python_백준BOJ_#2116_주사위 쌓기_review

접근 방법

Only one combination of the orientation exists for every orientation of the first block.

Thus, for each of the six first-block-orientation, sum up the max value of each "floor", then find the ultimate max value amongst the six sums.

 

최종 입력 코드

import sys

n = int(sys.stdin.readline())
tower = []
down_up_dic = {0:5, 1:3, 2:4, 3:1, 4:2, 5:0}

for i in range(n):
    tower.append(list(map(int, input().split())))

max_alpha = 0

for j in range(6):
    max_beta = 0
    first_dice = [1,2,3,4,5,6]

    down_num = tower[0][j]
    up_num = tower[0][down_up_dic[j]]
    
    first_dice.remove(down_num)
    first_dice.remove(up_num)
    max_beta += (max(first_dice))

    for k in range(1,n):
        next_dice = [1,2,3,4,5,6]

        down_index = tower[k].index(up_num)
        up_index = down_up_dic[down_index] #계속 업데이트됨 durng for 구문
        
        down_num = tower[k][down_index]
        up_num = tower[k][up_index]

        next_dice.remove(down_num)
        next_dice.remove(up_num)

        max_beta += max(next_dice)

    if max_alpha < max_beta:
        max_alpha = max_beta

print(max_alpha)

• down_up_dic를 dictionary 말고 list로 했어도 가능했을 것 같음
• max_alpha를 for loop가 진행됨에 따라 업데이트함

문제 유형: bruteforcing, implementaion | 브루트포스, 구현

문제 출처: 한국정보올림피아드 KOI 2003 초등부 2번, hwww.acmicpc.net/problem/2116