扫雷游戏
我天我给你大家发一下我最近做的扫雷，纯c写的，并且完成了两个功能，1、是第一次扫雷，不会被炸死。2、是每次可以扫出一大片。 希望大家可以从有所收获！ 源码呈上！
game.h
#ifndef GAME.H
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include<string.h>
#define ROWS ROW+2
#define COLS COL+2
#define ROW 9
#define COL 9
#define MINENUM 15
#define UN_MINENUM (ROW*COL - MINENUM)
void Init_broad(char arr_str[ROWS][COLS], int row, int col, char set);//初始化函数
void Print_broad(char arr_str[ROWS][COLS], int row, int col);//打印函数
void Desposit_Mine(char mine_broad[ROWS][COLS], int row, int col);//存放雷
void Mine_Clean(char test[ROWS][COLS], char show_broad[ROWS][COLS], char mine_broad[ROWS][COLS], int row, int col);//清理雷
static int Calculate_number(char show_broad[ROWS][COLS], int x, int y);//计算函数
void Judge_F_mine(char mine_broad[ROWS][COLS], int x, int y, int row, int col);//判断第一次是否为雷
void Clean_Mine(char show_broad[ROWS][COLS], char mine_broad[ROWS][COLS], int x, int y);//一大片清雷
int Computer_num(char test[ROWS][COLS], char show_broad[ROWS][COLS], int row, int col);//统计每次显示的个数
#endif // !GAME.H
game.c
#include"game.h"
//实现文件
void Init_broad(char arr_str[ROWS][COLS], int row, int col, char set)
{
int i = 0, j = 0;
for (i = 0; i < row; i++)
{
for (j = 0; j < col; j++)
{
arr_str[i][j] = set;//#号代表非雷
}
}
//memset(mine_broad, set, row*col * sizeof(mine_broad[0][0]));
}
void Print_broad(char arr_str[ROWS][COLS], int row, int col)
{
//打印表
system("cls");
int i = 0, j = 0;
for (i = 0; i <= row; i++)
{
printf("%d ", i);
}
printf("\n");
for (i = 1; i <= row; i++)
{
printf("%d ", i);
for (j = 1; j <= col; j++)
{
printf("%c ", arr_str[i][j]);
}
printf("\n");
}
}
void Desposit_Mine(char mine_broad[ROWS][COLS], int row, int col)
{
//生成随机数
int x = 0;
int y = 0;
int number = MINENUM;//雷的个数
//#-非雷 *-雷
//判断是否被占用
while (number)
{
x = rand() % row + 1;
y = rand() % col + 1;
if (mine_broad[x][y] == '0')
{
mine_broad[x][y] = '1';
number--;
}
}
}
void Mine_Clean(char test[ROWS][COLS], char show_broad[ROWS][COLS], char mine_broad[ROWS][COLS], int row, int col)
{
int x = 0, y = 0;//输入的值
int unmine_num = UN_MINENUM;//非雷的计数器
while (unmine_num)//查找循环
{
printf("请输入您想输入的坐标>");
scanf_s("%d%d", &x, &y);
if ((x >= 1 && x <= row) && (y >= 1 && y <= col))//判断输入合不合适
{
if (show_broad[x][y] == '*')
{
if (mine_broad[x][y] == '1' && unmine_num == UN_MINENUM)//防止第一次被炸死
{
Judge_F_mine(mine_broad, x, y, row, col);
unmine_num--;
}
if (mine_broad[x][y] == '1')//判断是否被雷炸死了！
{
printf("很遗憾，踩到地雷了！\n");
//Print_broad(mine_broad, ROW, COL);
break;
}
else
{
/*char number = Calculate_number(mine_broad, x, y);//正常运行的情况
show_broad[x][y] = number + '0';*/
