[C语言]推箱子
作者:小教学发布时间:2023-10-29分类:程序开发学习浏览:146
所需知识:C语言枚举,数组,代表循环,而循环,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 <stdlib.h>
#include <stdio.h>
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 <stdlib.h>
#include <stdio.h>
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");
}
}
将打印映射()在主中调用
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.修改打印地图函数为游戏原始()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,在使用交换机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0
printf(" "); //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1
printf("■");//墙
break;
case DEST://如果二维数组元素为2
printf("☆");//目的地
break;
case BOX://如果二维数组元素为3
printf("□");//箱子
break;
case PLAYER://如果二维数组元素为4
printf("♀");//玩家
break;
case PLAYER+DEST://如果二维数组元素为6
printf("♂");//玩家在目的地
break;
case BOX+DEST://如果二维数组元素为5
printf("★");箱子在目的地
break;
}
}
printf("\n");
}
}
对应数字和枚举变量类型对应上,播放器+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,将主函数中的打印地图用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
}
通过打印f(“%d%c”,ch,ch);
得到虚拟键值为//w 119 a 97 S 115 d 100,然后注释掉print f(“%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,如果玩家的坐标为映射[i][j];则玩家上面的的坐标就是映射[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;
}
如果映射[i][j]只有玩家的话,上移动玩家就会映射[i][j]=0,该位置变为空地,如果映射[i][j]是玩家加目的地,上移动玩家就会映射[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,地图[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
定义全局变量级别=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}
}
};
将所有映射[][]改成映射[级别][][];每通过一关,级别++;
8.通关判断
循环遍历二维数组,如果有地图[级别][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,当级别>;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 <stdlib.h>
#include <conio.h>
#include <stdio.h>
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 <graphics.h >
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.将图片图像文件放在.cpp文件同目录下
图像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Sprint,如果这里不知道Sprint f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岁;
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,所以也就不使用打印f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
如果级别>;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 <stdio.h>
#include <stdlib.h>
#include <conio.h>//_getch()函数头文件
#include <stdbool.h>//bool类型的头函数
#include <graphics.h >//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6
IMAGE 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 100
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] += 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
需要再定义一个和映射[]三维数组相同的数组用来保存每一关的初始情况重置映射[],
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按下我们遍历映射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 100
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] += 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
}
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