【c语言】推箱子

所需知识：C语言枚举，数组，for循环，while循环，switch,case语句，图形库相关函数

1.调整控制台窗口大小

#define  _CRT_SECURE_NO_WARNINGS#include #include int main(){system("mode con lines=15 cols=25");//调整窗口大小return 0;}

2.清掉控制台屏幕上的字

#define  _CRT_SECURE_NO_WARNINGS#include #include int main(){system("mode con lines=15 cols=25");system("cls");//清屏操作getchar();//不让程序退出，等待读字符return 0;}

3. 枚举类型定义地图中空地，墙，目的地，箱子，玩家

enum  Mine{SPACE,  //空地WALL,//墙DEST,  //目的地BOX,  //箱子PLAYER//玩家};

4.定义二维数组做地图，并且打印出来看看效果

//定义一个二维数组，做地图 空地0  墙1  目的地2  箱子3  玩家4    箱子在目的地 5  玩家在目的地6,与枚举类型对应上了int map[10][10] ={{0,0,0,0,0,0,0,0,0,0},{0,0,0,1,1,1,0,0,0,0},{0,0,0,1,2,1,1,1,1,0},{0,1,1,1,3,0,3,2,1,0},{0,1,2,3,4,0,1,1,1,0},{0,1,1,1,1,3,1,0,0,0},{0,0,0,0,1,2,1,0,0,0},{0,0,0,0,1,1,1,0,0,0},{0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0}};void printmap(){for(int i=0;i<10;i++){for (int j = 0; j < 10; j++){printf("%d ", map[i][j]);}printf("\n");}}

将printmap()在main中调用

为了让程序不会输入字符后退出，加上while循环

int main(){while (1){system("mode con lines=15 cols=25");system("cls");//清屏操作printmap();getchar();//不让程序退出，等待读字符}return 0;}

5.修改printmap函数为gamedraw()函数

为了保证游戏的美观性，我们将对应的数字转化为好看的图案
使用两层循环遍历二维数组，在使用switch已经将对应的数字用图案代替，此时我们要下载搜狗输入法

void gamedraw(){for (int i = 0; i < 10; i++){for (int j = 0; j < 10; j++){ switch (map[i][j]){case SPACE://如果二维数组元素为0printf("  ");  //空地        //一个中文字符相当于二个英文字符break;case WALL://如果二维数组元素为1printf("■");//墙break;case DEST://如果二维数组元素为2printf("☆");//目的地break;case BOX://如果二维数组元素为3printf("□");//箱子break;case PLAYER://如果二维数组元素为4printf("♀");//玩家break;case PLAYER+DEST://如果二维数组元素为6printf("♂");//玩家在目的地break;case BOX+DEST://如果二维数组元素为5printf("★");箱子在目的地break;                }}printf("\n");}}

对应数字和枚举变量类型对应上，PLAYER+DEST表示玩家如果出现在目的地的话对应数字为6，BOX+DEST表示箱子在目的地，将主函数中的printmap用gamedraw()换掉。

6.按键控制移动

要想使玩家移动，就先得确定玩家的坐标，通过遍历二维数组，找到数组中数字等于4的和4+2的，后者表示玩家在目的地，也要获取坐标，当找到玩家坐标时，要跳出循环，而break只能跳出一次循环，所以使用Goto函数，可以直接跳出多层循环.

int i = 0; int j = 0;//定义不要在for循环里面，要不然出作用域就会被销毁for (i = 0; i < 10; i++){for (j = 0; j < 10; j++){if (map[i][j] == PLAYER||map[i][j] == PLAYER+DEST){goto end;}}}end:;//找到直接来这里

在使用_getch()函数将按键的信息放入ch字符变量中，如果不知道上下左右对应的键值，我们可以打印出ch看看，我们把这些都封装成一个函数 keyevent()获取玩家坐标，并且读取键盘按下的信息

void keyevent(){int i = 0; int j = 0;for (i = 0; i < 10; i++){for (j = 0; j < 10; j++){if (map[i][j] == PLAYER || map[i][j] == PLAYER + DEST){goto end;}}}end:;char ch = _getch();printf("%d    %c", ch, ch);//w 119 a 97 s 115  d 100}

