Question

可能重复：
Determine if two rectangles overlap each other?

考虑到我有2个方格，我知道x和y位置，我也知道大小，如果我想看看物体是否相互碰撞，那么使用的公式是什么。

if(   ((shapeA->getX() - shapeA->getSize()) > (player->getX() - player->getSize())
    && (shapeA->getX() + shapeA->getSize()) < (player->getX() + player->getSize()))
    && (shapeA->getY() - shapeA->getSize()  > player->getY() - player->getSize()
    && (shapeA->getY() + shapeA->getSize()) < (player->getY() + player->getSize()))
                )

这很有效，但它很奇怪（并不是所有时间）。我一定错过了什么

Answer 1

检查矩形是否与另一个矩形相交或接触是非常容易的。看看下面的图片：

Rectangular intersection

如您所见，如果（[x，x + a]和[X，X + A]）和（[y，y + b]和[Y，Y + B]）之间的交点，则两个矩形相交两者都不是空的。

struct Rectangle{
    bool intersects(const Rectangle&);

    unsigned int a; //!< width of the rectangle
    unsigned int b; //!< height of the rectangle
    unsigned int x; //!< x position
    unsigned int y; //!< y position
};

bool Rectangle::intersects(const Rectangle& oRectangle){        
    return  (x < oRectangle.x + oRectangle.a) && // [x,x+a], [X,X+A] intersection
            (oRectangle.x < x + a)            && // [x,x+a], [X,X+A] intersection
            (y < oRectangle.y + oRectangle.b) && // [y,y+b], [Y,Y+B] intersection
            (oRectangle.y < y + b);              // [y,y+b], [Y,Y+B] intersection
}

所以你的代码应该是

if(((shapeA->getX() + shapeA->getSize()) > (player->getX()) // x intersection
    && (shapeA->getX() < (player->getX() + player->getSize())) // x intersection
    && (shapeA->getY() < player->getY() + player->getSize()  // y intersection
    && (shapeA->getY() + shapeA->getSize()) > player->getY())  // y intersection
)

Answer 2

假设getX / Y给出方块的左下角，

shapeMinX = shapeA; shapeMaxX = shapeB;
if (shapeA()->getX() > shapeB()->getX())
  swap (shapeMinX, shapeMaxX);
shapeMinY = shapeA; shapeMaxY = shapeB;
if (shapeA()->getY() > shapeB()->getY())
  swap (shapeMinY, shapeMaxY);

collision = (shapeMinX->getX()+shapeMinX->size() >= shapeMaxX()->getX) || (shapeMinX->getY()+shapeMinY->size() >= shapeMaxY()->getY);

Answer 3

你做错了测试，试试这个：

int left_bound_A=  shapeA->getX()-shapeA->getSize();
int right_bound_A= shapeA->getX()+shapeA->getSize();
int top_bound_A= shapeA->getY()-shapeA->getSize();
int bottom_bound_A= shapeA->getY()+shapeA->getSize();

int left_bound_B=  shapeB->getX()-shapeB->getSize();
int right_bound_B= shapeB->getX()+shapeB->getSize();
int top_bound_B= shapeB->getY()-shapeB->getSize();
int bottom_bound_B= shapeB->getY()+shapeB->getSize();

if( left_bound_A < right_bound_B &&
    right_bound_A > left_bound_B &&
    top_bound_A > bottom_bound_B &&
    bottom_bound_A < top_bound_B ) colide(shapeA,shapeB);

一般方法是测试形状交叉。如果实现Box或Rectangle类，代码将简化为：

 Box colision= intersect( shapeA->getBoundBox(), shapeB->getBoundBox() );
 if( colision.have_positive_area() )
     colide(shapeA,shapeB,colision);

Answer 4

是的，检查两个矩形是否简单的方法。正如一个建议，如果你想要计算矩形列表中矩形之间的所有可能的交集，通过增加它们的边界的x来预先排序它们然后开始利用这种关系的比较。这个问题可能会有所帮助

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