python实现简单的井字棋
本文实例为大家分享了python实现简单的井字棋的具体代码,供大家参考,具体内容如下
使用python实现井字棋游戏,没有具体算法,只是用随机下棋简单实现:
import random board = [['+','+','+'],['+','+','+'],['+','+','+']] def ma(board): if isempty(board): a = random.randint(0, 2) b = random.randint(0, 2) if board[a][b] != 'X' and board[a][b] != 'O': print("机器走:") board[a][b] = 'O' oput(board) else: ma(board) else: print("平局") def oput(board): print(" 0 1 2") for i in range(3): print(i, end=' ') for j in range(3): print(board[i][j], end=" ") print("") def winput(i,j): if board[i][j] == 'X': print("human win") else: print("machine win") return 1 def test(board): for i in range(3): for j in range(3): if board[i][j] != '+': if j == 0: if board[i][j] == board[i][j + 1] == board[i][j + 2]: return winput(i,j) if i == 0: if board[i][j] == board[i + 1][j] == board[i + 2][j]: return winput(i,j) if i == 0 and j == 0: if board[i][j] == board[i + 1][j + 1] == board[i + 2][j + 2]: return winput(i,j) if i == 2 and j == 0: if board[i][j] == board[i - 1][j + 1] == board[i - 2][j + 2]: return winput(i,j) def isempty(board): for i in range(3): for j in range(3): if board[i][j] == '+': return True return False def main(): print("初始棋盘:") oput(board) flag = 0 t = input("human first? Y/N human for X, machine for O\n") if t == 'Y': while isempty(board): print("人走: ") a, b = map(int, input("输入落子纵横坐标: a,b \n").split(',')) if board[a][b] == '+': board[a][b] = 'X' oput(board) flag = test(board) if flag == 1: break else: print("落子位置不对") continue ma(board) flag = test(board) if flag == 1: break if isempty(board) == 0 and flag == 0: print("平局") break elif t == 'N': while isempty(board): ma(board) flag = test(board) if isempty(board) == 0 and flag == 0: print("平局") break if flag == 1: break print("人走: ") a, b = map(int, input("输入落子纵横坐标: a,b \n").split(',')) if board[a][b] == '+': board[a][b] = 'X' oput(board) flag = test(board) if flag == 1: break else: print("落子位置不对") continue if __name__ == "__main__": main()
结果:
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