C语言详细讲解通过递归实现扫雷的展开

用户选择菜单

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

用户按1进入游戏

棋盘初始化

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

创建数组,并对其进行初始化

布置雷(随机布置)

void Setboard(char board[ROWS][COLS], int row, int col)
{
	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--;
		}
	}
}

用time函数产生随机值

打印棋盘

void Displayboard(char board[ROWS][COLS], int row, int col)
{
	int i, j;
	for (i = 0; i <= col; i++)
	{
		printf("%d ", i);
	}
	printf("
");
	for (i = 1; i <= row; i++)
	{
		printf("%d ", i);
		for (j = 1; j <= col; j++)
		{
			printf("%c ", board[i][j]);
		}
		printf("
");
	}
}

打印棋盘

玩家下棋

void Player(char board[ROWS][COLS], char show[ROWS][COLS], int row, int col)
{
	int x, y;
	int count = 0;
	while (1)
	{
		printf("请排雷:
");
		scanf("%d %d", &x, &y);
		if (x >= 1 && x <= 9 && y >= 1 && y <= 9)
		{
			if (board[x][y] == "0")
			{
				Openboard(show, board, x, y);
				Displayboard(show, ROW, COL);
			}
			else if (board[x][y] == "1")
			{
				printf("你死了
");
				break;
			}
		}
		else
		{
			printf("请重新输入");
		}
		int i, j;
		for (i = 1; i <= row; i++)
		{
			for (j = 1; j <= col; j++)
			{
				if (show[i][j] == "*")
				{
					count++;
				}
			}
		}
		if (count == Easy_count)
		{
			printf("成功
");   //这里的判断条件是遍历整个数组,统计雷的个数,如果雷的个数等于所剩余未排的*,说明排雷成功
			break;
		}
	}
}

用户输入值,并进行判断,如果该位置没有雷,我们进入展开函数

棋盘展开

void Openboard(char show[ROWS][COLS], char board[ROWS][COLS], int row, int col)
{
	if (row >= 1 && row <= ROW && col >= 1 && col <= COL)
	{
		int count=sum(board, row, col);
		if (count != 0)
		{
			show[row][col] = count + "0";
		}
		else if (show[row][col] != "_")
		{
			show[row][col] = "_";
			int i = 0, j = 0;
			for (i = row - 1; i <= row + 1; i++)
			{
				for (j = col - 1; j <= col + 1; j++)
				{
					Openboard(show, board, i,j);
				}
			}
		}
		else
		{
			return;
		}
	}
}

如果用户输入的这个位置没有雷,我们对其周围8个位置进行判断是否有雷,若有雷,我们把雷的个数显示在该位置上,若其周围8个位置没有雷并且不是下划线,我们把这个位置赋值为下划线,然后并对其8个位置进行同样的判断,如果周围没雷,而且周围的棋子也不是下划线,我们对其进行返回。

展开部分思维导图

展开函数最后一个else return 作用

这里我们show棋盘有三种情况,

1.该位置是*

2.该位置是下划线

3.该位置是雷的个数

else

return;

这里是作用:如果是下划线,我们就返回上一层函数。因为如果这里不是下划线,我们会在else return 之前的语句中进行判断,并对其周围8个位置进行操作,然后再对这8个棋子各个周围8个位置进行判断并操作,如果这里是下划线,就说明由这个位置为中心的周围8个棋子已经判断过了,并且以这8个位置为中心,已经递归过了,我们不需要再进行判断,所以直接返回就行