通过printf(“%d %c”, ch, ch);
得到虚拟键值为//w 119 a 97 s 115 d 100，然后注释掉printf(“%d %c”, ch, ch);然后通过swich,case 语句分别处理上下左右按键按下后的处理，

switch (ch){case 119:case 'w ':case 'W':break;case 97:case 'a ':case'A':break;case 115:case 's ':case'S':break;case 100:case 'd ':case'D':break;}

比如说按下w键，如果玩家上面是空地或者是目的地的话，玩家可以直接挪过去，因为没有障碍物阻碍，如果玩家的坐标为map[i][j];则玩家上面的的坐标就是map[i-1][j];要做的是

 if (map[i - 1][j] == SPACE||map[i - 1][j] == DEST) { map[i - 1][j] += PLAYER; map[i][j] -= PLAYER; }

如果map[i][j]只有玩家的话，上移动玩家就会map[i][j]=0，该位置变为空地，如果map[i][j]是玩家加目的地，上移动玩家就会map[i][j]=DEST，变为单纯的目的地.
如果玩家上面一个位置是箱子或者是箱子加目的地，就要看玩家上面的上面是什么了，如果是空地，或者是目的地，就可以推动，map[i - 2][j]是玩家上面的上面的坐标要做的是

 else  if(map[i-1][j]==BOX||map[i-1][j]==BOX+DEST) { if (map[i - 2][j] == SPACE || map[i - 2][j] == DEST) {//完成玩家上面有箱子，箱子的上面是空地或者是目的地都可以推动 map[i - 2][j] += BOX;//玩家上面的上面加一个箱子 map[i - 1][j] = map[i - 1][j] - BOX + PLAYER;//玩家的上面减去一个箱子加上一个玩家 map[i][j] -= PLAYER;//玩家消失在原来位置 } }

处理完的函数上键

case 119:case 'w ':case 'W':if (map[i - 1][j] == SPACE || map[i - 1][j] == DEST){map[i - 1][j] += 4;map[i][j] -= 4;}else  if (map[i - 1][j] == BOX || map[i - 1][j] == BOX + DEST){if (map[i - 2][j] == SPACE || map[i - 2][j] == DEST){map[i - 2][j] += BOX;map[i - 1][j] = map[i - 1][j] - BOX + PLAYER;map[i][j] -= PLAYER;//玩家消失在原来位置}}break;

向上推了一下，如果懂了上键怎么移动，别的也就会处理了

整体的keyevent()

void keyevent(){int i = 0; int j = 0;for (i = 0; i < 10; i++){for (j = 0; j < 10; j++){if (map[i][j] == PLAYER || map[i][j] == PLAYER + DEST){goto end;}}}end:;char ch = _getch();switch (ch){case 119:case 'w ':case 'W':if (map[i - 1][j] == SPACE || map[i - 1][j] == DEST){map[i - 1][j] += PLAYER;map[i][j] -=  PLAYER;}else  if (map[i - 1][j] == BOX || map[i - 1][j] == BOX + DEST){if (map[i - 2][j] == SPACE || map[i - 2][j] == DEST){map[i - 2][j] += BOX;map[i - 1][j] = map[i - 1][j] - BOX + PLAYER;map[i][j] -= PLAYER;//玩家消失在原来位置}}break;case 97:case 'a ':case'A':if (map[i][j - 1] == SPACE || map[i][j - 1] == DEST){map[i][j - 1] += PLAYER;map[i][j] -=  PLAYER;}else  if (map[i][j - 1] == BOX || map[i][j - 1] == BOX + DEST){if (map[i][j - 2] == SPACE || map[i][j - 2] == DEST){map[i][j - 2] += BOX;map[i][j - 1] = map[i][j - 1] - BOX + PLAYER;map[i][j] -= PLAYER;//玩家消失在原来位置}}break;case 115:case 's ':case'S':if (map[i + 1][j] == SPACE || map[i + 1][j] == DEST){map[i + 1][j] += PLAYER;map[i][j] -= PLAYER;}else  if (map[i + 1][j] == BOX || map[i + 1][j] == BOX + DEST){if (map[i + 2][j] == SPACE || map[i + 2][j] == DEST){map[i + 2][j] += BOX;map[i + 1][j] = map[i + 1][j] - BOX + PLAYER;map[i][j] -= PLAYER;//玩家消失在原来位置}}break;case 100:case 'd ':case'D':if (map[i][j + 1] == SPACE || map[i][j + 1] == DEST){map[i][j + 1] +=  PLAYER;map[i][j] -=  PLAYER;}else  if (map[i][j + 1] == BOX || map[i][j + 1] == BOX + DEST){if (map[i][j + 2] == SPACE || map[i][j + 2] == DEST){map[i][j + 2] += BOX;map[i][j + 1] = map[i][j + 1] - BOX + PLAYER;map[i][j] -= PLAYER;//玩家消失在原来位置}}break;}}

