POJ 2386 Lake Counting

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.

http://poj.org/problem?id=2386

#include <iostream>
using namespace std;

const int MAX_N = 100;
char field[MAX_N][MAX_N+1];
int n, m;

bool inRange(int x, int y){return x>=0 && x<n && y>=0 && y<m;}

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

	//移動する8方向をループ
	for(int dx = -1 ; dx <= 1 ; dx++){
		for(int dy = -1 ; dy <=1 ; dy++){
			int nx = x+dx;
			int ny = y+dy;

			if(inRange(nx,ny) && 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 << endl;
}

int main(){
	cin >> n >> m;
	for(int i=0 ; i<n ; i++){
		for(int j=0 ; j<m ; j++){
			cin >> field[i][j];
		}
	}
	solve();
}

深さ優先探索(DFS: Depth-First Search)の例。