio.h> const int maxn=100+10; char arr[maxn][maxn]; int vis[maxn][maxn]; int n,m; void dfs(int x,int y) { vis[x][y]=1;//或者可以令arr[i][j]='.'; for(int i=-1;i<=1;i++) { for(int j=-1;j<=1;j++) { int nx=x+i,ny=y+j; if(0<=nx && nx<n && 0<=ny && ny<m && arr[nx][ny]=='W' && !vis[nx][ny]) dfs(nx,ny); } } } int main() { scanf("%d%d",&n,&m); int i,j; for(i=0;i<n;i++) scanf("%s",arr[i]); int cnt=0; for(i=0;i<n;i++) { for(j=0;j<m;j++) { if(arr[i][j]=='W' && !vis[i][j]) { dfs(i,j); cnt++; } } } printf("%d\n",cnt); return 0; }