怎样判断两个矩形区域是否出现交叉或重叠现象

[gzq0712]

如上:

[yxz_lp]

CRect rect1,rect2;
...................
CRgn rgn;
rgn.CreateRectRgn(rect1.left,rect1.top,rect1.right,rect1.bottom);
if (rgn.RectInRegion(&rect2))
{
	//相交
}
else
{
	//不相交
}

[laoma_hbu]

CRect::IntersectRect
Nonzero if the intersection is not empty; 0 if the intersection is empty.

[pptor]

jf

[LouYongZhi]

bool IsCross( RECT *rect1, RECT *rect2 )
{
RECT rcTmp;
rcTmp.left = _max( rect1->left, rect2->left );
rcTmp.top = _max( rect1->top, rect2->top );
rcTmp.right = _min( rect1->right, rect2->right );
rcTmp.bottom = _min( rect1->bottom, rect2->bottom );
return (rcTmp.right > rcTmp.left && rcTmp.bottom > rcTmp.top);
}

[Chivalry]

mark

[lzf8977]

MARK 一下，，，以前也做过，不过方法比较笨，，

[tiger波波]

判断两个矩形区域是否出现交叉或重叠现象
你首先应该知道这2格矩形的rect值吧，对比下就行了

[np1234]

4楼的没有个别边的重叠的判断，比如第一个长方形的左边，与第二个的右边，应该是这样的
bool IsCross( RECT *rect1, RECT *rect2 )
{
RECT rcTmp;
rcTmp.left = _max( rect1-> left, rect2-> left );
rcTmp.top = _max( rect1-> top, rect2-> top );
rcTmp.right = _min( rect1-> right, rect2-> right );
rcTmp.bottom = _min( rect1-> bottom, rect2-> bottom );
return (rcTmp.right >= rcTmp.left && rcTmp.bottom >= rcTmp.top);
}