python命令行扫雷

4.编制一个可以在命令行模式下运行的扫雷游戏。
1)利用随机数进行布雷；
2)计算无雷格子的邻居中包含地雷的数目；
3)能够输入行号和列号表示要打开的格子；
4)能够进行成片的展开。

import random as r
import numpy as np


# 产生雷盘
def Leipan(n):
    # 产生比原定雷盘加2行2列是为了后面计算雷数时候的方便
    m = n + 2
    lei_pan = np.zeros((m, m), dtype=np.int0)
    lei_ran = []
    while len(lei_ran) != n:
        a = r.randint(1, n), r.randint(1, n)
        if a in lei_ran:
            continue
        else:
            lei_ran.append(a)
            lei_pan[a[0]][a[1]] = -1
    lei_pa = np.zeros((n + 2, n + 2), dtype=np.int0)
    for i in range(1, n + 1):
        for j in range(1, n + 1):
            if lei_pan[i][j] == -1:
                lei_pa[i][j] = -1
                continue
            else:
                lei_pa[i][j] = (-1) * int(
                    lei_pan[i][j + 1] + lei_pan[i][j - 1] + lei_pan[i + 1][j] + lei_pan[i - 1][j] + lei_pan[i + 1][
                        j + 1] + lei_pan[i + 1][j - 1] + lei_pan[i - 1][j + 1] + lei_pan[i - 1][j - 1])
    return lei_pa


# 判断是不是边界
def bianjie(lei_pan, booL_pan, r, c, n):
    if (lei_pan[r][c] == 0) and (booL_pan[r][c] == 0) and (r * c != 0) and (r != n + 1) and (c != n + 1):
        return True
    else:
        return False


# 进行递归探索雷盘
def game(lei_pan, booL_pan, a, b, n):
    for r in range(a - 1, a + 2):
        for c in range(b - 1, b + 2):
            if bianjie(lei_pan, bool_pan, r, c, n):
                booL_pan[r][c] = 1
                game(lei_pan, bool_pan, r, c, n)
            else:
                booL_pan[r][c] = 1


# 用户输入坐标操作
def user(lei_pan, bool_pan, n):
    while True:
        a, b = input("请输入要点击的坐标,号隔开：").split(",")
        a, b = int(a), int(b)
        if a > n or b > n or a < 1 or b < 1:
            print("坐标输入错误！")
        else:
            break
    # 游戏结束踩到雷了
    if lei_pan[a][b] == -1:
        return True
    elif lei_pan[a][b] != 0:
        bool_pan[a][b] = 1
        print_pan(lei_pan, bool_pan, n)
        return False
    else:
        game(lei_pan, bool_pan, a, b, n)
        print_pan(lei_pan, bool_pan, n)
        return False


# 显示过程雷盘
def print_pan(lei_pan, bool_pan, n):
    print("  ", end="")
    for a in range(1, n + 1):
        print(a, end=" ")
    print()
    for i in range(1, n + 1):
        if i >= 10:
            print(i, end="")
        else:
            print(i, end=" ")
        for j in range(1, n + 1):
            if bool_pan[i][j] == 1:
                print(lei_pan[i][j], end=" ")
            else:
                print("*", end=" ")
        print()


# 游戏结束雷盘
def print_end_pan(lei_pan, booL_pan, n):
    print("@代表雷")
    for i in range(1, n + 1):
        for j in range(1, n + 1):
            if lei_pan[i][j] == -1:
                print("@", end=" ")
            else:
                print(lei_pan[i][j], end=" ")
        print()


# 主函数
if __name__ == '__main__':
    while True:
        n = eval(input("请输入雷数:"))  # 雷的个数
        lei_pan = Leipan(n)
        bool_pan = [[0 for i in range(n + 2)] for j in range(n + 2)]  # 每个坐标是否访问
        print_pan(lei_pan, bool_pan, n)
        while True:
            if user(lei_pan, bool_pan, n):
                print("Game Over!")
                print_end_pan(lei_pan, bool_pan, n)
                break
            else:
                # 判断剩下的没有访问的坐标个数是否等于雷数
                cut = 0
                for i in range(1, n + 1):
                    for j in range(1, n + 1):
                        if bool_pan[i][j] == 0:
                            cut += 1
                # 未访问坐标数等于雷数游戏成功
                if cut == n:
                    print("恭喜你靓仔(妹)!成功通关！")
                    print_end_pan(lei_pan, bool_pan, n)
                    break
        s = input("再来一局输入1，退出输入其他字符！")
        if s != "1":
            break