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[이지민] - A와 B 2, 알고스팟, 회장뽑기, 입국심사 #189
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''' | ||
A와 B 문제를 풀고 아이디어를 얻음 | ||
(그냥 완탐시 시간초과) | ||
S -> T 이니까 반대로 T를 S로 만들도록 함 | ||
뒤가 'A'면 그냥 맨 끝을 자르고 | ||
앞이 'B'이면 앞을 하나 자르고 뒤집는다! | ||
''' | ||
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import sys | ||
sys.setrecursionlimit(10**6) | ||
a = list(sys.stdin.readline().rstrip()) | ||
b = list(sys.stdin.readline().rstrip()) | ||
result = 0 | ||
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def set_recursion(b): | ||
global result, a | ||
if len(a) == len(b): | ||
if a == b: | ||
result = 1 | ||
return | ||
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if b[-1] == 'A': | ||
set_recursion(b[:-1]) | ||
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if b[0] == 'B': | ||
set_recursion(b[:0:-1]) | ||
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set_recursion(b) | ||
print(result) |
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''' | ||
0-1 너비 우선 탐색 이용! | ||
''' | ||
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import sys | ||
from collections import deque | ||
m, n = map(int, sys.stdin.readline().split()) | ||
miros = [list(map(int, sys.stdin.readline().rstrip())) for _ in range(n)] | ||
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queue = deque([[0, 0, 0]]) | ||
visited = [[0 for _ in range(m)] for _ in range(n)] | ||
visited[0][0] = 1 | ||
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dx = [0, 0, -1, 1] | ||
dy = [-1, 1, 0, 0] | ||
result = -1 | ||
while queue: | ||
nx, ny, num = queue.popleft() | ||
if nx == n - 1 and ny == m - 1: | ||
result = num | ||
break | ||
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for i in range(4): | ||
if 0 <= nx + dx[i] < n and 0 <= ny + dy[i] < m and not visited[nx + dx[i]][ny + dy[i]]: | ||
new = miros[nx + dx[i]][ny + dy[i]] | ||
if new == 0: | ||
queue.appendleft([nx + dx[i], ny + dy[i], num]) | ||
else: | ||
queue.append([nx + dx[i], ny + dy[i], num + 1]) | ||
visited[nx + dx[i]][ny + dy[i]] = 1 | ||
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print(result) |
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import sys | ||
def solution(n, times): | ||
answer = sys.maxsize | ||
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front = 1 | ||
last = 1_000_000_000 * 1_000_000_000 | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 파이썬은 이런게 부럽네요... |
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while front <= last: | ||
mid = (front + last) // 2 | ||
num = 0 | ||
for t in times: | ||
num += mid // t | ||
if num < n: | ||
front = mid + 1 | ||
else: | ||
last = mid - 1 | ||
answer = mid | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. answer 를 여기서 최신화 시켜서 리턴 시키는 게 더 잘 읽히네요! |
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return answer |
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''' | ||
bfs를 활용해서 문제를 해결했다 | ||
''' | ||
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import sys | ||
from collections import deque | ||
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n = int(sys.stdin.readline()) | ||
friends = [[] for _ in range(n + 1)] | ||
info = [0 for _ in range(n + 1)] | ||
while True: | ||
a, b = map(int, sys.stdin.readline().split()) | ||
if a == -1 and b == -1: | ||
break | ||
friends[a].append(b) | ||
friends[b].append(a) | ||
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def bfs(target): | ||
global n, info | ||
queue = deque([[target, 0]]) | ||
visited = [0 for _ in range(n + 1)] | ||
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while queue: | ||
now, num = queue.popleft() | ||
for i in friends[now]: | ||
if i != target and not visited[i]: | ||
queue.append([i, num + 1]) | ||
visited[i] = num + 1 | ||
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return num | ||
Comment on lines
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to
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 엄청 깔끔하게 로직 잘 짜신 것 같아요..!! There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 오 점수를 이렇게 한 번에 구하신 게 좋네요! There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. num이 리턴되니 로직이 훨씬 깔끔한 것 같아요 멋집니당 |
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for i in range(1, n + 1): | ||
info[i] = bfs(i) | ||
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mini = min(info[1:]) | ||
print(mini, info.count(mini)) | ||
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for i in range(1, n + 1): | ||
if info[i] == mini: | ||
print(i, end=" ") |
There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
와 부럽다 파이썬...