7.多组地图的制作，修改map数组

定义全局变量level=0;表示关数，
将二维数组改为三维数组，三维数组的每一个元素就是二维数组，就是一个地图.

int map[3][10][10] ={{{0,0,0,0,0,0,0,0,0,0},{0,0,0,1,1,1,0,0,0,0},{0,0,0,1,2,1,1,1,1,0},{0,1,1,1,3,0,3,2,1,0},{0,1,2,3,4,0,1,1,1,0},{0,1,1,1,1,3,1,0,0,0},{0,0,0,0,1,2,1,0,0,0},{0,0,0,0,1,1,1,0,0,0},{0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0}},{{0,0,0,0,0,0,0,0,0,0},{0,0,1,1,0,0,1,1,0,0},{0,1,0,2,1,1,2,0,1,0},{1,0,0,0,3,0,0,0,0,1},{1,0,0,0,4,3,0,0,0,1},{0,1,0,0,3,3,0,0,1,0},{0,0,1,0,0,0,0,1,0,0},{0,0,0,1,2,2,1,0,0,0},{0,0,0,0,1,1,0,0,0,0},{0,0,0,0,0,0,0,0,0,0}},{{0,0,0,0,1,0,0,0,0,0},{0,0,0,1,0,1,0,0,0,0},{0,0,1,2,3,0,1,0,0,0},{0,1,0,0,0,0,0,1,0,0},{1,2,3,0,4,0,0,0,1,0},{0,1,0,0,0,0,0,3,2,1},{0,0,1,0,3,0,0,0,1,0},{0,0,0,1,2,0,0,1,0,0},{0,0,0,0,1,0,1,0,0,0},{0,0,0,0,0,1,0,0,0,0}}};

将所有map[][]改成map[level][][];每通过一关，level++;

8.通关判断

循环遍历二维数组，如果有map[level][i][j]==BOX，说明有箱子没有推到目的地，返回false ，循环结束如果没有的话，说明通过此关，返回true,该判断函数返回布尔类型的值

bool jude(){for (int i = 0; i < 10; i++){for (int j = 0; j < 10; j++){if (map[level][i][j] == BOX){return false;}}}}

if (jude()){level++;if (level > 2){printf("oioioioioioioioi奥哈呦学妹你通过了！");_getch();break;}}

