seyeonHello

import sysfrom collections import dequedirs = [(0, 1), (1, 0), (-1, 0), (0, -1)]board = []red,blue=[],[]N,M=0,0answer=sys.maxsizevisit=[]# 백트래킹def dfs(red,blue,count): global answer #종료지점 도착했는지 if board[red[0]][red[1]]==3 and board[blue[0]][blue[1]]==4: answer=min(answer,count) # 파란 수레가 이미 도착했다면 if board[blue[0]][blue[1]]==4: for i in range(4): rnr,rnc=red[0]+dir..

def solution(K, words): # 여기에 코드를 작성해주세요. answer = 1 cnt=0 for i in range(len(words)): if cnt+len(words[i])

2차원 배열을 사용하여 구현하였습니다. 벽 부수고 이동하기 2 : https://seyeon-hello.tistory.com/62 import sys from collections import deque input=sys.stdin.readline N,M,K=map(int,input().split()) arr=[list(map(int,input().strip())) for _ in range(N)] dirs=[(1,0),(0,1),(-1,0),(0,-1)] q=deque() q.append((0,0,0,1,1)) #가로,세로,벽,이동,낮 visit=[[sys.maxsize]*M for _ in range(N)] visit[0][0]=0 def bfs(): while q: r,c,cnt,move,day=q..

3차원 배열을 사용 안하는 방법으로 구현하였습니다. 아래 블로그를 참고했습니다. https://hongjw1938.tistory.com/163 import sys from collections import deque input=sys.stdin.readline n,m,k=map(int,input().split()) arr=[list(map(int,input().strip())) for _ in range(n)] dirs=[(1,0),(0,1),(-1,0),(0,-1)] visit=[[sys.maxsize for _ in range(m)] for _ in range(n)] q=deque() q.append((0,0,0,1)) visit[0][0]=1 answer=-1 while q: r,c,cnt,move..

import sys from collections import deque input=sys.stdin.readline dirs=[(1,0),(0,1),(-1,0),(0,-1)] N=int(input()) def bfs(): count=0 while True: count+=1 for j in range(len(fire)): r,c=fire.popleft() for i in range(4): nr,nc=r+dirs[i][0],c+dirs[i][1] if 0>nr or nr>=h or nc=w: continue if building[nr][nc]!='#' and building[nr][nc]!='*': building[nr][nc]='*' fire.append([nr,nc]) for j in range(len..

리스트에 방문했던 위치를 계속 누적했을 경우 메모리초과가 나와서, 딕셔너리를 통해 해당 위치의 전 위치를 저장하는 방법으로 구현하였습니다. import sys from collections import deque, defaultdict input=sys.stdin.readline N,K=map(int,input().split()) visit=[0]*100001 q=deque() dicts=defaultdict(int) q.append(N) while q: num=q.popleft() if num==K: print(visit[num]) answer=[] while num!=N: answer.append(num) num=dicts[num] answer.append(N) answer.reverse() prin..

import sys from collections import deque input=sys.stdin.readline N=int(input()) arr=[list(map(int,input().split())) for _ in range(N)] time=0 babySize=2 dirs=[(1,0),(0,1),(-1,0),(0,-1)] for r in range(N): for c in range(N): if arr[r][c]==9: baby=(r,c,0) answer=0 cnt=0 while True: q=deque() q.append(baby) visit=[[0]*N for _ in range(N)] visit[baby[0]][baby[1]]=1 remove=[] while q: r,c,time=q.pop..

필자는 통나무 모든 좌표를 방문체크를 했는데, 타 블로그 코드들을 확인해보니 중심좌표만 방문체크를 하고 있었다. 이 점을 개선해나가면 좋을 것 같다. import sys from collections import deque input=sys.stdin.readline N=int(input()) board=[list(input().strip()) for _ in range(N)] visit=[[[0]*2 for _ in range(N)] for _ in range(N)] dirs=[(1,0),(0,1),(-1,0),(0,-1),[(1,0),(0,0),(-1,0)],[(0,1),(0,0),(0,-1)]] slog=[] dlog=[] for r in range(N): for c in range(N): if b..

import sys from collections import deque input=sys.stdin.readline N,M=map(int,input().split()) arr=[list(map(int,input().split())) for _ in range(N)] dirs=[(1,0),(0,1),(-1,0),(0,-1)] score=0 def bfs(row,col,color): global removes, cmax visit = [[0] * N for _ in range(N)] q=deque([(row,col)]) colors=[(row,col)] visit[row][col]=1 remove=[(row,col)] cnt=0 mcnt=1 # 일반 블록 개수 while q: r,c=q.popleft() ..

3차원 배열 사용 import sys from collections import deque input=sys.stdin.readline dirs=[(1,0,0),(-1,0,0),(0,1,0),(0,0,1),(0,-1,0),(0,0,-1)] #L,R,C while True: L,R,C=map(int,input().split()) visit=[[[0]*C for _ in range(R)] for _ in range(L)] if L==0 and R==0 and C==0: break building=[] floor=[] for _ in range(L*R+L): tmp=list(input().strip()) if len(tmp)==0: building.append(floor) floor=[] else: floor..