$$\color{Red}{红与黑【DFS之联通性模型】}$$

这里附带打个广告——————我做的所有的题解

包括基础提高以及一些零散刷的各种各样的题

思路

深度优先遍历————>不开st数组就每次把遍历过的点变成条件不让遍历的点的类型即可。起点预处理一下也算黑色瓷砖。

java

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

    static int N = 25, n, m, x, y, cnt;
    static char[][] g = new char[N][N];
    static int[] dx = {-1, 0, 1, 0}, dy = {0, 1, 0, -1};
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    static void dfs(int x, int y) {
        if (g[x][y] == '#') return;
        if (g[x][y] == '.') {
            g[x][y] = '#';
            cnt ++;
        }
        for (int i = 0; i < 4; i++) {
            int a = x + dx[i], b = y + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= m) continue;
            if (g[a][b] == '#') continue;
            dfs(a, b);
        }
    }

    public static void main(String[] args) throws IOException {
        while (true) {
            String[] s1 = br.readLine().split(" ");
            m = Integer.parseInt(s1[0]);
            n = Integer.parseInt(s1[1]);
            if (n == 0 && m == 0) break;
            cnt = 0;
            for (int i = 0; i < n; i++) {
                String s2 = br.readLine();
                for (int j = 0; j < m; j++) {
                    g[i][j] = s2.charAt(j);
                    if (g[i][j] == '@') {
                        x = i;
                        y = j;
                        g[i][j] = '.';
                    }
                }
            }
            dfs(x, y);
            System.out.println(cnt);
        }
    }
}