主函数添加如果jude()返回1，然后就是通关，然后level++;
变成三维数组中的第二个元素，也就是换了地图，当level>2,表示通关，因为只设置了三个图

9.程序源码（无图形库）

#define  _CRT_SECURE_NO_WARNINGS#include #include #include enum  Mine{SPACE,  //空地WALL,//墙DEST,  //目的地BOX,  //箱子PLAYER//玩家};int level = 0;//定义一个二维数组，做地图 空地0  墙1  目的地2  箱子3  玩家4    箱子在目的地 5  玩家在目的地6,与枚举类型对应上了int map[3][10][10] ={{{0,0,0,0,0,0,0,0,0,0},{0,0,0,1,1,1,0,0,0,0},{0,0,0,1,2,1,1,1,1,0},{0,1,1,1,3,0,3,2,1,0},{0,1,2,3,4,0,1,1,1,0},{0,1,1,1,1,3,1,0,0,0},{0,0,0,0,1,2,1,0,0,0},{0,0,0,0,1,1,1,0,0,0},{0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0}},{{0,0,0,0,0,0,0,0,0,0},{0,0,1,1,0,0,1,1,0,0},{0,1,0,2,1,1,2,0,1,0},{1,0,0,0,3,0,0,0,0,1},{1,0,0,0,4,3,0,0,0,1},{0,1,0,0,3,3,0,0,1,0},{0,0,1,0,0,0,0,1,0,0},{0,0,0,1,2,2,1,0,0,0},{0,0,0,0,1,1,0,0,0,0},{0,0,0,0,0,0,0,0,0,0}},{{0,0,0,0,1,0,0,0,0,0},{0,0,0,1,0,1,0,0,0,0},{0,0,1,2,3,0,1,0,0,0},{0,1,0,0,0,0,0,1,0,0},{1,2,3,0,4,0,0,0,1,0},{0,1,0,0,0,0,0,3,2,1},{0,0,1,0,3,0,0,0,1,0},{0,0,0,1,2,0,0,1,0,0},{0,0,0,0,1,0,1,0,0,0},{0,0,0,0,0,1,0,0,0,0}}};void gamedraw(){for (int i = 0; i < 10; i++){for (int j = 0; j < 10; j++){ switch (map[level][i][j]){case SPACE:printf("  ");          //一个中文字符相当于二个英文字符break;case WALL:printf("■");break;case DEST:printf("☆");break;case BOX:printf("□");break;case PLAYER:printf("♀");break;case PLAYER+DEST:printf("♂");break;case BOX+DEST:printf("★");break;                }}printf("\n");}}void keyevent(){int i = 0; int j = 0;for (i = 0; i < 10; i++){for (j = 0; j < 10; j++){if (map[level][i][j] == PLAYER || map[level][i][j] == PLAYER + DEST){goto end;}}}end:;char ch = _getch();switch (ch){case 119:case 'w ':case 'W':if (map[level][i - 1][j] == SPACE || map[level][i - 1][j] == DEST){map[level][i - 1][j] += 4;map[level][i][j] -= 4;}else  if (map[level][i - 1][j] == BOX || map[level][i - 1][j] == BOX + DEST){if (map[level][i - 2][j] == SPACE || map[level][i - 2][j] == DEST){map[level][i - 2][j] += BOX;map[level][i - 1][j] = map[level][i - 1][j] - BOX + PLAYER;map[level][i][j] -= PLAYER;//玩家消失在原来位置}}break;case 97:case 'a ':case'A':if (map[level][i][j - 1] == SPACE || map[level][i][j - 1] == DEST){map[level][i][j - 1] += 4;map[level][i][j] -= 4;}else  if (map[level][i][j - 1] == BOX || map[level][i][j - 1] == BOX + DEST){if (map[level][i][j - 2] == SPACE || map[level][i][j - 2] == DEST){map[level][i][j - 2] += BOX;map[level][i][j - 1] = map[level][i][j - 1] - BOX + PLAYER;map[level][i][j] -= PLAYER;//玩家消失在原来位置}}break;case 115:case 's ':case'S':if (map[level][i + 1][j] == SPACE || map[level][i + 1][j] == DEST){map[level][i + 1][j] += PLAYER;map[level][i][j] -= PLAYER;}else  if (map[level][i + 1][j] == BOX || map[level][i + 1][j] == BOX + DEST){if (map[level][i + 2][j] == SPACE || map[level][i + 2][j] == DEST){map[level][i + 2][j] += BOX;map[level][i + 1][j] = map[level][i + 1][j] - BOX + PLAYER;map[level][i][j] -= PLAYER;//玩家消失在原来位置}}break;case 100:case 'd ':case'D':if (map[level][i][j + 1] == SPACE || map[level][i][j + 1] == DEST){map[level][i][j + 1] += 4;map[level][i][j] -= 4;}else  if (map[level][i][j + 1] == BOX || map[level][i][j + 1] == BOX + DEST){if (map[level][i][j + 2] == SPACE || map[level][i][j + 2] == DEST){map[level][i][j + 2] += BOX;map[level][i][j + 1] = map[level][i][j + 1] - BOX + PLAYER;map[level][i][j] -= PLAYER;//玩家消失在原来位置}}break;}}bool jude(){for (int i = 0; i < 10; i++){for (int j = 0; j < 10; j++){if (map[level][i][j] == BOX){return false;}}}}int main(){system("mode con lines=15 cols=25");//system("cls");//清屏操作while (1){gamedraw();//_getch();if (jude()){level++;if (level > 2){printf("oioioioioioioioi奥哈呦学妹你通过了！");_getch();break;}}keyevent();}getchar();//不让程序退出，等待读字符return 0;}

