C语言开发实现扫雷游戏

时间：2021-06-28 08:28:52|栏目：C代码|点击：次

本文实例为大家分享了C语言实现扫雷游戏的具体代码，供大家参考，具体内容如下

简单描述:

输入要要扫的位置.
如果是雷,则游戏失败.
如果将所有不是雷的位置都扫了一遍,则游戏胜利.

主要细节

1、初始化(玩家扫雷图,地雷布局图)

展示出来的地图(玩家能看到的地图)showMap

for (int row = 0; row < ROW; row++){

 for (int col = 0; col < COL; col++){
 showMap[row][col] = '*';
 }
}

地雷的布局地图(玩家看不到的地图)mineMap

for (int row = 0; row < ROW; row++){

 for (int col = 0; col < COL; col++){
 mineMap[row][col] = '*';
 }
 }
 int count = 0;
 while (count<MINECOUNT){
 int row = rand() % 9;
 int col = rand() % 9;
 mineMap[row][col] = 'L';
 count++;
 }
}

3、打印地图

void print(char Map[ROW][COL]){
 printf(" |0|1|2|3|4|5|6|7|8|\n");
 for (int row = 0; row < ROW; row++){
 printf("%d |", row);
 for (int col = 0; col < COL; col++){
 printf("%c ", Map[row][col]);
 }
 printf("\n");
 }
}

5、玩家输入要扫的位置信息(9*9格子中某一特定位置)

void Input(char showMap[ROW][COL], char mineMap[ROW][COL], int* row, int* col){
 while (1){
 printf("请输入您要点的位置坐标:");
 scanf("%d %d", row, col);
 if (((*row) < 0) || ((*row) >= ROW)
 || ((*col) < 0) || ((*col) >= COL)){
 printf("输入有误 ");
 continue;
 }
 if (showMap[*row][*col] == '*'){
 showMap[*row][*col] = ' ';
 break;
 }
 }
}

7、判断某一位置是不是地雷

//返回1表示雷;返回0表示不是雷
int isMine(char mineMap[ROW][COL], int row, int col){
 if (mineMap[row][col] == 'L'){
 return 1;
 }
 return 0;
}

9、某一位置附近地雷的个数

int MineNerborhood(char mineMap[ROW][COL], int row, int col){
 if (row < 0 || row >= ROW
 || col < 0 || col >= COL){
 return '0';
 }
 int count = 0;
 for (int r = row - 1; r <= row + 1; r++){
 for (int c = col - 1; c <= col + 1; c++){
 if (mineMap[r][c] == 'L'){
 count++;
 }
 }
 }
 return count + '0';
}

源代码:

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define ROW 9
#define COL 9
#define MINECOUNT 10

void Input(char showMap[ROW][COL], char mineMap[ROW][COL], int* row, int* col){
 while (1){
 printf("请输入您要点的位置坐标:");
 scanf("%d %d", row, col);
 if (((*row) < 0) || ((*row) >= ROW)
 || ((*col) < 0) || ((*col) >= COL)){
 printf("输入有误 ");
 continue;
 }
 if (showMap[*row][*col] == '*'){
 showMap[*row][*col] = ' ';
 break;
 }
 }
}

//返回1表示雷;返回0表示不是雷
int isMine(char mineMap[ROW][COL], int row, int col){
 if (mineMap[row][col] == 'L'){
 return 1;
 }
 return 0;
}
//
void init(char showMap[ROW][COL], char mineMap[ROW][COL]){

 for (int row = 0; row < ROW; row++){

 for (int col = 0; col < COL; col++){
 showMap[row][col] = '*';
 }
 }
 for (int row = 0; row < ROW; row++){

 for (int col = 0; col < COL; col++){
 mineMap[row][col] = '*';
 }
 }
 int count = 0;
 while (count<MINECOUNT){
 int row = rand() % 9;
 int col = rand() % 9;
 mineMap[row][col] = 'L';
 count++;
 }
}
void print(char Map[ROW][COL]){
 printf(" |0|1|2|3|4|5|6|7|8|\n");
 for (int row = 0; row < ROW; row++){
 printf("%d |", row);
 for (int col = 0; col < COL; col++){
 printf("%c ", Map[row][col]);
 }
 printf("\n");
 }
}
//(row,col)附近格子地雷的个数
int MineNerborhood(char mineMap[ROW][COL], int row, int col){
 if (row < 0 || row >= ROW
 || col < 0 || col >= COL){
 return '0';
 }
 int count = 0;
 for (int r = row - 1; r <= row + 1; r++){
 for (int c = col - 1; c <= col + 1; c++){
 if (mineMap[r][c] == 'L'){
 count++;
 }
 }
 }
 return count + '0';
}
//一局游戏
void game(){
 srand(time(0));
 //展示给玩家的地图
 char showMap[ROW][COL];
 //雷的布局图
 char mineMap[ROW][COL];
 //对两个地图进行初始化
 init(showMap, mineMap);
 //print(mineMap);
 //玩家已经扫到的不是雷的格子的数目
 int notMineCount = 0;
 printf("扫雷游戏开始啦!\n");
 while (1){
 //打印地图
 print(showMap);
 //输入地点
 int row, col;
 Input(showMap, mineMap, &row, &col);
 //判断是否是雷区
 if (isMine(mineMap, row, col) == 1){
 printf(" 大兄弟,踩雷咧!\n给您看看雷的布局\n");
 print(mineMap);
 break;
 }
 //判断附近雷的情况
 showMap[row][col]=MineNerborhood(mineMap, row, col);

 notMineCount++;
 //判断是否赢了
 if (notMineCount == ROW*COL - MINECOUNT){
 print(mineMap);
 printf("恭喜您赢啦!\n");
 break;
 }
 }
 
}
int menu(){
 printf("===================\n");
 printf("1.开始游戏\n");
 printf("0.结束游戏\n");
 printf("===================\n");
 printf("请输入您的选择:");
 int choice;
 scanf("%d", &choice);
 return choice;
}
int main(){
 while (1){
 int choice = menu();
 if (choice == 1){
 game();
 }else if (choice==0){
 break;
 }else{
 printf("输入有误,请重新输入\n");
 }
 }
 system("pause");
 return 0;
}

运行结果: