确定两个矩形是否相互重叠？ 我试图编写一个C++程序，该程序使用用户的下列输入来构造矩形（2到5）：高度、宽度、X-POS、Y-Po.所有这些矩形都将平行于x和y轴，即它们的所有边都有0或无穷大斜率。
我已经试着实现上面提到的问题，但是我运气不太好 我当前的实现执行以下操作：确定两个矩形是否相互重叠？ 我试图编写一个C++程序，该程序使用用户的下列输入来构造矩形（2到5）：高度、宽度、X-POS、Y-Po.所有这些矩形都将平行于x和y轴，即它们的所有边都有0或无穷大斜率。,c++,algorithm,geometry,overlap,rectangles,C++,Algorithm,Geometry,Overlap,Rectangles,我已经试着实现上面提到的问题，但是我运气不太好 我当前的实现执行以下操作： // 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 ar
// 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;
//获取矩形1的所有顶点并将它们存储在数组->arrRect1中
//点1X:arrRect1[0]，点1Y:arrRect1[1]等等。。。
//获取矩形2的所有顶点，并将它们存储在数组->arrRect2中
//点a的旋转边，矩形1
int rot_x，rot_y；
rot_x=-arrRect1[3]；
rot_y=arrRect1[2]；
//旋转边上的点
int pnt_x，pnt_y；
pnt_x=arrRect1[2]；
pnt_y=arrRect1[3]；
//测试点，a从rect 2开始
int tst_x，tst_y；
tst_x=arrRect2[0]；
tst_y=arrRect2[1]；
int值；
值=（rot_x*（tst_x-pnt_x））+（rot_y*（tst_y-pnt_y））；
库特
这相当于：
- A的左边缘到B的右边缘的左侧，[
RectA.Left
]，以及
- A的右边缘到B的左边缘的右侧，[
RectA.right>RectB.left
]，以及
- A的顶部高于B的底部，[
RectA.top>RectB.bottom
]，以及
- A的底部低于B的顶部[
RectA.bottom
]
注1：很明显，同样的原则可以扩展到任意数量的维度。
注2：只计算一个像素的重叠也应该是相当明显的，将该边界上的
更改为=
注3：当使用笛卡尔坐标（X，Y）时，此答案基于标准代数笛卡尔坐标（X从左到右增加，Y从下到上增加）。显然，如果计算机系统可能以不同方式机械化屏幕坐标（例如，从上到下增加Y，或从右到左增加X），则需要相应地调整语法
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;
}
{
int x；
int-y；
整数宽度；
内部高度；
};
布尔值范围（int值、int最小值、int最大值）
{Real[Value>MIN ] & &（value < P>）我已经实现了C版本，它很容易转换成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;
}
public bool相交（矩形矩形）
{
float ulx=数学最大值（x，rect.x）；
float uly=数学最大值（y，rect.y）；
float lrx=数学最小值（x+宽度，rect.x+rect.width）；
浮动lry=数学最小值（y+高度，矩形y+矩形高度）；
return ulx问自己一个相反的问题：我如何确定两个矩形是否完全不相交？显然，完全位于矩形B左侧的矩形a不相交。同样，如果a完全位于右侧。同样，如果a完全在B上方或完全在B下方。在任何其他情况下，a和B相交
以下内容可能有缺陷，但我对算法非常有信心：
struct Rectangle { int x; int y; int width; int height; };
bool is_left_of(Rectangle const & a, Rectangle const & b) {
if (a.x + a.width <= b.x) return true;
return false;
}
bool is_right_of(Rectangle const & a, Rectangle const & b) {
return is_left_of(b, a);
}
bool not_intersect( Rectangle const & a, Rectangle const & b) {
if (is_left_of(a, b)) return true;
if (is_right_of(a, b)) return true;
// Do the same for top/bottom...