10.演示1

20231002_124830

11.加图形库版本

12.头文件增加

#include 

13.定义保存空地，目的地，玩家，箱子，墙，箱子推到目的地图片的类

IMAGE  ima_all[6];

14.将图片image文件放在.cpp文件同目录下

image文件夹是自己创建的，用于放推箱子的素材，就是图片，图片可以在网上自己找推箱子的图片

15.加载图片函数

void loadimg(){for (int i = 0; i < 6; i++){char file[20] = "";sprintf(file,"./images/%d.png", i);loadimage(ima_all + i,file, 40, 40);}}

为什么这么加载图片，这里我们将照片命名为了0，1,2，3,4,5，要将这六个照片都加载进去，六个照片的相对路径里面只有照片名字不一样，我们可以循环将每个照片对应的相对路径的字符串放入到file字符串数组file中去，使用sprintf,如果这里不知道sprintf的用法，可以去看看我文件操作那一篇，文件操作，使用loadimage函数将六个图片加载进去，loadimage第一个参数是图片的首地址，第二个参数是该图片的相对路径，第三个，第四个参数是分辨率，也就是大小，后两个可以在
可以看到4241，这里统一用4040；

16. gamedraw函数的修改

因为使用图形库的话就不用之前的gamedraw函数打印图案，而是将图片贴上去

void gamedraw(){for (int i = 0; i < 10; i++){for (int j = 0; j < 10; j++){int x = j * 40;int y = i * 40;switch (map[level][i][j]){case SPACE:putimage(x, y, ima_all+2);         //一个中文字符相当于二个英文字符break;case WALL:putimage(x, y, ima_all+1);break;case DEST:putimage(x, y, ima_all+4);break;case BOX:putimage(x, y, ima_all+3);break;case PLAYER:putimage(x, y, ima_all);break;case PLAYER + DEST:putimage(x, y, ima_all);break;case BOX + DEST://就是箱子推到目的地putimage(x, y, ima_all+5);break;}}}}

由上图可知i,j和x,y反了过来，使用putimage函数将每个图片贴上去，putimage前两个参数为图片要贴在界面上左上角的坐标，第三个参数是要贴图片对应的地址，之前命名的照片在数组ima_all[6];是对应上的，就是说ima_all+3地址对应的就是对应的这个图片

17.主函数修改

int main(){initgraph(400, 400);loadimg();//system("mode con lines=15 cols=25");//system("cls");//清屏操作while (1){gamedraw();//_getch();if (jude()){level++;if (level > 2){//printf("oioioioioioioioi奥哈呦学妹你通过了！");_getch();break;}}keyevent();}getchar();//不让程序退出，等待读字符return 0;}

