BOJ - 숨바꼭질3(13549)

골드5 : BFS 문제이다.

풀이

cost에 따라 탐색하지 않을 노드를 분명히 정해야 한다.

Code

from collections import deque
import sys
 
 
input = sys.stdin.readline
 
MAXNUM = 1e6
N, K = map(int, input().split())
visited = [False] * int(MAXNUM + 1)
ans = 1e6
 
 
def BFS():
    global ans
    q = deque()
    q.append((N, 0))  # position, cost
 
    while q:
        pos, cost = q.popleft()
        visited[pos] = True
 
        if cost > ans:
            continue
 
        if pos == K:
            ans = min(ans, cost)
 
        nexts = [(pos - 1, cost + 1), (pos + 1, cost + 1), (2 * pos, cost)]
        for npos, ncost in nexts:
            if 0 <= npos <= MAXNUM and visited[npos] == False:
                q.append((npos, ncost))
 
 
BFS()
print(ans)

Reference

• 백준(13549번) - 숨바꼭질3