[백준, 10026번] 적록색약

풀이

• bfs를 사용해서 풀었으며 
• 적록색약의 경우는 visit1과 bfs1로 
• 아닐 경우엔 visit2와 bfs2로 풀 수 있도록 해결

 

import sys

from collections import deque

def bfs1():

    while q:

        y,x = q.popleft()

        for ny,nx in [[-1,0],[1,0],[0,-1],[0,1]]:

            dy,dx = y+ny,x+nx

            if 0<=dy<n and 0<=dx<n and not visit1[dy][dx] and arr[y][x] == arr[dy][dx]:

                visit1[dy][dx] = 1

                q.append([dy,dx])

    return

def bfs2():

    while q:

        y,x = q.popleft()

        for ny,nx in [[-1,0],[1,0],[0,-1],[0,1]]:

            dy,dx = y+ny,x+nx

            if 0<=dy<n and 0<=dx<n and not visit2[dy][dx]:

                if (arr[y][x] in ('R','G') and arr[dy][dx] in ('R','G')) or (arr[y][x] == 'B' and arr[dy][dx] == 'B'):

                    visit2[dy][dx] = 1

                    q.append([dy,dx])

    return

input = sys.stdin.readline

n = int(input())

arr = [list(input().rstrip('\n')) for _ in range(n)]

visit1 = [[0]*n for _ in range(n)]

visit2 = [[0]*n for _ in range(n)]

result1,result2 = 0,0

q = deque([])

for i in range(n):

    for j in range(n):

        if not visit1[i][j]: # 적록색약 x

            result1 += 1

            visit1[i][j] = 1

            q.append([i,j])

            bfs1()

        if not visit2[i][j]: # 적록색약 o

            result2 += 1

            visit2[i][j] = 1

            q.append([i,j])

            bfs2()

print(result1, end=' ')

print(result2)