不使用控制台显示地图，初始化界面，在界面上显示地图，由于一个图片是40乘40的，二维数组是10乘10的，所以界面行列都应该是400，所以initgraph(400, 400);由于不用控制台，所以也就不使用printf函数，也不用将控制台大小改变，也不要清屏.
当你此时开始编译运行的时候会出现下列错误

解决方案：调试-》属性-》高级-》字符集-》多字符集

18.游戏通关显示

如果level>2

settextcolor(BLACK);//字体颜色settextstyle(25, 0, "微软雅黑");//字体风格setbkmode(TRANSPARENT);//字体背景透明outtextxy(100, 100, "oioioioioioioioi奥哈呦学妹你通过了！");//字体显示位置，以及内容_getch();break;

19.程序源码（带图形库版）

#define  _CRT_SECURE_NO_WARNINGS#include #include #include //_getch()函数头文件#include //bool类型的头函数#include //图形库头文件//定义一个二维数组，做地图//空地0  墙1  目的地2  箱子3  玩家4    箱子在目的地 5  玩家在目的地6IMAGE  ima_all[6];int level = 0;void loadimg(){for (int i = 0; i < 6; i++){char file[20] = "";sprintf(file,"./images/%d.png", i);loadimage(ima_all + i,file, 40, 40);}}enum  Mine{SPACE,  //空地WALL,//墙DEST,  //目的地BOX,  //箱子PLAYER//玩家};int map[3][10][10] ={{{0,0,0,0,0,0,0,0,0,0},{0,0,0,1,1,1,0,0,0,0},{0,0,0,1,2,1,1,1,1,0},{0,1,1,1,3,0,3,2,1,0},{0,1,2,3,4,0,1,1,1,0},{0,1,1,1,1,3,1,0,0,0},{0,0,0,0,1,2,1,0,0,0},{0,0,0,0,1,1,1,0,0,0},{0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0}},{{0,0,0,0,0,0,0,0,0,0},{0,0,1,1,0,0,1,1,0,0},{0,1,0,2,1,1,2,0,1,0},{1,0,0,0,3,0,0,0,0,1},{1,0,0,0,4,3,0,0,0,1},{0,1,0,0,3,3,0,0,1,0},{0,0,1,0,0,0,0,1,0,0},{0,0,0,1,2,2,1,0,0,0},{0,0,0,0,1,1,0,0,0,0},{0,0,0,0,0,0,0,0,0,0}},{{0,0,0,0,1,0,0,0,0,0},{0,0,0,1,0,1,0,0,0,0},{0,0,1,2,3,0,1,0,0,0},{0,1,0,0,0,0,0,1,0,0},{1,2,3,0,4,0,0,0,1,0},{0,1,0,0,0,0,0,3,2,1},{0,0,1,0,3,0,0,0,1,0},{0,0,0,1,2,0,0,1,0,0},{0,0,0,0,1,0,1,0,0,0},{0,0,0,0,0,1,0,0,0,0}}};void gamedraw(){for (int i = 0; i < 10; i++){for (int j = 0; j < 10; j++){int x = j * 40;int y = i * 40;switch (map[level][i][j]){case SPACE:putimage(x, y, ima_all+2);         //一个中文字符相当于二个英文字符break;case WALL:putimage(x, y, ima_all+1);break;case DEST:putimage(x, y, ima_all+4);break;case BOX:putimage(x, y, ima_all+3);break;case PLAYER:putimage(x, y, ima_all);break;case PLAYER + DEST:putimage(x, y, ima_all);break;case BOX + DEST:putimage(x, y, ima_all+5);break;}}}}void keyevent(){int i = 0; int j = 0;for (i = 0; i < 10; i++){for (j = 0; j < 10; j++){if (map[level][i][j] == PLAYER||map[level][i][j] == PLAYER+DEST){goto end;}}}end:;char ch = _getch();//printf("%d    %c", ch, ch);//w 119 a 97 s 115  d 100switch (ch){case 119: case 'w ': case 'W': if (map[level][i - 1][j] == SPACE||map[level][i - 1][j] == DEST) { map[level][i - 1][j] += 4; map[level][i][j] -= 4; } else  if(map[level][i-1][j]==BOX||map[level][i-1][j]==BOX+DEST) { if (map[level][i - 2][j] == SPACE || map[level][i - 2][j] == DEST) { map[level][i - 2][j] += BOX; map[level][i - 1][j] = map[level][i - 1][j] - BOX + PLAYER; map[level][i][j] -= PLAYER;//玩家消失在原来位置 } }break;case 97:case 'a ':case'A':if (map[level][i][j-1] == SPACE || map[level][i][j-1] == DEST){map[level][i][j-1] += 4;map[level][i][j] -= 4;}else  if (map[level][i][j-1] == BOX || map[level][i][j-1] == BOX + DEST){if (map[level][i][j-2] == SPACE || map[level][i][j-2] == DEST){map[level][i][j-2] += BOX;map[level][i][j-1] = map[level][i][j-1] - BOX + PLAYER;map[level][i][j] -= PLAYER;//玩家消失在原来位置}}break;case 115:case 's ':case'S':if (map[level][i+1][j] == SPACE || map[level][i+1][j] == DEST){map[level][i+1][j] += 4;map[level][i][j] -= 4;}else  if (map[level][i+1][j] == BOX || map[level][i+1][j] == BOX + DEST){if (map[level][i+2][j] == SPACE || map[level][i+2][j] == DEST){map[level][i+2][j] += BOX;map[level][i+1][j] = map[level][i+1][j] - BOX + PLAYER;map[level][i][j] -= PLAYER;//玩家消失在原来位置}}break;case 100:case 'd ':case'D':if (map[level][i][j+1] == SPACE || map[level][i][j+1] == DEST){map[level][i][j+1] += 4;map[level][i][j] -= 4;}else  if (map[level][i][j+1] == BOX || map[level][i][j+1] == BOX + DEST){if (map[level][i][j+2] == SPACE || map[level][i][j +2] == DEST){map[level][i][j+2] += BOX;map[level][i][j+1] = map[level][i][j+1] - BOX + PLAYER;map[level][i][j] -= PLAYER;//玩家消失在原来位置}}break;}}bool jude(){for (int i = 0; i < 10; i++){for (int j = 0; j < 10; j++){if (map[level][i][j] == BOX){return false;}}}}int main(){initgraph(10 * 40, 10 * 40);loadimg();system("mode con lines=15 cols=25");//调整窗口大小while (1){//system("cls");gamedraw();if (jude()){level++;if (level > 2){settextcolor(BLACK);settextstyle(25, 0, "微软雅黑");setbkmode(TRANSPARENT);outtextxy(100, 100, "oioioioioioioioi奥哈呦学妹你通过了！");_getch();break;}}keyevent();}getchar();//不让程序退出return 0;}

