【C语言】简易版扫雷（附源码）

目录

一、函数实现

1.打印菜单

2.初始化棋盘

3.打印棋盘

4.电脑布置雷

5.玩家排查雷

二、游戏测试

三、源码

1.game.h

2.test.c

3.game.c

四、游戏优化

---

一、函数实现

1.打印菜单

void menu()
{
	printf("*****************************\n");
	printf("*********  1. play  *********\n");
	printf("*********  0. exit  *********\n");
	printf("*****************************\n");
}

2.初始化棋盘

void InitBoard(char board[ROWS][COLS], int rows, int cols, char set)
{
	int i;
	for (i = 0; i < rows; i++)
	{
		int j;
		for (j = 0; j < cols; j++)
		{
			board[i][j] = set;
		}
	}
}

3.打印棋盘

void DisplayBoard(char board[ROWS][COLS], int row, int col)
{
	int i;
	printf("------扫雷游戏------\n");
	for (i = 0; i <= col; i++)
	{
		printf("%d ", i);
	}
	printf("\n");
	for (i = 1; i <= row; i++)
	{
		printf("%d ", i);
		int j;
		for (j = 1; j <= col; j++)
		{
			printf("%c ", board[i][j]);
		}
		printf("\n");
	}
	printf("--------------------\n");
}

4.电脑布置雷

void SetMine(char board[ROWS][COLS], int row, int col)
{
	//布置10个雷
	//生成随机的坐标，布置雷
	int count = EASY_COUNT;
	while (count)
	{
		int x = rand() % row + 1;
		int y = rand() % col + 1;
		if (board[x][y] == '0')
		{
			board[x][y] = '1';
			count--;
		}
	}
}

5.玩家排查雷

// (x-1,y-1) (x-1, y ) (x-1,y+1)   |   (1,1) (1,2) (1,3)
// ( x ,y-1) ( x , y ) ( x ,y+1)   |   (2,1) (2,2) (2.3)
// (x+1,y-1) (x+1, y ) (x+1,y+1)   |   (3,1) (3,2) (3,3)
int GetMineCount(char mine[ROWS][COLS], int x, int y)
{
	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 FindMine(char mine[ROWS][COLS], char show[ROWS][COLS], int row, int col)
{
	int x, y;
	int win = 0;
	while (win < row * col - EASY_COUNT)
	{
		printf("请输入要排查的坐标:");
		scanf("%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 if(mine[x][y] == '0' && show[x][y] == '*')
			{
				//该位置不是雷，就统计这个坐标周围有几个雷
				int count = GetMineCount(mine, x, y);
				show[x][y] = count + '0';
				DisplayBoard(show, ROW, COL);
				win++;
				printf("您已排查%d块区域，还有%d块区域待排查\n", win, row * col - EASY_COUNT - win);
			}
			else
			{
				printf("该坐标已被排查，请重新输入\n");
			}
		}
		else
		{
			printf("坐标非法，请重新输入\n");
		}
	}
	if (win == row * col - EASY_COUNT)
	{
		printf("你赢了！\n");
		DisplayBoard(mine, ROW, COL);
	}
}

二、游戏测试

我完整的玩了一把，应该是没有bug的，如果有问题请在评论区留言。

三、源码

编译器：Visual Studio 2022

1.game.h

#pragma once

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#define EASY_COUNT 10	//雷的个数

#define ROW 9	//雷区的行数
#define COL 9	//雷区的列数

//二维数组的大小，比雷区大一圈方便计算
#define ROWS ROW+2
#define COLS COL+2

//初始化棋盘
void InitBoard(char board[ROWS][COLS], int rows, int cols, char set);

//打印棋盘
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
void FindMine(char mine[ROWS][COLS], char show[ROWS][COLS], int row, int col);

