BOJ 1926 - 그림

www.acmicpc.net/problem/1926

1926번: 그림

BFS 문제 풀어봤습니다.

힘들어!

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class boj1926 {

    static int[][] arr;
    static boolean[][] visited;
    static int[] dx = {-1,1,0,0};
    static int[] dy = {0,0,-1,1};
    static int max = 0;
    static int picutre = 0;
    static int R;
    static int C;
    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        R = sc.nextInt();
        C = sc.nextInt();
        arr = new int[R][C];
        visited = new boolean[R][C];
        for(int i=0; i<R; i++){
            for(int j=0; j<C; j++){
                arr[i][j] = sc.nextInt();
            }
        }

        for(int i=0; i<R; i++){
            for(int j=0; j<C; j++){
                if(visited[i][j] == false && arr[i][j] == 1) {
                    visited[i][j] = true;
                    Pair pair = new Pair(i,j);
                    BFS(pair);
                    picutre++;
                }
            }
        }
        System.out.println(picutre);
        System.out.println(max);
    }

    private static void BFS(Pair pair) {

        Queue<Pair> queue = new LinkedList<>();
        queue.add(pair);
        int area = 1;

        while (!queue.isEmpty()){
            Pair cur = queue.poll();
//            System.out.println("CUR X " + cur.x + " CUR Y " + cur.y);
            for(int dir=0; dir<4; dir++){
                int nx = cur.x + dx[dir];
                int ny = cur.y + dy[dir];
                if(nx <0 || nx >= R || ny<0 || ny >= C) continue;
                if(visited[nx][ny] == true || arr[nx][ny] == 0) continue;
                visited[nx][ny] = true;
                queue.add(new Pair(nx,ny));
                area++;
            }
        }
        max = Math.max(area, max);
    }

    static class Pair{
        int x;
        int y;

        Pair(int x, int y){
            this.x = x;
            this.y = y;
        }
    }

}