Clean_Mine(show_broad, mine_broad, x, y);
unmine_num -= Computer_num(test, show_broad, row, col);
}
Print_broad(show_broad, ROW, COL);
}
else
{
printf("坐标被占用，请重新输入！\n");
}
}
else
{
printf("坐标错误，请重新输入！\n");
}
//Print_broad(show_broad, ROW, COL);//打印一下
}
if (unmine_num == 0)
{
printf("恭喜您，勇士，顺利通关！\n");
}
}
static int Calculate_number(char mine_broad[ROWS][COLS], int x, int y)
{
return mine_broad[x - 1][y] + mine_broad[x - 1][y - 1] \
+ mine_broad[x][y - 1] + mine_broad[x + 1][y - 1] \
+ mine_broad[x + 1][y] + mine_broad[x + 1][y + 1] \
+ mine_broad[x][y + 1] + mine_broad[x - 1][y + 1] \
- 8 * '0' ;//但是都是字符相加的，
}
void Judge_F_mine(char mine_broad[ROWS][COLS], int x, int y, int row, int col)
{
if (mine_broad[x][y] == '1')
{
mine_broad[x][y] = '0';
int a = 0, b = 0;
while (1)
{
a = rand() % row + 1;
b = rand() % col + 1;
if (mine_broad[a][b] == '0' && a != x)
{
mine_broad[a][b] = '1';
break;
}
}
}
return;
}
void Clean_Mine(char show_broad[ROWS][COLS], char mine_broad[ROWS][COLS], int x, int y)
{
if (mine_broad[x][y] == '0' && x >= 0 && y >= 0 && show_broad[x][y] == '*')//0
{
show_broad[x][y] = Calculate_number(mine_broad, x, y) + '0';//show_broad[x][y]是存储字符型的，而函数的返回值却为整数
}
if (mine_broad[x - 1][y] == '0' && x >= 0 && y >= 0 && show_broad[x - 1][y] == '*')//1
{
show_broad[x - 1][y] = Calculate_number(mine_broad, x - 1, y) + '0';
if (Calculate_number(mine_broad, x - 1, y) == 0)
{
Clean_Mine(show_broad, mine_broad, x - 1, y);
}
}
if (mine_broad[x - 1][y - 1] == '0' && x >= 0 && y >= 0 && show_broad[x - 1][y - 1] == '*')//2
{
show_broad[x - 1][y - 1] = Calculate_number(mine_broad, x - 1, y - 1) + '0';
if (Calculate_number(mine_broad, x - 1, y - 1) == 0)
{
Clean_Mine(show_broad, mine_broad, x - 1, y - 1);
}
}
if (mine_broad[x][y - 1] == '0' && x >= 0 && y >= 0 && show_broad[x][y - 1] == '*')//3
{
show_broad[x][y - 1] = Calculate_number(mine_broad, x, y - 1) + '0';
if (Calculate_number(mine_broad, x, y - 1) == 0)
{
Clean_Mine(show_broad, mine_broad, x, y - 1);
}
}
if (mine_broad[x + 1][y - 1] == '0' && x >= 0 && y >= 0 && show_broad[x + 1][y - 1] == '*')//4
{
show_broad[x + 1][y - 1] = Calculate_number(mine_broad, x + 1, y - 1) + '0';
if (Calculate_number(mine_broad, x + 1, y - 1) == 0)
{
Clean_Mine(show_broad, mine_broad, x + 1, y - 1);
}
}
if (mine_broad[x + 1][y] == '0' && x >= 0 && y >= 0 && show_broad[x + 1][y] == '*')//5
{
show_broad[x + 1][y] = Calculate_number(mine_broad, x + 1, y) + '0';
if (Calculate_number(mine_broad, x + 1, y) == 0)
{
Clean_Mine(show_broad, mine_broad, x + 1, y);
}
}
