[JAVA] 백준 적록색약

dfs 방식으로 접근하면 풀리는 문제이다.

적록색약인 사람일때의 탐색용 배열은 R과 G를 한쪽으로 통일 시켜서 배열을 저장해서 dfs해주면 된다.(replaceAll 사용)

 

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

public class Main {
​​​​static int N;
​​​​static char[][] org, rg;
​​​​static boolean[][] visited;

​​​​public static void main(String[] args) throws IOException {
​​​​​​​​BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
​​​​​​​​StringTokenizer st = new StringTokenizer(br.readLine());
​​​​​​​​N = Integer.parseInt(st.nextToken());

​​​​​​​​org = new char[N][N];
​​​​​​​​rg = new char[N][N];
​​​​​​​​visited = new boolean[N][N];
​​​​​​​​for (int i = 0; i < N; i++) {
​​​​​​​​​​​​String s = br.readLine();
​​​​​​​​​​​​String rgs = s.replaceAll("G", "R");
​​​​​​​​​​​​for (int j = 0; j < N; j++) {
​​​​​​​​​​​​​​​​org[i][j] = s.charAt(j);
​​​​​​​​​​​​​​​​rg[i][j] = rgs.charAt(j);
​​​​​​​​​​​​}
​​​​​​​​}

​​​​​​​​int first = dfs(org);
​​​​​​​​visited = new boolean[N][N];
​​​​​​​​int second = dfs(rg);
​​​​​​​​System.out.println(first + " " +second);
​​​​}

​​​​private static int dfs(char[][] arr) {
​​​​​​​​int count = 0;
​​​​​​​​for (int i = 0; i < N; i++) {
​​​​​​​​​​​​for (int j = 0; j < N; j++) {
​​​​​​​​​​​​​​​​if (!visited[i][j]) {
​​​​​​​​​​​​​​​​​​​​go(i, j, arr[i][j],arr);
​​​​​​​​​​​​​​​​​​​​count++;
​​​​​​​​​​​​​​​​}
​​​​​​​​​​​​}
​​​​​​​​}

​​​​​​​​return count;
​​​​}

​​​​private static void go(int r, int c, char alpha,char[][] arr) {
​​​​​​​​int[] dr = {-1, 0, 1, 0};
​​​​​​​​int[] dc = {0, -1, 0, 1};
​​​​​​​​visited[r][c]=true;
​​​​​​​​for (int d = 0; d <4; d++) {
​​​​​​​​​​​​int nr = r+dr[d];
​​​​​​​​​​​​int nc = c+dc[d];

​​​​​​​​​​​​if(boundaryCheck(nr,nc)&&!visited[nr][nc]&&arr[nr][nc]==alpha) {
​​​​​​​​​​​​​​​​visited[nr][nc]=true;
​​​​​​​​​​​​​​​​go(nr,nc,alpha, arr);
​​​​​​​​​​​​}
​​​​​​​​}
​​​​​​​​return;
​​​​}

​​​​private static boolean boundaryCheck(int nr, int nc) {
​​​​​​​​return (nr>=0)&&(nr<N)&&(nc>=0)&&(nc<N);
​​​​}


}