광물캐기

def cal_score(pick, mineral): ''' mineral: ['diamond', 'iron', 'stone'] ''' if len(mineral) == 0: return 0 my_dict = {'diamond': 0, 'iron': 1, 'stone': 2} # 곡괭이와 광물을 index로 바꾸기위한 dictionary fatigue_table = [[1, 1, 1], [5, 1, 1], [25, 5, 1]] # fatigue 점수 fatigue = 0 for m in mineral: fatigue += fatigue_table[my_dict[pick]][my_dict[m]] return fatigue def solution(picks, minerals): minerals = minerals[:sum(picks)*5] # 캘 수 있는 양만 남기기 * 중요! batch = [minerals[5*i:5*(i+1)] for i in range(len(minerals)//5+1)] # 5개씩 자르기 batch.sort(key=lambda x: (-x.count('diamond'), -x.count('iron'), -x.count('stone'))) # diamond, iron, stone 갯수 순으로 sort temp = ['diamond', 'iron', 'stone'] my_picks = [] for i in range(3): my_picks.extend([temp[i]]*picks[i]) # 곡괭이를 나열 fatigue = 0 for mineral in batch: if len(my_picks) == 0: break pick = my_picks.pop(0) # 곡괭이 사용 fatigue += cal_score(pick, mineral) return fatigue

def solution(picks, minerals): def solve(picks, minerals, fatigue): if sum(picks) == 0 or len(minerals) == 0: return fatigue result = [float('inf')] for i, fatigues in enumerate(({"diamond": 1, "iron": 1, "stone": 1}, {"diamond": 5, "iron": 1, "stone": 1}, {"diamond": 25, "iron": 5, "stone": 1},)): if picks[i] > 0: temp_picks = picks.copy() temp_picks[i] -= 1 result.append( solve(temp_picks, minerals[5:], fatigue + sum(fatigues[mineral] for mineral in minerals[:5]))) return min(result) return solve(picks, minerals, 0)