문제

풀이

1. 정상인 경우와 색약인 경우의 그림을 따로 할당.
2. bfs 진행하여 구간 개수 구하기.

코드

import java.io.*;
import java.util.*;

public class Main {
    static char[][] map1,map2;
    static int[][] v1,v2;
    static int[][] d=new int[][]{{-1,0},{1,0},{0,-1},{0,1}};
    static int n;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        n=Integer.parseInt(br.readLine());
        map1=new char[n][n];
        map2=new char[n][n];
        v1=new int[n][n];
        v2=new int[n][n];
        for(int i=0;i<n;i++){
            String s=br.readLine();
            for(int j=0;j<n;j++){
                if(s.charAt(j)=='R'){
                    map1[i][j]=s.charAt(j);
                    map2[i][j]=s.charAt(j);
                }
                else if(s.charAt(j)=='G'){
                    map1[i][j]=s.charAt(j);
                    map2[i][j]='R';
                }
                else {
                    map1[i][j]=s.charAt(j);
                    map2[i][j]=s.charAt(j);
                }
            }
        }

        int red=0,green=0,blue1=0,redGreen=0,blue2=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(v1[i][j]==0){
                    bfs(i,j,map1,v1);
                    if(map1[i][j]=='R')red++;
                    else if(map1[i][j]=='G')green++;
                    else blue1++;
                }
                if(v2[i][j]==0){
                    bfs(i,j,map2,v2);
                    if(map2[i][j]=='R')redGreen++;
                    else blue2++;
                }
            }
        }

        System.out.println((red+green+blue1)+" "+(redGreen+blue2));
    }

    static void bfs(int y,int x,char[][] map,int[][] v){
        char c=map[y][x];
        Queue<int[]> q=new LinkedList<>();
        q.add(new int[]{y,x});
        v[y][x]=1;
        while(!q.isEmpty()){
            y=q.peek()[0];
            x=q.peek()[1];
            q.poll();
            for(int i=0;i<4;i++){
                int dy=y+d[i][0];
                int dx=x+d[i][1];
                if(dy<0||dx<0||dy>=n||dx>=n)continue;
                if(map[dy][dx]==c && v[dy][dx]==0){
                    q.add(new int[]{dy,dx});
                    v[dy][dx]=1;
                }
            }
        }
    }
}

#bfs