20.演示2

20231002_153153

21.关卡的增加

只需要将定义的全局变量map[3][10][10]，中3改成你想要的关卡数，然后在三维数组中增加像上面的二维数组即可，全部通关的level需要>关卡数-1即可

22.关卡重开

需要再定义一个和map[][][]三维数组相同的数组用来保存每一关的初始情况resetmap[][][],

int mapreset[3][10][10] ={{{0,0,0,0,0,0,0,0,0,0},{0,0,0,1,1,1,0,0,0,0},{0,0,0,1,2,1,1,1,1,0},{0,1,1,1,3,0,3,2,1,0},{0,1,2,3,4,0,1,1,1,0},{0,1,1,1,1,3,1,0,0,0},{0,0,0,0,1,2,1,0,0,0},{0,0,0,0,1,1,1,0,0,0},{0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0}},{{0,0,0,0,0,0,0,0,0,0},{0,0,1,1,0,0,1,1,0,0},{0,1,0,2,1,1,2,0,1,0},{1,0,0,0,3,0,0,0,0,1},{1,0,0,0,4,3,0,0,0,1},{0,1,0,0,3,3,0,0,1,0},{0,0,1,0,0,0,0,1,0,0},{0,0,0,1,2,2,1,0,0,0},{0,0,0,0,1,1,0,0,0,0},{0,0,0,0,0,0,0,0,0,0}},{{0,0,0,0,1,0,0,0,0,0},{0,0,0,1,0,1,0,0,0,0},{0,0,1,2,3,0,1,0,0,0},{0,1,0,0,0,0,0,1,0,0},{1,2,3,0,4,0,0,0,1,0},{0,1,0,0,0,0,0,3,2,1},{0,0,1,0,3,0,0,0,1,0},{0,0,0,1,2,0,0,1,0,0},{0,0,0,0,1,0,1,0,0,0},{0,0,0,0,0,1,0,0,0,0}}};

