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
描述
由于最近的降雨，水汇集在Farmer John's田地的不同地方，其由N×M（1 <= N <= 100; 1 <= M <= 100）的正方形矩形表示。每个方格包含水（'W'）或旱地（'。'）。农民约翰想弄清楚他的田地里有多少个池塘。池塘是一组连接的正方形，其中有水，其中一个正方形被认为与其所有八个邻居相邻。
给出农夫约翰的田地图，确定他有多少池塘。
#include<iostream>
#define MAX_N 100
#define MAX_M 100
using namespace std;
int N,M;
char field[MAX_N][MAX_M+1];
int dfs(int x,int y)
{
field[x][y]='.';
for(int dx=-1;dx<=1;dx++)
{
for(int dy=-1;dy<=1;dy++)
{
int nx=x+dx,ny=y+dy;
if(0<=nx&&nx<=N&&ny>=0&&ny<=M&&field[nx][ny]=='W')
dfs(nx,ny);
}
}
}
void solve()
{
int res=0;
for(int i=0;i<N;i++)
{
for(int j=0;j<M;j++)
{
if(field[i][j]=='W')
{
dfs(i,j);
res++;
}
}
}
cout<<res;
}
int main()
{
cin>>N>>M;
for(int i=0;i<N;i++)
{
cin>>field[i];
}
solve();
}