递增序列 python实现1111

答案：180414

问题

对于一个字母矩阵，我们称矩阵中的一个三升序列是指在矩阵中找到三个字母，它们在同一行，同一列，或者在同一 45 度的斜线上，这三个字母从左向右看、或者从上向下看是递增的。

例如，如下矩阵中
YQPD
BKEZ
AFYV

有BKZ、BEZ、AFY、AFV、AKP、DEF 等 6 个三升序列。注意当三个字母是从左下到右上排列时，从左向右看和从上向下看是不同的顺序。

对于下面的 30 行 50 列的矩阵，请问总共有多少个三升序列？

VLPWJVVNNZSWFGHSFRBCOIJTPYNEURPIGKQGPSXUGNELGRVZAG
SDLLOVGRTWEYZKKXNKIRWGZWXWRHKXFASATDWZAPZRNHTNNGQF
ZGUGXVQDQAEAHOQEADMWWXFBXECKAVIGPTKTTQFWSWPKRPSMGA
BDGMGYHAOPPRRHKYZCMFZEDELCALTBSWNTAODXYVHQNDASUFRL
YVYWQZUTEPFSFXLTZBMBQETXGXFUEBHGMJKBPNIHMYOELYZIKH
ZYZHSLTCGNANNXTUJGBYKUOJMGOGRDPKEUGVHNZJZHDUNRERBU
XFPTZKTPVQPJEMBHNTUBSMIYEGXNWQSBZMHMDRZZMJPZQTCWLR
ZNXOKBITTPSHEXWHZXFLWEMPZTBVNKNYSHCIQRIKQHFRAYWOPG
MHJKFYYBQSDPOVJICWWGGCOZSBGLSOXOFDAADZYEOBKDDTMQPA
VIDPIGELBYMEVQLASLQRUKMXSEWGHRSFVXOMHSJWWXHIBCGVIF
GWRFRFLHAMYWYZOIQODBIHHRIIMWJWJGYPFAHZZWJKRGOISUJC
EKQKKPNEYCBWOQHTYFHHQZRLFNDOVXTWASSQWXKBIVTKTUIASK
PEKNJFIVBKOZUEPPHIWLUBFUDWPIDRJKAZVJKPBRHCRMGNMFWW
CGZAXHXPDELTACGUWBXWNNZNDQYYCIQRJCULIEBQBLLMJEUSZP
RWHHQMBIJWTQPUFNAESPZHAQARNIDUCRYQAZMNVRVZUJOZUDGS
PFGAYBDEECHUXFUZIKAXYDFWJNSAOPJYWUIEJSCORRBVQHCHMR
JNVIPVEMQSHCCAXMWEFSYIGFPIXNIDXOTXTNBCHSHUZGKXFECL
YZBAIIOTWLREPZISBGJLQDALKZUKEQMKLDIPXJEPENEIPWFDLP
HBQKWJFLSEXVILKYPNSWUZLDCRTAYUUPEITQJEITZRQMMAQNLN
DQDJGOWMBFKAIGWEAJOISPFPLULIWVVALLIIHBGEZLGRHRCKGF
LXYPCVPNUKSWCCGXEYTEBAWRLWDWNHHNNNWQNIIBUCGUJYMRYW
CZDKISKUSBPFHVGSAVJBDMNPSDKFRXVVPLVAQUGVUJEXSZFGFQ
IYIJGISUANRAXTGQLAVFMQTICKQAHLEBGHAVOVVPEXIMLFWIYI
ZIIFSOPCMAWCBPKWZBUQPQLGSNIBFADUUJJHPAIUVVNWNWKDZB
HGTEEIISFGIUEUOWXVTPJDVACYQYFQUCXOXOSSMXLZDQESHXKP
FEBZHJAGIFGXSMRDKGONGELOALLSYDVILRWAPXXBPOOSWZNEAS
VJGMAOFLGYIFLJTEKDNIWHJAABCASFMAKIENSYIZZSLRSUIPCJ
BMQGMPDRCPGWKTPLOTAINXZAAJWCPUJHPOUYWNWHZAKCDMZDSR
RRARTVHZYYCEDXJQNQAINQVDJCZCZLCQWQQIKUYMYMOVMNCBVY
ABTCRRUXVGYLZILFLOFYVWFFBZNFWDZOADRDCLIRFKBFBHMAXX

思路代码

a = []
with open(r'C:\Users\76390\Desktop\1.txt', 'r') as fp:
    for line in fp.readlines():
        a.append(line.strip())
    fp.close()
ans = 0
dirs = [(0, 1), (1, 0), (1, 1), (1, -1), (-1, 1)]
for i in range(len(a)):
    for j in range(len(a[0])):
        for dir in dirs:
            x1, y1 = i, j
            while True:
                x1 += dir[0]
                y1 += dir[1]
                x2, y2 = x1, y1
                if x1 >= len(a) or y1 >= len(a[0]) or x1 < 0 or y1 < 0:
                    break

                while True:
                    x2 += dir[0]
                    y2 += dir[1]

                    if x2 >= len(a) or y2 >= len(a[0]) or x2 < 0 or y2 < 0:
                        break
                    if a[i][j] < a[x1][y1] < a[x2][y2]:
                        ans += 1

print(ans)