사탕 게임 풀이

문제 링크: https://www.acmicpc.net/problem/3085

백준 알고리즘 기초 2/2 500에서 2번째 - 3085번 사탕 게임을 풀어보았다.

 

풀이: https://velog.io/@yibangwon/%EB%B0%B1%EC%A4%80-3085-%EC%82%AC%ED%83%95-%EA%B2%8C%EC%9E%84-%ED%8C%8C%EC%9D%B4%EC%8D%AC 를 참고했다.

 

C++

 

Python

def check(arr):
	n=len(arr)
	ans=1
	for i in range(n):
		cnt=1
		for j in range(1,n):
			if arr[i][j]==arr[i][j-1]: cnt+=1
			else: cnt=1
			if cnt>ans: ans=cnt
		cnt=1
		for j in range(1,n):
			if arr[j][i]==arr[j-1][i]: cnt+=1
			else: cnt=1
			if cnt>ans: ans=cnt
	return ans

n=int(input())
inp=[list(input()) for i in range(n)]
res=0
for i in range(n):
	for j in range(n):
		if j+1<n:
			inp[i][j],inp[i][j+1]=inp[i][j+1],inp[i][j]
			temp=check(inp)
			if temp>res: res=temp
			inp[i][j],inp[i][j+1]=inp[i][j+1],inp[i][j]
		if i+1<n:
			inp[i][j],inp[i+1][j]=inp[i+1][j],inp[i][j]
			temp=check(inp)
			if temp>res: res=temp
			inp[i][j],inp[i+1][j]=inp[i+1][j],inp[i][j]
print(res)

Java