C language to realize mine sweeping

  Minesweeping is a very classic game . Today I use c Language to implement it .

  First, clarify the idea of mine sweeping :

  1. Use a two-dimensional array to store the location of Mines （arr2）, Use another two-dimensional array to store the chessboard （arr1）.

  2. I typed arr1 Coordinates of a certain position in , The computer calculates the number of mines around this position , And print this point into thunder number . If it is 0, Print blank , And take the surrounding points as a new location point , To calculate , This can realize the function of expanding blank .

Then I have to finish so much , You need to implement these functions ：1. Print function , Print chessboard .2. Buried lightning function , Computers bury mines randomly .3. Receive instruction function , Let the computer know that I want to clear mines , Or mark ray , Or unmark ray .4. Minesweeping function , Give the computer a coordinate , To determine whether they were killed , How many mines are there in this place without being killed , If it is 0, Then expand outward .5. Mark ray function , Give the coordinates , The computer marks this thunder , This point can no longer be scanned .6. Unmark ray function , Is to unmark .7. Extension function , The expansion function expands the blank , Because you need to scan around , So in leipan （arr1） On , The best we can do is scan the corner , Then it will scan how many mines there are around , This exceeds the size of the thunder disk , So our arr2 Need to be compared with arr1 A big circle , Then use function recursion to expand .

First , Write the menu function

void menu()
{
	system("title  Mine clearance ");
	system("color f0");
	system("date /T");
	system("TIME /T");
	printf("-----------------------------\n");
	printf("******  1. Start the game     *******\n");
	printf("******  0. Quit the game     *******\n");
	printf("-----------------------------\n");
	int input = 0;
	char arr1[HANG][LIE] = { 0 };
	char arr2[HANG + 2][LIE + 2] = { 0 };
	printf(" Please select :");
	do
	{
		scanf("%d", &input);
		switch (input)
		{
		case 1:
			game(arr1,arr2,HANG,LIE);
			break;
		case 0:
			printf(" Quit the game ");
			return;
		default:
			printf(" error , Re input :");
		}
	}while (input);
}

Then write two initialization functions , Yes arr1 and arr2 To initialize

void chushihua2(char arr[HANG+2][LIE+2],int hang,int lie,char a)
{
	int i = 0;
	int j = 0;
	for (i = 0; i < hang; i++)
		for (j = 0; j < lie; j++)
			arr[i][j] = a;
}
void chushihua(char arr[HANG][LIE], int hang, int lie,char a)
{
	int i = 0;
	int j = 0;
	for (i = 0; i < hang; i++)
		for (j = 0; j < lie; j++)
			arr[i][j] = a;
}

Then write the print function

void display(char arr[HANG][LIE], int hang, int lie)
{
	//system("cls");
	int i = 0;
	int j = 0;
	for (i = 0; i < hang; i++)
		printf("  %d ", i + 1);
	printf("\n");
	for (i = 0; i < hang; i++)
	{
		printf("%d", i + 1);
		for (j = 0; j < lie - 1; j++)
		{
			printf(" %c |", arr[i][j]);
		}
		printf(" %c\n ", arr[i][j]);
		if (i < hang - 1)
		{
			for (j = 0; j < lie - 1; j++)
				printf("---|");
			printf("---\n");
		}
	}
}

Then write the theme of the game ,game function

void game(char arr1[HANG][LIE],char arr2[HANG+2][LIE+2],int hang,int lie)
{
	flag = 0;
	// arr1  The board  arr2  answer 
	chushihua(arr1, HANG, LIE,'#');
	chushihua2(arr2, HANG + 2, LIE + 2,'0');
	put_lei(arr2, HANG + 2, LIE + 2);
	while (1)
	{
		display(arr1,hang,lie);
		jieguo(arr1,arr2,hang,lie);
		put_order(arr1,arr2,hang,lie);
	}
}

Then there is the filling function

void put_lei(char arr[HANG + 2][LIE + 2], int hang, int lie)
{
	int count = 0;
	printf(" The computer is burying mines \n");
	while (count < NUM)
	{
		int i = rand() % (hang - 2) + 1;
		srand((unsigned int)(time(NULL)));
		int j = rand() % (lie - 2) + 1;
		if (arr[i][j] == '0')
		{
			arr[i][j] = '1';
			count++;
		}
	}
}