周围雷个数判断

int sum(char board[ROWS][COLS], int x, int y)
{
	return (board[x - 1][y - 1] +
		board[x - 1][y] +
		board[x - 1][y + 1] +
		board[x][y - 1] +
		board[x][y + 1] +
		board[x + 1][y - 1] +
		board[x + 1][y] +
		board[x + 1][y + 1] - 8 * "0");
}

test.c

#include"game.h"
void menu()
{
	printf("****************************
");
	printf("********  1.play  **********
");
	printf("********  0.exit  **********
");
	printf("****************************
");
}
void game()
{
	char board[ROWS][COLS] = { 0 };
	char show[ROWS][COLS] = { 0 };
	Itnboard(board, ROWS, COLS,"0");  //初始化棋盘
	Itnboard(show, ROWS, COLS, "*");
	Setboard(board, ROW, COL);   
	Displayboard(board, ROW, COL);
	Player(board, show, ROW, COL);            //玩家输入
}
int main()
{
	int input=1;
  srand((unsigned int)time(NULL));
	do{
		menu();
	scanf("%d", &input);
	switch (input)
	{
	case 1:
		game();
		break;
	case 0:
		break;
	default:
		printf("输入错误请重新输入:
 ");
	}
	} while (input);
}

game.c

#include"game.h"
void Itnboard(char board[ROWS][COLS], int rows, int cols,char c)
{
	int i, j;
	for (i = 0; i < rows; i++)
	{
		for (j = 0; j <cols; j++)
		{
			board[i][j] = c;
		}
	}
}
void Displayboard(char board[ROWS][COLS], int row, int col)
{
	int i, j;
	for (i = 0; i <= col; i++)
	{
		printf("%d ", i);
	}
	printf("
");
	for (i = 1; i <= row; i++)
	{
		printf("%d ", i);
		for (j = 1; j <= col; j++)
		{
			printf("%c ", board[i][j]);
		}
		printf("
");
	}
}
void Setboard(char board[ROWS][COLS], int row, int col)
{
	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--;
		}
	}
}
int sum(char board[ROWS][COLS], int x, int y)
{
 
	return (board[x - 1][y - 1] +
		board[x - 1][y] +
		board[x - 1][y + 1] +
		board[x][y - 1] +
		board[x][y + 1] +
		board[x + 1][y - 1] +
		board[x + 1][y] +
		board[x + 1][y + 1] - 8 * "0");
}
void Player(char board[ROWS][COLS], char show[ROWS][COLS], int row, int col)
{
	int x, y;
	int count = 0;
	while (1)
	{
		printf("请排雷:
");
		scanf("%d %d", &x, &y);
		if (x >= 1 && x <= 9 && y >= 1 && y <= 9)
		{
			if (board[x][y] == "0")
			{
				Openboard(show, board, x, y);
				Displayboard(show, ROW, COL);
			}
			else if (board[x][y] == "1")
			{
				printf("你死了
");
				break;
			}
		}
		else
		{
			printf("请重新输入");
		}
		int i, j;
		for (i = 1; i <= row; i++)
		{
			for (j = 1; j <= col; j++)
			{
				if (show[i][j] == "*")
				{
					count++;
				}
			}
		}
		if (count == Easy_count)
		{
			printf("成功
");
			break;
		}
	}
}
void Openboard(char show[ROWS][COLS], char board[ROWS][COLS], int row, int col)
{
 
	if (row >= 1 && row <= ROW && col >= 1 && col <= COL)
	{
		int count=sum(board, row, col);
		if (count != 0)
		{
			show[row][col] = count + "0";
		}
		else if (show[row][col] != "_")
		{
			show[row][col] = "_";
			int i = 0, j = 0;
			for (i = row - 1; i <= row + 1; i++)
			{
				for (j = col - 1; j <= col + 1; j++)
				{
					
					Openboard(show, board, i,j);
				}
			}
		}
		else
		{
			return;
		}
	}
}

game.h

#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
void Itnboard(char board[ROWS][COLS], int rows, int cols,char c);
void Displayboard(char board[ROWS][COLS], int row, int col);
void Setboard(char board[ROWS][COLS], int row, int col);
void Player(char board[ROWS][COLS], char show[ROWS][COLS], int row, int col);
void Openboard(char show[ROWS][COLS], char board[ROWS][COLS], int row, int col);

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原文地址:https://blog.csdn.net/weixin_49449676/article/details/124616964