搜索入门——DFS之走迷宫

DFS也叫深度优先遍历，是一种可以优化的暴力搜索方法，可用于图的遍历、树的遍历、走迷宫等等。这一小节，我们实现一下用DFS走迷宫。DFS可以找到一条出迷宫的路，但是不一定是最短的，这点需要注意。在走迷宫问题上，DFS可以判断该迷宫是否有解以及找出其中一个解，也可以判断一个迷宫有多少连通区域。
接下来让我们看看poj上第2386号问题：

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

接下来请看中文翻译：
在这里插入图片描述
这个问题看似不是走迷宫，确跟走迷宫近乎一样。我们可以从园子左上角开始遍历，一发DFS，我们把所有能遍历到的’W’置为’.’ 。看至少多少次DFS能将遍历完所有，这个次数就是水洼个数。

//
// Created by 程勇 on 2018/10/6.
//

#include <iostream>
using namespace std;

const int maxn = 101, maxm = 101;
int N, M;
char filed[maxn][maxm];

void dfs(int x, int y)
{
    filed[x][y] = '.';  // 将遍历到的点置为'.'
    // 9个方向循环遍历（包含本身）
    for(int dx=-1; dx<=1; dx++)
    {
        for(int dy=-1; dy<=1; dy++)
        {
            int nx = x+dx;
            int ny = y+dy;
            if(0<=nx && nx<N && 0<=ny && ny<M && filed[nx][ny]=='W')
                dfs(nx, ny);
        }
    }
}

int main()
{
    cin >> N >> M;
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < M; ++j) {
            cin >> filed[i][j];
        }
    }
    int res = 0;
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < M; ++j) {
            if(filed[i][j] == 'W')
            {
                dfs(i, j);
                res++;
            }
        }
    }
    cout << res << endl;
    return 0;
}