}
bool intersect(Rectangle const & a, Rectangle const & b) {
return !not_intersect(a, b);
}
结构矩形{intx；inty；intwidth；intheight；}；
布尔是（矩形常数a、矩形常数b）的左{
如果（a.x+a.widthstruct Rect
{
矩形（整数x1，整数x2，整数y1，整数y2）
：x1（x1）、x2（x2）、y1（y1）、y2（y2）
{
断言（x1r2.x2||
r2.x1>r1.x2||
r1.y1>r2.y2||
r2.y1>r1.y2；
返回！noOverlap；
}
在问题中，您可以链接到矩形处于任意旋转角度时的数学。但是，如果我理解问题中有关角度的部分，我会解释所有矩形都相互垂直
了解重叠面积公式的一般方法是：
使用以下示例：
1 2 3 4 5 6
1 +---+---+
| |
2 + A +---+---+
| | B |
3 + + +---+---+
| | | | |
4 +---+---+---+---+ +
| |
5 + C +
| |
6 +---+---+
1 2 3 4 5 6
1 +---+---+
| |
2+A+---+---+
|| B|
3 + + +---+---+
| | | | |
4 +---+---+---+---+ +
| |
5+C+
| |
6 +---+---+
1） 将所有的x坐标（包括左坐标和右坐标）收集到一个列表中，然后对其进行排序并删除重复项
1 3 4 5 6
1 2 3 4 6
1 3 4 5 6
2） 将所有y坐标（顶部和底部）收集到一个列表中，然后对其进行排序并删除重复项
1 3 4 5 6
1 2 3 4 6
1 2 3 4 6
3） 通过唯一x坐标之间的间距数量*唯一y坐标之间的间距数量创建二维阵列
4 * 4
4 * 4
4） 将所有矩形绘制到此网格中，增加其上出现的每个单元格的计数：
1 3 4 5 6
1 +---+
| 1 | 0 0 0
2 +---+---+---+
| 1 | 1 | 1 | 0
3 +---+---+---+---+
| 1 | 1 | 2 | 1 |
4 +---+---+---+---+
0 0 | 1 | 1 |
6 +---+---+
1 3 4 5 6
1 +---+
| 1 | 0 0 0
2 +---+---+---+
| 1 | 1 | 1 | 0
3 +---+---+---+---+
| 1 | 1 | 2 | 1 |
4 +---+---+---+---+
0 0 | 1 | 1 |
6 +---+---+
5） 绘制矩形时，很容易截取重叠。struct Rect
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;
}
{
矩形（整数x1，整数x2，整数y1，整数y2）
：x1（x1）、x2（x2）、y1（y1）、y2（y2）
{
断言（x1bool bOverlap = !((A.Left >= B.Right || B.Left >= A.Right)
&& (A.Bottom >= B.Top || B.Bottom >= A.Top));
如果矩形重叠，则重叠区域将大于零。现在让我们找到重叠区域：
如果它们重叠，则重叠矩形的左边缘将为max（r1.x1，r2.x1）
，右边缘将为min（r1.x2，r2.x2）
4 * 4
1 3 4 5 6
1 +---+
| 1 | 0 0 0
2 +---+---+---+
| 1 | 1 | 1 | 0
3 +---+---+---+---+
| 1 | 1 | 2 | 1 |
4 +---+---+---+---+
0 0 | 1 | 1 |
6 +---+---+
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;
}
bool bOverlap = !((A.Left >= B.Right || B.Left >= A.Right)
&& (A.Bottom >= B.Top || B.Bottom >= A.Top));
area = (max(r1.x1, r2.x1) - min(r1.x2, r2.x2)) * (max(r1.y1, r2.y1) - min(r1.y2, r2.y2))
(r1.x + r1.width < r2.x)
(r1.x > r2.x + r2.width)
(r1.y + r1.height < r2.y)
(r1.y > r2.y + r2.height)
function checkOverlap(r1, r2) : Boolean
{
return !(r1.x + r1.width < r2.x || r1.y + r1.height < r2.y || r1.x > r2.x + r2.width || r1.y > r2.y + r2.height);
}
public boolean intersects(Rectangle r) {
int tw = this.width;
int th = this.height;
int rw = r.width;
int rh = r.height;
if (rw <= 0 || rh <= 0 || tw <= 0 || th <= 0) {
return false;
}
int tx = this.x;
int ty = this.y;
int rx = r.x;
int ry = r.y;
rw += rx;
rh += ry;
tw += tx;
th += ty;
// overflow || intersect
return ((rw < rx || rw > tx) &&
(rh < ry || rh > ty) &&
(tw < tx || tw > rx) &&
(th < ty || th > ry));
}
if(!(dx > Wa+Wb)||!(dy > Ha+Hb)) returns true
