구슬탈출
이 문제와 동일한데 구슬탈출 문제는 마지막에 1을 리턴했다면 이 문제는 갯수를 리턴해주기만 하면 되고 , 0 대신 -1 으로 바꾸면 됩니다.
import sys
from collections import deque
input = sys.stdin.readline
n, m = map(int, input().split())
graph = []
for i in range(n):
graph.append(list(sys.stdin.readline().rstrip()))
dx =[-1, 1, 0, 0]
dy = [0, 0, -1, 1]
def bfs(rx, ry, bx, by, cnt):
cnt = 0
visited[rx][ry][bx][by] = True
while q:
for _ in range(len(q)):
rx, ry, bx, by, cnt = q.popleft()
if cnt > 10:
return -1
if graph[rx][ry] =='O':
return cnt
for i in range(4):
rnx = rx
rny = ry
bnx = bx
bny = by
while True:
rnx += dx[i]
rny += dy[i]
if graph[rnx][rny]=='#':
rnx -= dx[i]
rny -= dy[i]
break
if graph[rnx][rny] == 'O':
break
while True:
bnx += dx[i]
bny += dy[i]
if graph[bnx][bny]=='#':
bnx -= dx[i]
bny -= dy[i]
break
if graph[bnx][bny] == 'O':
break
if graph[bnx][bny] == 'O':
continue
if rnx == bnx and rny == bny:
if abs(rnx - rx) + abs(rny - ry) > abs(bnx - bx) + abs(bny - by):
rnx -= dx[i]
rny -= dy[i]
else:
bnx -= dx[i]
bny -= dy[i]
if not visited[rnx][rny][bnx][bny]:
visited[rnx][rny][bnx][bny] = True
q.append((rnx, rny, bnx, bny, cnt+1))
return -1
red = []
blue = []
for i in range(n):
for j in range(m):
if graph[i][j] == 'R':
red.append((i, j))
if graph[i][j] == 'B':
blue.append((i, j))
q = deque()
q.append((red[0][0], red[0][1], blue[0][0], blue[0][1], 0))
visited = [[[[False for _ in range(m)]for _ in range(n)]for _ in range(m)]for _ in range(n)]
print(bfs(red[0][0], red[0][1], blue[0][0], blue[0][1], 0))