Table des matières
4. Mine de mise en page de l'ordinateur
1. Réalisation de la fonction
1. Menu Imprimer
void menu()
{
printf("*****************************\n");
printf("********* 1. play *********\n");
printf("********* 0. exit *********\n");
printf("*****************************\n");
}2. Initialiser la carte
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. Imprimez l'échiquier
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. Mine de mise en page de l'ordinateur
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. Les joueurs vérifient les mines
// (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);
}
}2. Test de jeu
J'y ai joué complètement et il ne devrait y avoir aucun bug. Si vous avez des questions, veuillez laisser un message dans la zone de commentaire.
3. Code source
Compilateur : Visual Studio 2022
1.jeu.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.jeu.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);
}
}4. Optimisation du jeu
1. Réaction en chaîne du dragueur de mines (mise à jour)
[Langage C] Chain Reaction of Minesweeper Game (extension récursive) - Programmeur recherché
2. Vous ne marcherez pas sur le tonnerre pour la première fois (restez à l'écoute)
......