超级简单推箱子小游戏 (简单代码和思路)

推箱子

游戏演示

代码展示

#include<stdio.h>
#include<stdlib.h>
#include<getch.h>

int main(int argc,const char*argv[])
{
	int cut=0;
	int a[8][8]={
		          {0,0,3,3,3,3,0,0},  
		          {0,0,3,5,5,3,0,0},  
		          {0,0,3,0,5,3,0,0},  
		          {0,3,0,0,4,5,3,0},  
		          {3,3,0,4,0,0,3,3},  
		          {3,0,0,3,4,4,0,3},  
		          {3,0,0,2,0,0,0,3},  
		          {3,3,3,3,3,3,3,3}};  
   int x=6,y=3;
    for(;;)
	{
		int cunt=0;
		system("clear");
	   for(int i=0;i<8;i++)
	   {
		   for(int j=0;j<8;j++)
		   {
			   if(a[i][j]==9)
			   {
				   cunt++;
			   }
			   switch(a[i][j])
			   {
				   case 0:printf("  ");break;
				   case 2:printf("@ ");break;//人
				   case 3:printf("# ");break;//墙
				   case 4:printf("$ ");break;//箱子
				   case 5:printf("o ");break;//目标位置
				   case 7:printf("@ ");break;
				   case 9:printf("$ ");break;
			   }
		   }
		   printf("\n");
	   }
	   if(4==cunt)
	   {
		   printf("成功\n步数：%d",cut);
		   return 0;
	   }
	  //根据数据的大小来判断当前的坐标上的情况
	  switch(getch())
	  {
		  case 183:
		       if(0!=x&&3==a[x-1][y])
			   {
				   a[x][y]=2;
			   }
			   else if(5==a[x-1][y]||0==a[x-1][y])
			   {
				   a[x][y]-=2;
				   a[x-1][y]+=2;
				   x--;
				   cut++;
			   }
			   else if((5==a[x-2][y]||0==a[x-2][y])&&(4==a[x-1][y]||9==a[x-1][y]))
			   {
				   a[x-2][y]+=4;
				   a[x-1][y]-=4;
				   a[x-1][y]+=2;
				   a[x][y]-=2;
				   x--;cut++;
			   }break;
		  case 184:
			   if(0!=x&&3==a[x+1][y])
			   {
				   a[x][y]=2;
			   }
			   else if((5==a[x+1][y])||(0==a[x+1][y]))
			   {
				   a[x][y]-=2;
				   a[x+1][y]+=2;
				   x++;cut++;
			   }
			   else if((5==a[x+2][y]||0==a[x+2][y])&&(4==a[x+1][y]||9==a[x+1][y]))
			   {
				   a[x+2][y]+=4;
				   a[x+1][y]-=4;
				   a[x+1][y]+=2;
				   a[x][y]-=2;
				   x++;cut++;
			   }break;
		  case 185:
		       if(0!=x&&3==a[x][y+1])
			   {
				   a[x][y]=2;
			   }
			   else if(5==a[x][y+1]||0==a[x][y+1])
			   {
				   a[x][y]-=2;
				   a[x][y+1]+=2;
				   y++;cut++;
			   }
			   else if((0==a[x][y+2]||5==a[x][y+2])&&(4==a[x][y+1]||9==a[x][y+1]))
			   {
				   a[x][y+2]+=4;
				   a[x][y+1]-=4;
				   a[x][y+1]+=2;
				   a[x][y]-=2;
				   y++;cut++;
			   }break;
		  case 186:
		       if(0!=x&&3==a[x][y-1])
			   {
				   a[x][y]=2;
			   }
			   else if(5==a[x][y-1]||0==a[x][y-1])
			   {
				   a[x][y]-=2;
				   a[x][y-1]+=2;
				   y--;cut++;
			   }
			   else if((0==a[x][y-2]||5==a[x-1][y])&&(4==a[x][y-1]||9==a[x][y-1]))
			   {
				   a[x][y-2]+=4;
				   a[x][y-1]-=4;
				   a[x][y-1]+=2;
				   a[x][y]-=2;
				   y--;cut++;
			   }break;
	  }
}		
}

本文地址：https://blog.csdn.net/weixin_49198344/article/details/108173105