碰撞检测其实就是有预见性；

假设下面有障碍物；

然后判断；

**************************************************************************************************

//定义容器数组
int cheng[20][14]={1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,0,0,0,0,0,0,0,0,0,0,0,0,1,
                   1,1,1,1,1,1,1,1,1,1,1,1,1,1};

//方块下方碰撞检测；
int detection(int x,int i)
{
 int m,n,ce=0;
    for(n=0;n<=3;n++)
 {
  for(m=0;m<=3;m++)
  {
   if(cheng[i+1-n][x/20+m-2]==b[3-n][m])
   {
    if(b[3-n][m]==1)
    {
     ce=1;
     break;
    }
   }
   
  }
  if(ce==1)break;
 } 
 return ce;
}

//左右检测；
int jiance(int x,int i)
{
 int ce=0,n,m;

for(n=0;n<=3;n++)
{
 for(m=0;m<=3;m++)
  {
    if(cheng[i-3+m][x/20+1-n]==b[m][3-n])
  {
   if(b[m][3-n]==1)
   {
   ce=1;
   break;
   }
  }
  }
 if(ce==1)break;
}
 
    return ce;
}

////////////////////////////////////////////摞叠////////////////////////////////////////////////

//下方的检测由于方块的固定；

//如果检测有方块跳出画图并赋值；
   if(detection(x,i)==1)
   {
    for(n=0;n<=3;n++)
    {
     for(m=0;m<=3;m++)
     {
      if(b[n][m]==1)
       cheng[i-3+n][x/20-2+m]=b[n][m];//赋值以便下次检测到有方块；
     }
    }
    break;
   }

////////////////////////////////////检测左右移动///////////////////////////////////////////

case 'a':
     x-=20;
     if (jiance(x,i)==1)x+=20;break;
    case 'd':
     x+=20;
     if (jiance(x,i)==1)x-=20;break;

///////////////////////////////////////////////检测旋转////////////////////////////////////////

//其实用到的还是左右检测；

//因为旋转检测就是判断移动后是否会有阻碍；

//跟左右一样；

case 'w':
     //图形旋转；
     //旋转检测；
     for(m=0;m<=3;m++)
     {
      for(n=0;n<=3;n++)
      {
       store2[m][n]=b[m][n];
      }
     }
     for(n=0;n<=2;n++)
      store[n]=b[0][n];
     
     store[3]=b[1][1];
     for(n=0;n<=2;n++)
      b[0][n]=b[3-n][0];
     for(n=0;n<=2;n++)
      b[3-n][0]=b[3][3-n];
     for(n=0;n<=2;n++)
      b[3][3-n]=b[n][3];
     for(n=0;n<=2;n++)
      b[n][3]=store[n];
     
     b[1][1]=b[2][1];
     b[2][1]=b[2][2];
     b[2][2]=b[1][2];
     b[1][2]=store[3];
     if(jiance(x,i)==1)//若有方块，画图数组恢复原值；
     {
      for(m=0;m<=3;m++)
      {
       for(n=0;n<=3;n++)
       {
        b[m][n]=store2[m][n];
       }
      }
     }
     break;
     ////////////////////////////////////////////////////////////////////////////////////////

 感觉这种检测效率太低；

为了提高效率在下方的检测我已加入了break；

使他快速跳出；

后面的还没有加；

感觉运行起来还是有点卡；

不够灵活；

、、、、