Then it gives the instruction function For convenience of measurement bug Add an additional debugging mode

void put_order(char arr1[HANG][LIE],char arr2[HANG+2][LIE+2],int hang,int lie)
{
	if (max == 1)
	{
		int i = 1;
		while (i)
		{
			int x = 0;
			int y = 0;
			printf(" Print chessboard (1) or  Print mine disk (2) or  Print the coordinates of all mines (3) or  Exit debugging (4):");
			scanf("%d", &i);
			switch (i)
			{
			case 1:
				for (x = 0; x < HANG; x++)
				{
					for (y = 0; y < LIE; y++)
						printf("%c ", arr1[x][y]);
					printf("\n");
				}
				break;
			case 2:
				for (x = 0; x < HANG+2; x++)
				{
					for (y = 0; y < LIE+2; y++)
						printf("%c ", arr2[x][y]);
					printf("\n");
				}
				break;
			case 3:
				for (x = 0; x < HANG + 2; x++)
				{
					for (y = 0; y < LIE + 2; y++)
					{
						if (arr2[x][y] == '1')
							printf("%d ,%d \n", x, y);
					}
				}
				break;
			case 4:
				max = 0;
				return;
			default:
				printf(" error , Re input :");
			}
		}
	}
	if (max == 0)
	{
		printf(" Mine clearance (1) or  Mark ray (2) or  Unmark ray (3) or  Tips (4) or  Debug mode (5):");
		int input = 0;
		while (1)
		{
			scanf("%d", &input);
			switch (input)
			{
			case 1:
				saolei(arr1, arr2, hang, lie);
				return;
			case 2:
				biaojilei(arr1, hang, lie);
				return;
			case 3:
				quxiaobiaojilei(arr1, hang, lie);
				return;
			case 4:
				tishi(arr1, arr2, hang, lie);
				return;
			case 5:
				max = 1;
				return;
			default:
				printf(" error , Re input :");
			}
		}
	}
}

Then write the mark Cancel mark Prompt function . The tag function needs to create a global variable flag（ Marked quantity ）, If flag=0, Then you can't unmark , Because it has not been marked . If flag = NUM（ The number of ray ）, Then you can't mark any more , Because if you mark the whole chessboard , That way, we can win directly .

void biaojilei(char arr[HANG][LIE],int hang,int lie)
{
	if (flag == NUM)
	{
		printf(" The number of marked mines has reached the online \n");
		return;
	}
	int i = 0;
	int j = 0;
	printf(" Enter the marking coordinates :");
	while (1)
	{
		scanf("%d %d", &i, &j);
		if (i<1 || i>hang || j<1 || j> lie)
		{
			printf(" error , Re input :");
			continue;
		}
		if (arr[i - 1][j - 1] =='@')
		{
			printf(" This coordinate has been marked , Re input :");
			continue;
		}
		if (arr[i - 1][j - 1] != '#')
		{
			printf(" This coordinate has been scanned , Re input :");
			continue;
		}
		arr[i - 1][j - 1] = '@';
		flag++; 
		break;
	}
}
void quxiaobiaojilei(char arr[HANG][LIE],int hang,int lie)
{
	if (flag == 0)
	{
		printf(" Not marked ray , Cannot unmark \n");
		return;
	}
	int i = 0;
	int j = 0;
	printf(" Enter the coordinates of the mark :");
	while (1)
	{
		scanf("%d %d", &i, &j);
		if (i<1 || i>hang || j<1 || j> lie)
		{
			printf(" error , Re input :");
			continue;
		}
		if(arr[i-1][j-1] != '@')
		{
			printf(" This position is not marked , Cannot unmark , Re input :");
			continue;		
		}
		else
		{
			arr[i - 1][j - 1] = '#';
			flag--;
			break;
		}
	}
}
void tishi(char arr1[HANG][LIE], char arr2[HANG + 2][LIE + 2], int hang, int lie)
{
	int i = 0;
	int j = 0;
	for (i = 0; i < hang; i++)
		for (j = 0; j < lie; j++)
		{
			if (arr2[i+1][j+1] == '1' && arr1[i][j] == '#')
			{
				arr1[i][j] = '@';
				flag++;
				return;
			}
		}
}

