백준 10026 적록색약] 파이썬 풀이 🦄

• [백준 10026 적록색약] 파이썬 풀이 🦄
  • 문제
  • Keypoints
  • Solution
  • References

[백준 10026 적록색약] 파이썬 풀이 🦄

문제

동일한 값에 대해서 bfs/dfs를 통하여 탐색하는 기본 문제

Keypoints

• 두 가지 버전에 대해서 탐색해야 하므로, board 를 Copy한다.

Solution

dx = [0,0,1,-1]
dy = [1,-1,0,0]

def bfs3(board, x,y):
    color = board[y][x]
    board[y][x]=0
    queue = [(x,y)]
    while queue:
        x,y = queue.pop(0)
        for i in range(4):
            nx, ny = x+dx[i], y+dy[i]
            if 0<=nx<n and 0<=ny<n and board[ny][nx] == color:
                queue.append((nx,ny))
                board[ny][nx]=0

def bfs2(board, x,y):
    color = board[y][x]
    board[y][x]=0
    if color=="G" or color=="R":
        color_list=["R", "G"]
    else:
        color_list =["B"]
    queue = [(x,y)]
    while queue:
        x,y = queue.pop(0)
        for i in range(4):
            nx, ny = x+dx[i], y+dy[i]
            if 0<=nx<n and 0<=ny<n and board[ny][nx] in color_list:
                    queue.append((nx,ny))
                    board[ny][nx]=0


def count(board, bfs):
    count = 0
    for x in range(n):
        for y in range(n):
            if board[y][x] != 0:
                bfs(board, x,y)
                count +=1
    return count

import copy 
n = int(input())
board = [list(input().strip()) for i in range(n)]
board2 = copy.deepcopy(board)

a = count(board,bfs3)
b = count(board2,bfs2)
print(a, b)

References

Problem Link