オフラインリアルタイムどう書くE04 の問題をPythonで解く

問題と他者の回答はこちら
http://qiita.com/mtsmfm/items/6d9112fcc568908caaba

path は登山道。["A"～"H"] 毎の [頂上～麓の階段(横道の数+1)] の2次元配列。
path の内容は、真っすぐ上る階段は 0、右の登山道に移動する階段は +1、左の登山道に移動する階段は -1、落石で通れない階段は None。
step は階段の位置。頂上は [0]、麓は [-1](最後の要素)。
direction は横道の方角。1～8。
across は横道の順序および方角を表す文字列。
place および rock は登山道の位置。"A"～"H"。
index は登る/降りる位置。0～7 ("A"～"H"に対応するインデックス番号)。

# -*- coding:utf-8 -*-

PATH = "ABCDEFGH"

def make_path(across):
    path = [[0] * (len(across) + 1) for i in xrange(len(PATH))]
    for step, direction in enumerate(across):
        index = int(direction) % len(PATH)
        path[index+0][step] = -1
        path[index-1][step] = +1
    return path

def fall_rock(path, index):
    for step in xrange(len(path[0])):
        next_index = (index + path[index][step]) % len(PATH)
        path[index][step] = None
        index = next_index

def climb(path, index):
    for step in xrange(len(path[0]) - 1, -1, -1):
        if path[index][step] is None:
            return False
        index = (index + path[index][step]) % len(PATH)
    return True

def solve(data):
    across, rock = data.split(':')
    path = make_path(across)
    fall_rock(path, PATH.index(rock))
    climbable_places = [place
                        for index, place in enumerate(PATH)
                        if climb(path, index)]
    return ''.join(climbable_places)

if __name__ == '__main__':
    testdata = (
        (0, "2512:C", "DEFGH"),
        (1, "1:A", "CDEFGH"),
        (2, ":C", "ABDEFGH"),
        (3, "2345:B", "AGH"),
        (4, "1256:E", "ABCDH"),
        (5, "1228:A", "ADEFG"),
        (6, "5623:B", "AEFGH"),
        (7, "8157:C", "ABDEFGH"),
        (8, "74767:E", "ABCFGH"),
        (9, "88717:D", "ABCEFGH"),
        (10, "148647:A", "ACDEFH"),
        (11, "374258:H", "BCDEFH"),
        (12, "6647768:F", "ABCDEH"),
        (13, "4786317:E", "ABFGH"),
        (14, "3456781:C", ""),
        (15, "225721686547123:C", "CEF"),
        (16, "2765356148824666:F", "ABCDEH"),
        (17, "42318287535641783:F", "BDE"),
        (18, "584423584751745261:D", "FGH"),
        (19, "8811873415472513884:D", "CFG"),
        (20, "74817442725737422451:H", "BCDEF"),
        (21, "223188865746766511566:C", "ABGH"),
        (22, "2763666483242552567747:F", "ABCG"),
        (23, "76724442325377753577138:E", "EG"),
        (24, "327328486656448784712618:B", ""),
        (25, "4884637666662548114774288:D", "DGH"),
        (26, "84226765313786654637511248:H", "DEF"),
        (27, "486142154163288126476238756:A", "CDF"),
        (28, "1836275732415226326155464567:F", "BCD"),
        (29, "62544434452376661746517374245:G", "G"),
        (30, "381352782758218463842725673473:B", "A"),
        )
    def test():
        for no, data, correct in testdata:
            answer = solve(data)
            print '%s #%d "%s" "%s" -> "%s"' % (('NG!!','OK  ')[answer == correct], no, data, correct, answer)
    test()

横道設定が "2512"、落石位置が C の場合の解説

path は登山道。["A"～"H"] 毎の [頂上～麓の階段(横道の数+1)] の2次元配列。
真っすぐ上る場合は +0 、右にずれる場合は +1、左にずれる場合は -1、落石で通行できない場合は None。

初期状態の登山道。すべて真っすぐ上る道。

横道の設定が 2512