這篇文章主要講解了python如何實現(xiàn)五子棋程序，內(nèi)容清晰明了，對此有興趣的小伙伴可以學習一下，相信大家閱讀完之后會有幫助。

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五子棋游戲相信大部分人都玩過，今天我們用python來實現(xiàn)一次

構(gòu)建五子棋棋盤

from collections import namedtuple

Chessman = namedtuple('Chessman', 'Name Value Color')
Point = namedtuple('Point', 'X Y')

BLACK_CHESSMAN = Chessman('黑子', 1, (45, 45, 45))
WHITE_CHESSMAN = Chessman('白子', 2, (219, 219, 219))

offset = [(1, 0), (0, 1), (1, 1), (1, -1)]


class Checkerboard:
  def __init__(self, line_points):
    self._line_points = line_points
    self._checkerboard = [[0] * line_points for _ in range(line_points)]

  def _get_checkerboard(self):
    return self._checkerboard

  checkerboard = property(_get_checkerboard)

  # 判斷是否可落子
  def can_drop(self, point):
    return self._checkerboard[point.Y][point.X] == 0

  def drop(self, chessman, point):
    """
    落子
    :param chessman:
    :param point:落子位置
    :return:若該子落下之后即可獲勝，則返回獲勝方，否則返回 None
    """
    print(f'{chessman.Name} ({point.X}, {point.Y})')
    self._checkerboard[point.Y][point.X] = chessman.Value

    if self._win(point):
      print(f'{chessman.Name}獲勝')
      return chessman

  # 判斷是否贏了
  def _win(self, point):
    cur_value = self._checkerboard[point.Y][point.X]
    for os in offset:
      if self._get_count_on_direction(point, cur_value, os[0], os[1]):
        return True

  def _get_count_on_direction(self, point, value, x_offset, y_offset):
    count = 1
    for step in range(1, 5):
      x = point.X + step * x_offset
      y = point.Y + step * y_offset
      if 0 <= x < self._line_points and 0 <= y < self._line_points and self._checkerboard[y][x] == value:
        count += 1
      else:
        break
    for step in range(1, 5):
      x = point.X - step * x_offset
      y = point.Y - step * y_offset
      if 0 <= x < self._line_points and 0 <= y < self._line_points and self._checkerboard[y][x] == value:
        count += 1
      else:
        break

    return count >= 5

當前名稱：python如何實現(xiàn)五子棋程序-創(chuàng)新互聯(lián)
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