2636-치즈

https://www.acmicpc.net/problem/2636

2636번: 치즈

빙산 문제와 마찬가지로, BFS 알고리즘을 활용해 치즈에 닿은 면이 있으면 그 치즈를 없애준 후, 남아있는 치즈를 계산한다. 그 과정을 반복하여 코드를 작성하면 아래와 같은 코드로 만들 수 있다.

package BFS;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Collections;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;

public class N2636 {

    public static class Pos
    {
        int r, c, dist;
        public Pos(int r, int c,int dist){
           this.r = r;
           this.c = c;
           this.dist = dist;
        }
    }

    private static final int[] DR = {1, 0, -1, 0};
    private static final int[] DC = {0, 1, 0, -1};
    public static void Bfs(Queue<Pos> q, int R, int C, String[][] maze)
    {
        while (!q.isEmpty())
        {
            Pos Cur = q.poll();
            for (int i = 0; i < DR.length; i++) {
                int nextR = Cur.r + DR[i];
                int nextC = Cur.c + DC[i];

                if(nextR < 0 || nextC <0 || nextC >= C || nextR >=R)
                    continue;
                if (Integer.parseInt(maze[nextR][nextC]) == 1) {
                    maze[nextR][nextC] = String.valueOf(Integer.parseInt(maze[nextR][nextC]) - 1);
                }
            }
        }
    }

    public static void Bfs2(Queue<Pos> q, int R, int C, String[][] maze, boolean[][] visited)
    {
        while (!q.isEmpty())
        {
            Pos Cur = q.poll();
            for (int i = 0; i < DR.length; i++) {
                int nextR = Cur.r + DR[i];
                int nextC = Cur.c + DC[i];

                if(nextR < 0 || nextC <0 || nextC >= C || nextR >=R)
                    continue;
                if (Integer.parseInt(maze[nextR][nextC]) != 1 && !visited[nextR][nextC]) {
                    visited[nextR][nextC] = true;
                    q.add(new Pos(nextR,nextC,0));
                }
            }
        }
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] a = br.readLine().split(String.valueOf(' '));
        int R = Integer.parseInt(a[0]);
        int C = Integer.parseInt(a[1]);
        String[][] maze = new String[R][C];
        boolean[][] visited = new boolean[R][C];
        Queue<Pos> q = new LinkedList<>();
        PriorityQueue<Integer> p = new PriorityQueue<>();
        int time = 0;
        int count = 0;
        for (int i = 0; i < R; i++) {
            String curRow = br.readLine();
            for (int j = 0; j < C; j++) {
                String curCol[] = curRow.split(" ");
                maze[i][j] = curCol[j];
            }
        }

        while(true){
            count = 0;
            boolean isFinish = true;
            for (int i = 0; i <R ; i++) {
                for (int j = 0; j <C ; j++) {
                    if(maze[i][j].equals("1")){
                        count++;
                    }
                    visited[i][j] = false;
                }
            }
            q.add(new Pos(0,0,0));
            visited[0][0] = true;
            Bfs2(q,R,C,maze,visited);

            for (int i = 0; i <R ; i++) {
                for (int j = 0; j <C ; j++) {
                    if(visited[i][j]){
                        q.add(new Pos(i,j,0));
                        continue;
                    }
                    isFinish = false;
                }
            }
            if(isFinish){
                System.out.println(time);
                System.out.println(p.poll());
                break;
            }
            Bfs(q,R,C,maze);
            time++;
            p.add(count);
        }


    }
}