Python基于生成器迭代實現(xiàn)的八皇后問題示例
本文實例講述了Python基于生成器迭代實現(xiàn)的八皇后問題。分享給大家供大家參考,具體如下:
問題:有一個棋盤和8個要放到上面的皇后,唯一的要求是皇后之間不能形成威脅。也就是說,必須把他們防止成每個皇后都不能吃掉其他皇后的狀態(tài)。
# -*- coding: utf-8 -*- #python 2.7.13 __metaclass__ = type def confict(state, nextX): nextY = len(state) for i in range(nextY): if abs(state[i] - nextX) in (0, nextY - i): return True return False def queens(num=8, state=()): for pos in range(num): if not confict(state, pos): if len(state) == num -1: yield (pos,) else: for result in queens(num, state + (pos,)): yield (pos,) + result print list(queens()) #打印輸出
運行結(jié)果:
[(0, 4, 7, 5, 2, 6, 1, 3), (0, 5, 7, 2, 6, 3, 1, 4), (0, 6, 3, 5, 7, 1, 4, 2), (0, 6, 4, 7, 1, 3, 5, 2), (1, 3, 5, 7, 2, 0, 6, 4), (1, 4, 6, 0, 2, 7, 5, 3), (1, 4, 6, 3, 0, 7, 5, 2), (1, 5, 0, 6, 3, 7, 2, 4), (1, 5, 7, 2, 0, 3, 6, 4), (1, 6, 2, 5, 7, 4, 0, 3), (1, 6, 4, 7, 0, 3, 5, 2), (1, 7, 5, 0, 2, 4, 6, 3), (2, 0, 6, 4, 7, 1, 3, 5), (2, 4, 1, 7, 0, 6, 3, 5), (2, 4, 1, 7, 5, 3, 6, 0), (2, 4, 6, 0, 3, 1, 7, 5), (2, 4, 7, 3, 0, 6, 1, 5), (2, 5, 1, 4, 7, 0, 6, 3), (2, 5, 1, 6, 0, 3, 7, 4), (2, 5, 1, 6, 4, 0, 7, 3), (2, 5, 3, 0, 7, 4, 6, 1), (2, 5, 3, 1, 7, 4, 6, 0), (2, 5, 7, 0, 3, 6, 4, 1), (2, 5, 7, 0, 4, 6, 1, 3), (2, 5, 7, 1, 3, 0, 6, 4), (2, 6, 1, 7, 4, 0, 3, 5), (2, 6, 1, 7, 5, 3, 0, 4), (2, 7, 3, 6, 0, 5, 1, 4), (3, 0, 4, 7, 1, 6, 2, 5), (3, 0, 4, 7, 5, 2, 6, 1), (3, 1, 4, 7, 5, 0, 2, 6), (3, 1, 6, 2, 5, 7, 0, 4), (3, 1, 6, 2, 5, 7, 4, 0), (3, 1, 6, 4, 0, 7, 5, 2), (3, 1, 7, 4, 6, 0, 2, 5), (3, 1, 7, 5, 0, 2, 4, 6), (3, 5, 0, 4, 1, 7, 2, 6), (3, 5, 7, 1, 6, 0, 2, 4), (3, 5, 7, 2, 0, 6, 4, 1), (3, 6, 0, 7, 4, 1, 5, 2), (3, 6, 2, 7, 1, 4, 0, 5), (3, 6, 4, 1, 5, 0, 2, 7), (3, 6, 4, 2, 0, 5, 7, 1), (3, 7, 0, 2, 5, 1, 6, 4), (3, 7, 0, 4, 6, 1, 5, 2), (3, 7, 4, 2, 0, 6, 1, 5), (4, 0, 3, 5, 7, 1, 6, 2), (4, 0, 7, 3, 1, 6, 2, 5), (4, 0, 7, 5, 2, 6, 1, 3), (4, 1, 3, 5, 7, 2, 0, 6), (4, 1, 3, 6, 2, 7, 5, 0), (4, 1, 5, 0, 6, 3, 7, 2), (4, 1, 7, 0, 3, 6, 2, 5), (4, 2, 0, 5, 7, 1, 3, 6), (4, 2, 0, 6, 1, 7, 5, 3), (4, 2, 7, 3, 6, 0, 5, 1), (4, 6, 0, 2, 7, 5, 3, 1), (4, 6, 0, 3, 1, 7, 5, 2), (4, 6, 1, 3, 7, 0, 2, 5), (4, 6, 1, 5, 2, 0, 3, 7), (4, 6, 1, 5, 2, 0, 7, 3), (4, 6, 3, 0, 2, 7, 5, 1), (4, 7, 3, 0, 2, 5, 1, 6), (4, 7, 3, 0, 6, 1, 5, 2), (5, 0, 4, 1, 7, 2, 6, 3), (5, 1, 6, 0, 2, 4, 7, 3), (5, 1, 6, 0, 3, 7, 4, 2), (5, 2, 0, 6, 4, 7, 1, 3), (5, 2, 0, 7, 3, 1, 6, 4), (5, 2, 0, 7, 4, 1, 3, 6), (5, 2, 4, 6, 0, 3, 1, 7), (5, 2, 4, 7, 0, 3, 1, 6), (5, 2, 6, 1, 3, 7, 0, 4), (5, 2, 6, 1, 7, 4, 0, 3), (5, 2, 6, 3, 0, 7, 1, 4), (5, 3, 0, 4, 7, 1, 6, 2), (5, 3, 1, 7, 4, 6, 0, 2), (5, 3, 6, 0, 2, 4, 1, 7), (5, 3, 6, 0, 7, 1, 4, 2), (5, 7, 1, 3, 0, 6, 4, 2), (6, 0, 2, 7, 5, 3, 1, 4), (6, 1, 3, 0, 7, 4, 2, 5), (6, 1, 5, 2, 0, 3, 7, 4), (6, 2, 0, 5, 7, 4, 1, 3), (6, 2, 7, 1, 4, 0, 5, 3), (6, 3, 1, 4, 7, 0, 2, 5), (6, 3, 1, 7, 5, 0, 2, 4), (6, 4, 2, 0, 5, 7, 1, 3), (7, 1, 3, 0, 6, 4, 2, 5), (7, 1, 4, 2, 0, 6, 3, 5), (7, 2, 0, 5, 1, 4, 6, 3), (7, 3, 0, 2, 5, 1, 6, 4)]
輸出列表長度:
print len(list(queens()))# 輸出:92
更多關(guān)于Python相關(guān)內(nèi)容感興趣的讀者可查看本站專題:《Python數(shù)學(xué)運算技巧總結(jié)》、《Python數(shù)據(jù)結(jié)構(gòu)與算法教程》、《Python函數(shù)使用技巧總結(jié)》、《Python字符串操作技巧匯總》、《Python入門與進(jìn)階經(jīng)典教程》及《Python文件與目錄操作技巧匯總》
希望本文所述對大家Python程序設(shè)計有所幫助。
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