14503 로봇 청소기

https://www.acmicpc.net/problem/14503

14503번: 로봇 청소기

n,m = map(int,input().split())
r,c,d = map(int,input().split())
graph = [list(map(int,input().split())) for _ in range(n)]
visited = [[0]*m for _ in range(n)]
#처음 빈칸은 전부 청소되지 않은 상태이다.
#순서대로 구현하자!!
#청소하는칸 카운트
cnt=0

#방향 조건에 맞게 상우하좌 1씩 빼면 반시계 90도 0일때는 3으로
dx=[-1,0,1,0]
dy=[0,1,0,-1]
while 1:
    flag = 0
    if visited[r][c]==0:
        visited[r][c]=1
        cnt+=1
    #현재 칸의 주변 4칸 중 청소되지 않은 빈 칸이 없는 경우
    for i in range(4):
        nx=r+dx[i]
        ny=c+dy[i]
        #범위 안에 있으며, 방문하지 않았고, 벽이 아니라 빈칸이면
        if 0<=nx<n and 0<=ny<m and visited[nx][ny]==0 and graph[nx][ny]==0:
            #청소안한 빈칸 있는거
            flag=1
    #빈칸없어 뒤로 후진해야됨
    if flag==0:
        if d==0:
            nx = r + dx[2]
            ny = c + dy[2]
            if 0 <= nx < n and 0 <= ny < m and visited[nx][ny] == 0 and graph[nx][ny] == 0:
                cnt+=1
                visited[nx][ny] = 1
                r,c=nx,ny
            elif 0 <= nx < n and 0 <= ny < m and visited[nx][ny] == 1 and graph[nx][ny] == 0:
                r, c = nx, ny
            elif 0 <= nx < n and 0 <= ny < m and graph[nx][ny] == 1:
                print(cnt)
                break
        elif d==1:
            nx = r + dx[3]
            ny = c + dy[3]
            if 0 <= nx < n and 0 <= ny < m and visited[nx][ny] == 0 and graph[nx][ny] == 0:
                cnt+=1
                visited[nx][ny] = 1
                r, c = nx, ny
            elif 0 <= nx < n and 0 <= ny < m and visited[nx][ny] == 1 and graph[nx][ny] == 0:
                r, c = nx, ny
            elif 0 <= nx < n and 0 <= ny < m and graph[nx][ny] == 1:
                print(cnt)
                break
        elif d==2:
            nx = r + dx[0]
            ny = c + dy[0]
            if 0 <= nx < n and 0 <= ny < m and visited[nx][ny] == 0 and graph[nx][ny] == 0:
                cnt+=1
                visited[nx][ny] = 1
                r, c = nx, ny
            elif 0 <= nx < n and 0 <= ny < m and visited[nx][ny] == 1 and graph[nx][ny] == 0:
                r, c = nx, ny
            elif 0 <= nx < n and 0 <= ny < m and graph[nx][ny] == 1:
                print(cnt)
                break
        elif d==3:
            nx = r + dx[1]
            ny = c + dy[1]
            if 0 <= nx < n and 0 <= ny < m and visited[nx][ny] == 0 and graph[nx][ny] == 0:
                cnt+=1
                r, c = nx, ny
            elif 0 <= nx < n and 0 <= ny < m and visited[nx][ny] == 1 and graph[nx][ny] == 0:
                r, c = nx, ny
            elif 0 <= nx < n and 0 <= ny < m and graph[nx][ny] == 1:
                print(cnt)
                break

    elif flag==1:
        if d==0:
            #90도로 방향돌리기
            d=3

        else:
            #90도로 방향돌리기
            d-=1
        nx = r + dx[d]
        ny = c + dy[d]
        if 0 <= nx < n and 0 <= ny < m and visited[nx][ny] == 0 and graph[nx][ny] == 0:
            #방문표시
            visited[nx][ny]=1
            r, c = nx, ny
            cnt+=1

리펙토링이 필요하긴하다.. 

실전에서는?? 함수화하는게 관리가 편하려나