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[2023-09-29] wooyeol #271 #295
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늦어서 죄송합니다 여러분들 ㅠㅠㅠㅠ |
zsmalla
approved these changes
Sep 30, 2023
| answer_dict[text] += 1 | ||
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| # 문자열의 길이가 5 이상이라면 더 이상 검색하지 않는다. | ||
| if len(text) >= 5: |
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큐에 삽입할 때(52번 라인) 이 조건문을 추가해주면 연산 횟수도 줄고 조금 더 깔끔해지지 않을까 싶습니다!
| N, M, K = map(int, input().split()) | ||
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| # N by M 의 격자 | ||
| table = [] |
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별 다른 추가 연산이 없기 때문에 57 ~ 59번 라인을 table = [input().rstrip() for _ in range(N)] 으로 가능! 파이써닉하게 가봅시다
limstonestone
approved these changes
Sep 30, 2023
| answer_dict = defaultdict(int) | ||
| for row_i, row in enumerate(table): | ||
| for col_i, col in enumerate(row): | ||
| bfs(row_i, col_i) |
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우열님 이부분은 range(len()) 을 사용하셔도 괜찮을 것 같습니다!!
제 로컬 IDE 에서 돌려보니 약 1.75 배정도 속도차이가 나네요!
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도현님이 코멘트 남겨주신 것처럼, 위에서 table의 크기인 N과 M 주어지기 때문에
for row_i in range(N):
for col_i in range(M):
bfs(row_i, col_i)
로 쓰시면 더 깔끔하고 빠를것 같아요~!!!
ksumini
approved these changes
Oct 4, 2023
| answer_dict = defaultdict(int) | ||
| for row_i, row in enumerate(table): | ||
| for col_i, col in enumerate(row): | ||
| bfs(row_i, col_i) |
There was a problem hiding this comment.
도현님이 코멘트 남겨주신 것처럼, 위에서 table의 크기인 N과 M 주어지기 때문에
for row_i in range(N):
for col_i in range(M):
bfs(row_i, col_i)
로 쓰시면 더 깔끔하고 빠를것 같아요~!!!
| # 상, 하, 좌, 우, 대각선 왼쪽 위, 대각선 오른쪽 위, 대각선 왼쪽 아래, 대각선 오른쪽 아래 | ||
| directions = ((-1,0),(1,0),(0,-1),(0,1),(-1,-1),(-1,1),(1,-1),(1,1)) | ||
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| def bfs(x, y): |
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저는 DFS로 풀이했는데, 우열님께서 BFS로 깔끔하게 풀이해주신것 같아요~!! 고생하셨습니다 :)
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PR Summary
문자열 지옥에 빠진 호석
https://www.acmicpc.net/problem/20166
풀이시간
12:42 ~ (문제 풀이 실패)
문제 조건
3 <= N,M <= 10
1 <= K <= 1,000
1 <= 신이 좋아하는 문자열 <= 5
시간 복잡도 :
O(N * M * 8^5 + K)
O(3,277,800)
접근법
무슨 알고리즘으로 풀이 할 수 있을까? -> BFS 탐색