숨바꼭질 4 풀이

문제 링크: https://www.acmicpc.net/problem/13913

백준 알고리즘 기초 2/2 610에서 2번째 - 13913번 숨바꼭질 4를 풀어보았다.

 

풀이: https://velog.io/@ms269/%EB%B0%B1%EC%A4%80-13913-%EC%88%A8%EB%B0%94%EA%BC%AD%EC%A7%88-4-%ED%8C%8C%EC%9D%B4%EC%8D%AC-Python 를 참고했다.

 

C++

 

Python

from collections import deque

def path(a):
	arr=[]
	tmp=a
	for i in range(visited[a]+1):
		arr.append(tmp)
		tmp=course[tmp]
	return arr[::-1]

def bfs(a):
	q=deque([a])
	while q:
		b=q.popleft()
		if b==k: 
			return visited[b]
		for i in (b-1,b+1,2*b):
			if 0<=i<=100000 and not visited[i]:
				visited[i]=visited[b]+1
				q.append(i)
				course[i]=b

n,k=map(int,input().split())
visited=[0 for i in range(100001)]
course=[0 for i in range(100001)]
print(bfs(n))
print(*path(k))

Java