2.test.c

#define _CRT_SECURE_NO_WARNINGS 1

#include "game.h"

void menu()
{
	printf("*****************************\n");
	printf("*********  1. play  *********\n");
	printf("*********  0. exit  *********\n");
	printf("*****************************\n");
}

void game()
{
	char mine[ROWS][COLS];//存放布置好的雷
	char show[ROWS][COLS];//存放排查出的雷的信息
	//初始化棋盘
	//1.mine数组最开始是全'0'
	//2.show数组最开始是全'*'
	InitBoard(mine, ROWS, COLS, '0');
	InitBoard(show, ROWS, COLS, '*');
	//打印棋盘
	//DisplayBoard(mine, ROW, COL);
	DisplayBoard(show, ROW, COL);
	//1.布置雷
	SetMine(mine, ROW, COL);
	//DisplayBoard(mine, ROW, COL);
	//2.排查雷
	FindMine(mine, show, ROW, COL);
}

int main()
{
	int input;
	srand((unsigned int)time(NULL));
	do
	{
		menu();
		printf("请选择:");
		scanf("%d", &input);
		switch (input)
		{
		case 1:
			game();
			break;
		case 0:
			printf("退出游戏\n");
			break;
		default:
			printf("选择错误，请重新选择\n");
			break;
		}
	} while (input);
	return 0;
}

3.game.c

#define _CRT_SECURE_NO_WARNINGS 1

#include "game.h"

void InitBoard(char board[ROWS][COLS], int rows, int cols, char set)
{
	int i;
	for (i = 0; i < rows; i++)
	{
		int j;
		for (j = 0; j < cols; j++)
		{
			board[i][j] = set;
		}
	}
}

void DisplayBoard(char board[ROWS][COLS], int row, int col)
{
	int i;
	printf("------扫雷游戏------\n");
	for (i = 0; i <= col; i++)
	{
		printf("%d ", i);
	}
	printf("\n");
	for (i = 1; i <= row; i++)
	{
		printf("%d ", i);
		int j;
		for (j = 1; j <= col; j++)
		{
			printf("%c ", board[i][j]);
		}
		printf("\n");
	}
	printf("--------------------\n");
}

void SetMine(char board[ROWS][COLS], int row, int col)
{
	//布置10个雷
	//生成随机的坐标，布置雷
	int count = EASY_COUNT;
	while (count)
	{
		int x = rand() % row + 1;
		int y = rand() % col + 1;
		if (board[x][y] == '0')
		{
			board[x][y] = '1';
			count--;
		}
	}
}

// (x-1,y-1) (x-1, y ) (x-1,y+1)   |   (1,1) (1,2) (1,3)
// ( x ,y-1) ( x , y ) ( x ,y+1)   |   (2,1) (2,2) (2.3)
// (x+1,y-1) (x+1, y ) (x+1,y+1)   |   (3,1) (3,2) (3,3)
int GetMineCount(char mine[ROWS][COLS], int x, int y)
{
	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 FindMine(char mine[ROWS][COLS], char show[ROWS][COLS], int row, int col)
{
	int x, y;
	int win = 0;
	while (win < row * col - EASY_COUNT)
	{
		printf("请输入要排查的坐标:");
		scanf("%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 if(mine[x][y] == '0' && show[x][y] == '*')
			{
				//该位置不是雷，就统计这个坐标周围有几个雷
				int count = GetMineCount(mine, x, y);
				show[x][y] = count + '0';
				DisplayBoard(show, ROW, COL);
				win++;
				printf("您已排查%d块区域，还有%d块区域待排查\n", win, row * col - EASY_COUNT - win);
			}
			else
			{
				printf("该坐标已被排查，请重新输入\n");
			}
		}
		else
		{
			printf("坐标非法，请重新输入\n");
		}
	}
	if (win == row * col - EASY_COUNT)
	{
		printf("你赢了！\n");
		DisplayBoard(mine, ROW, COL);
	}
}

四、游戏优化

1.扫雷的连锁反应（已更新）

【C语言】扫雷游戏的连锁反应（递归展开）_字节连结的博客-CSDN博客

2.第一次一定不会踩到雷（敬请期待）

......