RGB거리 2 풀이

문제 링크: https://www.acmicpc.net/problem/17404

백준 알고리즘 기초 1/2 402에서 2번째 - 17404번 RGB거리 2를 풀어보았다.

 

풀이: https://dalseoin.tistory.com/entry/%EB%B0%B1%EC%A4%80-%ED%8C%8C%EC%9D%B4%EC%8D%AC-17404-RGB%EA%B1%B0%EB%A6%AC-2 를 참고했다.

 

C++

 

Python

n=int(input())
arr=[]
inf=10000
res=inf
for i in range(n):
	arr.append(list(map(int,input().split())))
dp=[[0]*3 for i in range(2)]
for i in range(3):
	dp[0][0],dp[0][1],dp[0][2]=inf,inf,inf
	dp[0][i]=arr[0][i]
	for j in range(1,n):
		dp[1][0]=min(dp[0][1],dp[0][2])+arr[j][0]
		dp[1][1]=min(dp[0][0],dp[0][2])+arr[j][1]
		dp[1][2]=min(dp[0][0],dp[0][1])+arr[j][2]
		dp[0][0],dp[0][1],dp[0][2]=dp[1][0],dp[1][1],dp[1][2]
	res=min(res,min(dp[0][(i+1)%3],dp[0][(i+2)%3]))
print(res)

Java