注意还是全局变量，我们设置当r键按下重置本关，所以我们要修改 keyevent()函数，当r按下我们遍历map数组将他赋值为原来的地图

void keyevent(){int i = 0; int j = 0;for (i = 0; i < 10; i++){for (j = 0; j < 10; j++){if (map[level][i][j] == PLAYER || map[level][i][j] == PLAYER + DEST){goto end;}}}end:;char ch = _getch();//printf("%d    %c", ch, ch);//w 119 a 97 s 115  d 100switch (ch){case 119:case 'w ':case 'W':if (map[level][i - 1][j] == SPACE || map[level][i - 1][j] == DEST){map[level][i - 1][j] += 4;map[level][i][j] -= 4;}else  if (map[level][i - 1][j] == BOX || map[level][i - 1][j] == BOX + DEST){if (map[level][i - 2][j] == SPACE || map[level][i - 2][j] == DEST){map[level][i - 2][j] += BOX;map[level][i - 1][j] = map[level][i - 1][j] - BOX + PLAYER;map[level][i][j] -= PLAYER;//玩家消失在原来位置}}break;case 97:case 'a ':case'A':if (map[level][i][j - 1] == SPACE || map[level][i][j - 1] == DEST){map[level][i][j - 1] += 4;map[level][i][j] -= 4;}else  if (map[level][i][j - 1] == BOX || map[level][i][j - 1] == BOX + DEST){if (map[level][i][j - 2] == SPACE || map[level][i][j - 2] == DEST){map[level][i][j - 2] += BOX;map[level][i][j - 1] = map[level][i][j - 1] - BOX + PLAYER;map[level][i][j] -= PLAYER;//玩家消失在原来位置}}break;case 115:case 's ':case'S':if (map[level][i + 1][j] == SPACE || map[level][i + 1][j] == DEST){map[level][i + 1][j] += 4;map[level][i][j] -= 4;}else  if (map[level][i + 1][j] == BOX || map[level][i + 1][j] == BOX + DEST){if (map[level][i + 2][j] == SPACE || map[level][i + 2][j] == DEST){map[level][i + 2][j] += BOX;map[level][i + 1][j] = map[level][i + 1][j] - BOX + PLAYER;map[level][i][j] -= PLAYER;//玩家消失在原来位置}}break;case 100:case 'd ':case'D':if (map[level][i][j + 1] == SPACE || map[level][i][j + 1] == DEST){map[level][i][j + 1] += 4;map[level][i][j] -= 4;}else  if (map[level][i][j + 1] == BOX || map[level][i][j + 1] == BOX + DEST){if (map[level][i][j + 2] == SPACE || map[level][i][j + 2] == DEST){map[level][i][j + 2] += BOX;map[level][i][j + 1] = map[level][i][j + 1] - BOX + PLAYER;map[level][i][j] -= PLAYER;//玩家消失在原来位置}}break;case 'r':///新增case'R' :///新增for (int i = 0; i < 10; i++)///新增{///新增for (int j = 0; j < 10; j++)///新增{map[level][i][j] = mapreset[level][i][j];///新增}///新增}///新增break;///新增}///新增}

来源地址：https://blog.csdn.net/yyqzjw/article/details/133484321