Next is minesweeping and expanding the blank function

int jisuanleishu(char arr[HANG+2][LIE+2],int hang,int lie,int i,int j)
{
	int sz= arr[i - 1][j - 1] + arr[i - 1][j] + arr[i - 1][j + 1] + arr[i][j - 1] + arr[i][j + 1] + arr[i + 1][j - 1] + arr[i + 1][j] + arr[i + 1][j + 1] - 8 * '0';
	return sz;
}
void panduan(char arr1[HANG][LIE], char arr2[HANG + 2][LIE + 2], int hang, int lie,int i ,int j)
{
	int sz = jisuanleishu(arr2, hang, lie, i, j);
	if (sz == 0)
	{
		arr1[i - 1][j - 1] = ' ';
		if (i - 1 >= 1 && i - 1 <= hang && j - 1 >= 1 && j - 1 <= lie && arr1[i - 2][j - 2] == '#')
			panduan(arr1, arr2, hang, lie, i - 1, j - 1);
		if (i  >= 1 && i  <= hang && j - 1 >= 1 && j - 1 <= lie && arr1[i - 1][j - 2] == '#')
			panduan(arr1, arr2, hang, lie, i , j - 1);
		if (i + 1 >= 1 && i + 1 <= hang && j - 1 >= 1 && j - 1 <= lie && arr1[i][j - 2] == '#')
			panduan(arr1, arr2, hang, lie, i + 1, j - 1);
		if (i - 1 >= 1 && i - 1 <= hang && j  >= 1 && j  <= lie && arr1[i - 2][j - 1] == '#')
			panduan(arr1, arr2, hang, lie, i - 1, j );
		if (i + 1 >= 1 && i + 1 <= hang && j >= 1 && j <= lie && arr1[i][j - 1] == '#')
			panduan(arr1, arr2, hang, lie, i + 1, j);
		if (i - 1 >= 1 && i - 1 <= hang && j + 1 >= 1 && j + 1 <= lie && arr1[i - 2][j] == '#')
			panduan(arr1, arr2, hang, lie, i - 1, j + 1);
		if (i >= 1 && i <= hang && j + 1 >= 1 && j + 1 <= lie && arr1[i - 1][j] == '#')
			panduan(arr1, arr2, hang, lie, i , j + 1);
		if (i + 1 >= 1 && i + 1 <= hang && j + 1 >= 1 && j + 1 <= lie && arr1[i][j] == '#')
			panduan(arr1, arr2, hang, lie, i + 1, j + 1);
	}
	else
		arr1[i - 1][j - 1] = '0' + sz;
}
void saolei(char arr1[HANG][LIE],char arr2[HANG+2][LIE+2],int hang,int lie)
{
	int i = 0;
	int j = 0;
	printf(" Enter the coordinates of the scan :");
	while (1)
	{
		scanf("%d %d", &i, &j);
		if (i<1 || i>hang || j<1 || j> lie)
		{
			printf(" error , Re input :");
			continue;
		}
		if (arr1[i - 1][j - 1] != '#')
		{
			printf(" This coordinate cannot be scanned , Because it's not an unknown point \n");
			printf(" Re input :");
			continue;
		}
		if (arr2[i][j] == '1')
		{
			printf(" unfortunately , You're killed in the blast \n");
			printf(" Return in three seconds \n");
			Sleep(3000);
			system("cls");
			menu();
		}
		else
		{
			panduan(arr1, arr2, hang, lie, i, j);
			return;
		}
	}
}

Finally, we need a function to judge whether the minesweeping is successful

void jieguo(char arr1[HANG][LIE], char arr2[HANG+2][HANG+2],int hang,int lie)
{
	int i = 0;
	int j = 0;
	int x = 0;
	int y = 0;
	for (i = 0; i < hang; i++)
	{
		for (j = 0; j < lie; j++)
		{
			if (arr1[i][j] == '#' && arr2[i + 1][j + 1] != '1')
			{
				y = 0;
				break;
			}
			else
				y++;
		}
		if (y == 0)
			break;
	}
	for (i = 0; i < hang; i++)
	{
		for (j = 0; j < lie; j++)
		{
			if (arr1[i][j] == '@' && arr2[i + 1][j + 1] == '1')
				x++;
			if (x + y == NUM)
			{
				printf(" Mine clearance succeeded ！\n");
				printf(" Return in three seconds \n");
				Sleep(3000);
				system("cls");
				menu();
				return;
			}
		}
	}
}

thus , Most of the demining has been completed

The test diagram is as follows

 

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