6.Python类与对象优质课件.ppt

* def conflict(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 conflict(state,pos): if len(state)==num-1: yield (pos,) else: for result in queens(num,state+(pos,)): yield (pos,)+result list(queens(4)) [(1, 3, 0, 2), (2, 0, 3, 1)] for solution in queens(8): print(solution) (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,