[BOJ]#13913. 숨박꼭질4/골드4/1hour

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

Author: zaqquum

Hongjoo/백준/숨박꼭질4.py

@@ -0,0 +1,48 @@
+import sys
+from collections import deque
+input = sys.stdin.readline
+N, K = map(int, input().split())
+MAX = 100000
+# answer = [] 
+#[1] 최단 시간 찾기 
+visited = [MAX]* (MAX+1) # level 기입
+t= 0 
+next_node = [N]
+q = deque([N])
+visited[N] = t
+while q :
+  t+=1 
+  next_node = len(q)
+  for i in range(next_node) : # 현재 level의 node개수만큼 반복
+    cx = q.popleft()
+    # 만약 목적 달성시 , 끝
+    for nx in [cx -1 , cx+1 , cx*2]:
+      if 0<= nx <= MAX and visited[nx]>= MAX : 
+        q.append(nx)
+        visited[nx] = t   
+  # 현재 q -> 다음 level 의 노드만 남아있는 상태
+  # 만약 K을 도달한 경우-> 최단 시간 저장
+  if visited[K] <MAX : 
+    # answer.append(t)
+    break
+
+#[2]역추적K-> N - backtracking , DFS
+re_visited = []
+stack = deque([[K,t]])
+while stack : 
+  cx, ct = stack.pop()
+  if cx not in re_visited : 
+    re_visited.append(cx)
+    if cx == N : 
+      break
+    if cx %2==0 : #짝수면
+      ad_li = [cx-1,cx+1 , cx//2]
+    else : 
+      ad_li = [cx-1, cx+1]
+    # print("##",ad_li , type(ad_li))
+    for nx in ad_li :
+      if 0<=nx <=MAX and nx not in re_visited and visited[nx] == ct-1: 
+        stack.append([nx,ct-1]) 
+print(t)
+print(" ".join(map(str,list(reversed(re_visited)))))
+