【C语言】函数递归实现——扫雷小游戏（详细步骤）（三）

#pragma once
#include<stdio.h>
#include <stdlib.h>
#include <time.h>
#define ROW 9
#define COL 9
#define ROWS ROW+2
#define COLS COL+2
#define EASY_COUNT 10
//初始化棋盘
void InitBoard(char board[ROWS][COLS], int rows, int cols, char set);
//显示棋盘
void DisplayBoard(char board[ROWS][COLS], int row, int col);
//布置雷
void SetMine(char board[ROWS][COLS], int row, int col);
//排查雷的
void FindMine(char mine[ROWS][COLS], char show[ROWS][COLS], int row, int col);
//特殊情况大片展开
void Clear(char mine[ROWS][COLS], char show[ROWS][COLS], int x, int y, int row, int col, int* win);

#include"game.h"
void InitBoard(char board[ROWS][COLS], int rows, int cols, char set)
{
  int i = 0;
  for (i = 0; i < rows; i++)
  {
  int j = 0;
  for (j = 0; j < cols; j++)
    board[i][j] = set;
  }
}
void SetMine(char board[ROWS][COLS], int row, int col)
{
  int count = EASY_COUNT;
  while (count)//一共设置了10个雷，没设置一颗雷，count就减一
  {
  //生成随机下标，这里利用时间戳和rand（）函数
  int x = rand() % row + 1;
  int y = rand() % col + 1;
  if (board[x][y] == '0')
  {
    board[x][y] = '1';
    count--;
  }
  }
}
 void DisplayBoard(char board[ROWS][COLS], int row, int col)
{
  int i = 0;
  printf("------------------------\n");
  for (i = 0; i <= 9; i++)//方便观看棋盘，给每行每列标上序号，标行号
  {
  printf("%d ", i);
  }
  printf("\n");
  for (i = 1; i <= row; i++)
  {
  int j = 0;
  printf("%d ", i);//标列号
  for (j = 1; j <= col; j++)
  {
    printf("%c ", board[i][j]);
  }
  printf("\n");
  }
  printf("------------------------\n");
}
 int GetMineCount(char mine[ROWS][COLS], int x, int y)//因为雷是‘1’，所以将排查点周围的点相加
                                                   //因为‘1’-‘0’=1，Ascii码
                                                   //所以周围八个字符相加减去8*‘0’等于几
               //说明有几颗雷在旁边
 {
  return (mine[x - 1][y] +
   mine[x - 1][y - 1] +
   mine[x][y - 1] +
   mine[x + 1][y - 1] +
   mine[x + 1][y] +
   mine[x + 1][y + 1] +
   mine[x][y + 1] +
   mine[x - 1][y + 1] - 8 * '0');
 }
 void Clear(char mine[ROWS][COLS], char show[ROWS][COLS] , int x , int y , int row , int col, int*win)
 {
  if (x >= 1 && x <= row && y >= 1 && y <= col)
  {
   if (show[x][y] == ' ' || show[x][y]!='*')
    return;
   else if (GetMineCount(mine, x, y) != 0)
   {
    show[x][y] = GetMineCount(mine, x, y) + '0';
    (*win)++;
    return;
   }
   else
   {
    show[x][y] = ' ';
    (*win)++;
    Clear(mine, show, x - 1, y, ROW, COL, win);
    Clear(mine, show, x + 1, y, ROW, COL, win);
    Clear(mine, show, x, y - 1, ROW, COL, win);
    Clear(mine, show, x, y + 1, ROW, COL, win);
   }
  }return;
 }
 void FindMine(char mine[ROWS][COLS],char show[ROWS][COLS],int row,int col)
 {
  int x = 0;
  int y = 0;
  int win = 0;
  while (win < row * col - EASY_COUNT)//一共10颗雷，只要排过的点<所有格子减去10个雷，就继续
  {
   printf("请输入要排查的坐标:>");
   scanf_s("%d %d", &x, &y);
   if (x >= 1 && x <= row && y >= 1 && y <= col)
   {
    if (mine[x][y] == '1')
    {
     printf("很遗憾，你被炸死了\n");
     DisplayBoard(mine, ROW, COL);
     break;
    }
    else
    {
     Clear(mine, show, x, y, ROW, COL, &win);
     DisplayBoard(show, ROW, COL);
     printf("已排除位置数= %d\n", win);
   }
   else
   {
    printf("坐标非法，重新输入\n");
   }
  }
  if (win == row * col - EASY_COUNT)
  {
   printf("恭喜你，排雷成功\n");
   DisplayBoard(mine, ROW, COL);
  }
 }

#include"game.h"
void menu()
{
  printf("****作者：敲代码的布莱恩特******\n");
  printf("*********   1. play     ********\n");
  printf("*********   0. exit     ********\n");
  printf("********************************\n");
}
void game()
{
  char mine[ROWS][COLS] = { 0 };//初始化两个二维数组，一个有用于存放雷的信息
  char show[ROWS][COLS] = { 0 };//一个用于存放排查出的雷的信息
  //对棋盘进行初始化
  InitBoard(mine, ROWS, COLS, '0');
  InitBoard(show, ROWS, COLS, '*');
  ///布置雷
  SetMine(mine, ROW, COL);
  //打印布置好了雷的棋盘，将设置好雷的棋盘显示出来
  DisplayBoard(show, ROW, COL);
  //排雷
  FindMine(mine, show, ROW, COL);
}
int main()
{
  srand((unsigned int)time(NULL));
  int input = 0;
  do
  {
  menu();
  scanf_s("%d", &input);
  switch (input)
  {
  case 1:
    game();
    break;
  case 0:
    printf("退出游戏\n");
    break;
  default:
    printf("没有这个选项，重新选择\n");
    break;
  }
  } while (1);
  return 0;
}