if (mine_broad[x + 1][y + 1] == '0' && x >= 0 && y >= 0 && show_broad[x + 1][y + 1] == '*')//6
{
show_broad[x + 1][y + 1] = Calculate_number(mine_broad, x + 1, y + 1) + '0';
if (Calculate_number(mine_broad, x + 1, y + 1) == 0)
{
Clean_Mine(show_broad, mine_broad, x + 1, y + 1);
}
}
if (mine_broad[x][y + 1] == '0' && x >= 0 && y >= 0 && show_broad[x][y + 1] == '*')//7
{
show_broad[x][y + 1] = Calculate_number(mine_broad, x, y + 1) + '0';
if (Calculate_number(mine_broad, x, y + 1) == 0)
{
Clean_Mine(show_broad, mine_broad, x, y + 1);
}
}
if (mine_broad[x - 1][y + 1] == '0' && x >= 0 && y >= 0 && show_broad[x - 1][y + 1] == '*')//8
{
show_broad[x - 1][y + 1] = Calculate_number(mine_broad, x - 1, y + 1) + '0';
if (Calculate_number(mine_broad, x - 1, y + 1) == 0)
{
Clean_Mine(show_broad, mine_broad, x - 1, y + 1);
}
}
}
int Computer_num(char test[ROWS][COLS], char show_broad[ROWS][COLS], int row, int col)
{
int count = 0;
int i = 1, j = 0;
for (i = 1; i <= row; i++)
{
for (j = 1; j <= col; j++)
{
if (show_broad[i][j] != '*')
{
if (test[i][j] != 'a')
{
count++;
test[i][j] = 'a';//只有代表雷
}
}
}
}
return count;
}
test.c
#include"game.h"
//测试文件
void Tips()
{
printf("\t 游戏玩法介绍\n");
printf("一、按1游戏开始，0退出游戏！\n");
printf("二、游戏需要输入坐标才可以扫雷！\n");
printf("三、扫出所有雷后取得游戏胜利，碰到雷游戏结束！\n");
printf("------------------------------------\n\n");
}
void menu()
{
printf("************************\n");
printf("****** 1、play *******\n");
printf("****** 0、exit *******\n");
printf("************************\n");
}
void game()
{
char mine_broad[ROWS][COLS] = { '0' };//雷区的数组
char show_broad[ROWS][COLS] = { '0' };//打印的数组
char test[ROWS][COLS] = { '0' };
//初始化
Init_broad(mine_broad, ROWS, COLS, '0');
Init_broad(show_broad, ROWS, COLS, '*');
Init_broad(test, ROWS, COLS, 'M');//用来统计非雷个数--效率变低
Print_broad(show_broad, ROW, COL);
//存放雷
Desposit_Mine(mine_broad, ROW, COL);
//Print_broad(mine_broad, ROW, COL);//test1
//排雷
Mine_Clean(test, show_broad, mine_broad, ROW, COL);
}
void test()
{
Tips();
srand((unsigned int)time(NULL));//设置种子
int input = 0;
do
{
menu();//菜单
printf("请选择>");
scanf_s("%d", &input);
switch (input)
{
case 1:
//游戏大循环体
game();//游戏
break;
case 0:
//退出
printf("退出游戏\n");
break;
default:
printf("选择错误，请重新选择!\n");
break;
}
} while (input);
}
int main()
{
test();//game()函数是函数运行的函数
system("pause");
return 0;
}
在这里，我说一下，显示一大片我们采用的是递归的思想，我们输入一个坐标(x,y), 假设不为雷, 我们对一个不为雷(x,y),判断它的周围是否有雷，有雷的不管，没雷的我们显示它的数字，如果该数是零，就把它作为新的(x,y)，继续判断他的周围，以此类推，直到我们申请的二维数组的0和MAX位置（i>=0&&i<=ROWS, j>=0&&j<=COLS），当然你打印大一大片里面判断成功条件就要有所改变。这个不难，方法也特别多。嘿嘿，我才用的方法比较笨拙，浪费内存。大家可以做出更好的优化方法。 谢谢大家，就分享到这了，希望对您有帮助
