Jinho Tech

기본적인 BFS 문제이다. B에서 C로 가는 최단거리의 길이를 출력해주면 된다.

from collections import deque

def bfs(y, x):
    q = deque()
    q.append((y, x))
    graph[y][x] = 0
    while q:
        y, x = q.popleft()
        for dy, dx in d:
            Y, X = y+dy, x+dx
            if (0 <= Y < R) and (0 <= X < C) and graph[Y][X] in ['.', 'C']:
                if graph[Y][X] == 'C':
                    return graph[y][x]+1
                q.append((Y, X))
                graph[Y][X] = graph[y][x]+1

R, C = map(int, input().split())
graph = [list(input()) for _ in range(R)]
d = [(-1, 0), (1, 0), (0, -1), (0, 1)]
for i in range(R):
    for j in range(C):
        if graph[i][j] == 'B':
            print(bfs(i, j))
            break