c/c++语言开发共享C++实现简单推箱子小游戏

  #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;   }  }   }