結果

問題 No.1045 直方体大学
ユーザー tcltktcltk
提出日時 2021-05-19 05:42:19
言語 PyPy3
(7.3.15)
結果
TLE  
実行時間 -
コード長 1,937 bytes
コンパイル時間 349 ms
コンパイル使用メモリ 82,176 KB
実行使用メモリ 194,180 KB
最終ジャッジ日時 2024-10-09 09:58:45
合計ジャッジ時間 26,229 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 7 TLE * 10
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#region Header
#!/usr/bin/env python3
# from typing import *
import sys
import io
import math
import collections
import decimal
import itertools
import bisect
import heapq
def input():
return sys.stdin.readline()[:-1]
# sys.setrecursionlimit(1000000)
#endregion
# _INPUT = """2
# 2 7 3
# 5 1 4
# """
# sys.stdin = io.StringIO(_INPUT)
def is_loadable(lower1, lower2, upper1, upper2):
return max(lower1, lower2) >= max(upper1, upper2) and min(lower1, lower2) >= min(upper1, upper2)
def main():
N = int(input())
Box = []
for i in range(N):
a, b, c = map(int, input().split())
Box.append((a, b, c))
Loadable = [[[[False] * 3 for _ in range(N)] for _ in range(3)] for _ in range(N)]
for i1 in range(N):
for k1 in range(3):
for i2 in range(N):
if i2 == i1:
continue
for k2 in range(3):
if is_loadable(Box[i1][(k1+1)%3], Box[i1][(k1+2)%3], Box[i2][(k2+1)%3], Box[i2][(k2+2)%3]):
Loadable[i1][k1][i2][k2] = True
dp = [[[0] * 3 for _ in range(N)] for _ in range(1<<N)]
for i in range(N):
dp[1<<i][i][0] = Box[i][0]
dp[1<<i][i][1] = Box[i][1]
dp[1<<i][i][2] = Box[i][2]
for s1 in range(1, 1<<N):
for i1 in range(N):
if not(s1 & (1<<i1)):
continue
for i2 in range(N):
if s1 & (1<<i2):
continue
s2 = s1 | (1<<i2)
for k1 in range(3):
dp[s2][i1][k1] = max(dp[s2][i1][k1], dp[s1][i1][k1])
for k2 in range(3):
if Loadable[i1][k1][i2][k2]:
dp[s2][i2][k2] = max(dp[s2][i2][k2], dp[s1][i1][k1] + Box[i2][k2])
ans = max(max(dp[(1<<N)-1][i]) for i in range(N))
print(ans)
if __name__ == '__main__':
main()
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