判断两个矩形是否互相重叠?

我试图写一个c++程序，从用户获取以下输入来构造矩形(2和5之间):高度，宽度，x-pos, y-pos。所有这些矩形都平行于x轴和y轴，也就是说它们所有边的斜率都是0或无穷大。

我试图实现这个问题中提到的东西，但我没有太多的运气。

我目前的实现如下:

// Gets all the vertices for Rectangle 1 and stores them in an array -> arrRect1
// point 1 x: arrRect1[0], point 1 y: arrRect1[1] and so on...
// Gets all the vertices for Rectangle 2 and stores them in an array -> arrRect2

// rotated edge of point a, rect 1
int rot_x, rot_y;
rot_x = -arrRect1[3];
rot_y = arrRect1[2];
// point on rotated edge
int pnt_x, pnt_y;
pnt_x = arrRect1[2]; 
pnt_y = arrRect1[3];
// test point, a from rect 2
int tst_x, tst_y;
tst_x = arrRect2[0];
tst_y = arrRect2[1];

int value;
value = (rot_x * (tst_x - pnt_x)) + (rot_y * (tst_y - pnt_y));
cout << "Value: " << value;

然而，我不太确定(a)我是否已经正确地实现了我链接的算法，或者如果我确实如何解释这一点?

有什么建议吗?

当前回答

struct rect
{
    int x;
    int y;
    int width;
    int height;
};

bool valueInRange(int value, int min, int max)
{ return (value >= min) && (value <= max); }

bool rectOverlap(rect A, rect B)
{
    bool xOverlap = valueInRange(A.x, B.x, B.x + B.width) ||
                    valueInRange(B.x, A.x, A.x + A.width);

    bool yOverlap = valueInRange(A.y, B.y, B.y + B.height) ||
                    valueInRange(B.y, A.y, A.y + A.height);

    return xOverlap && yOverlap;
}

2008-11-20 18:36:20

其他回答

struct Rect
{
   Rect(int x1, int x2, int y1, int y2)
   : x1(x1), x2(x2), y1(y1), y2(y2)
   {
       assert(x1 < x2);
       assert(y1 < y2);
   }

   int x1, x2, y1, y2;
};

//some area of the r1 overlaps r2
bool overlap(const Rect &r1, const Rect &r2)
{
    return r1.x1 < r2.x2 && r2.x1 < r1.x2 &&
           r1.y1 < r2.y2 && r2.x1 < r1.y2;
}

//either the rectangles overlap or the edges touch
bool touch(const Rect &r1, const Rect &r2)
{
    return r1.x1 <= r2.x2 && r2.x1 <= r1.x2 &&
           r1.y1 <= r2.y2 && r2.x1 <= r1.y2;
}

2008-11-20 19:31:23

设这两个矩形是矩形A和矩形b，设它们的中心为A1和B1 (A1和B1的坐标很容易求出来)，设高为Ha和Hb，宽为Wa和Wb，设dx为A1和B1之间的宽度(x)， dy为A1和B1之间的高度(y)。

现在我们可以说我们可以说A和B重叠，当

if(!(dx > Wa+Wb)||!(dy > Ha+Hb)) returns true

2012-06-28 15:16:15

对于那些使用中心点和一半大小的矩形数据的人，而不是典型的x,y,w,h或x0,y0,x1,x1，下面是你可以这样做:

#include <cmath> // for fabsf(float)

struct Rectangle
{
    float centerX, centerY, halfWidth, halfHeight;
};

bool isRectangleOverlapping(const Rectangle &a, const Rectangle &b)
{
    return (fabsf(a.centerX - b.centerX) <= (a.halfWidth + b.halfWidth)) &&
           (fabsf(a.centerY - b.centerY) <= (a.halfHeight + b.halfHeight)); 
}

2017-08-13 18:27:11

我已经实现了c#版本，它很容易转换为c++。

public bool Intersects ( Rectangle rect )
{
  float ulx = Math.Max ( x, rect.x );
  float uly = Math.Max ( y, rect.y );
  float lrx = Math.Min ( x + width, rect.x + rect.width );
  float lry = Math.Min ( y + height, rect.y + rect.height );

  return ulx <= lrx && uly <= lry;
}

2008-11-20 18:46:04

struct rect
{
    int x;
    int y;
    int width;
    int height;
};

bool valueInRange(int value, int min, int max)
{ return (value >= min) && (value <= max); }

bool rectOverlap(rect A, rect B)
{
    bool xOverlap = valueInRange(A.x, B.x, B.x + B.width) ||
                    valueInRange(B.x, A.x, A.x + A.width);

    bool yOverlap = valueInRange(A.y, B.y, B.y + B.height) ||
                    valueInRange(B.y, A.y, A.y + A.height);

    return xOverlap && yOverlap;
}

2008-11-20 18:36:20

判断两个矩形是否互相重叠?

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