[백준] 1012 유기농 배추 - JAVA

이번에도 상당히 전형적인 문제이다.

나는 dfs를 사용했고 특별히 어려운 부분은 없었다. 더 어려운 그래프 문제로 넘어가도 될듯??

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

public class Main {

    static boolean[][] visited;
    static int[][] arr;
    static int T, N, M, K, cabbage_size;

    static int[] dx = {0, 0, -1, 1};
    static int[] dy = {-1, 1, 0, 0};

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        T = Integer.parseInt(br.readLine());
        for (int t = 0; t < T; t++) {
            cabbage_size = 0;
            StringTokenizer st = new StringTokenizer(br.readLine());
            N = Integer.parseInt(st.nextToken());
            M = Integer.parseInt(st.nextToken());
            K = Integer.parseInt(st.nextToken());

            visited = new boolean[N][M];
            arr = new int[N][M];

            for (int i = 0; i < K; i++) {
                st = new StringTokenizer(br.readLine());
                int n = Integer.parseInt(st.nextToken());
                int m = Integer.parseInt(st.nextToken());
                arr[n][m] = 1;
            }
            for (int a = 0; a < N; a++) {
                for (int b = 0; b < M; b++) {
                    if (arr[a][b] == 1 && !visited[a][b]) {
                        cabbage_size++;
                        dfs(a, b);
                    }
                }
            }
            System.out.println(cabbage_size);
        }

    }
    private static void dfs(int x, int y) {
        visited[x][y] = true;
        for (int i = 0; i < 4; i++) {
            int nx = x + dx[i];
            int ny = y + dy[i];
            if (nx >= 0 && ny >= 0 && nx < N && ny < M) {
                if (!visited[nx][ny] && arr[nx][ny] == 1) {
                    dfs(nx, ny);
                }
            }
        }
    }
}