four points of A be (xAleft,yAtop),(xAleft,yAbottom),(xAright,yAtop),(xAright,yAbottom)
four points of A be (xBleft,yBtop),(xBleft,yBbottom),(xBright,yBtop),(xBright,yBbottom)
A.width = abs(xAleft-xAright);
A.height = abs(yAleft-yAright);
B.width = abs(xBleft-xBright);
B.height = abs(yBleft-yBright);
C.width = max(xAleft,xAright,xBleft,xBright)-min(xAleft,xAright,xBleft,xBright);
C.height = max(yAtop,yAbottom,yBtop,yBbottom)-min(yAtop,yAbottom,yBtop,yBbottom);
A and B does not overlap if
(C.width >= A.width + B.width )
OR
(C.height >= A.height + B.height)
/**
* Check if two rectangles collide
* x_1, y_1, width_1, and height_1 define the boundaries of the first rectangle
* x_2, y_2, width_2, and height_2 define the boundaries of the second rectangle
*/
boolean rectangle_collision(float x_1, float y_1, float width_1, float height_1, float x_2, float y_2, float width_2, float height_2)
{
return !(x_1 > x_2+width_2 || x_1+width_1 < x_2 || y_1 > y_2+height_2 || y_1+height_1 < y_2);
}
class Vector2D
{
public:
Vector2D(int x, int y) : x(x), y(y) {}
~Vector2D(){}
int x, y;
};
bool DoRectanglesOverlap( const Vector2D & Pos1,
const Vector2D & Size1,
const Vector2D & Pos2,
const Vector2D & Size2)
{
if ((Pos1.x < Pos2.x + Size2.x) &&
(Pos1.y < Pos2.y + Size2.y) &&
(Pos2.x < Pos1.x + Size1.x) &&
(Pos2.y < Pos1.y + Size1.y))
{
return true;
}
return false;
}
DoRectanglesOverlap(Vector2D(3, 7),
Vector2D(8, 5),
Vector2D(6, 4),
Vector2D(9, 4));
if ((Pos1.x < Pos2.x + Size2.x) &&
(Pos1.y < Pos2.y + Size2.y) &&
(Pos2.x < Pos1.x + Size1.x) &&
(Pos2.y < Pos1.y + Size1.y))
↓
if (( 3 < 6 + 9 ) &&
( 7 < 4 + 4 ) &&
( 6 < 3 + 8 ) &&
( 4 < 7 + 5 ))
import java.util.Scanner;
public class ProgrammingEx3_28 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out
.print("Enter r1's center x-, y-coordinates, width, and height:");
double x1 = input.nextDouble();
double y1 = input.nextDouble();
double w1 = input.nextDouble();
double h1 = input.nextDouble();
w1 = w1 / 2;
h1 = h1 / 2;
System.out
.print("Enter r2's center x-, y-coordinates, width, and height:");
double x2 = input.nextDouble();
double y2 = input.nextDouble();
double w2 = input.nextDouble();
double h2 = input.nextDouble();
w2 = w2 / 2;
h2 = h2 / 2;
// Calculating range of r1 and r2
double x1max = x1 + w1;
double y1max = y1 + h1;
double x1min = x1 - w1;
double y1min = y1 - h1;
double x2max = x2 + w2;
double y2max = y2 + h2;
double x2min = x2 - w2;
double y2min = y2 - h2;
if (x1max == x2max && x1min == x2min && y1max == y2max
&& y1min == y2min) {
// Check if the two are identicle
System.out.print("r1 and r2 are indentical");
} else if (x1max <= x2max && x1min >= x2min && y1max <= y2max
&& y1min >= y2min) {
// Check if r1 is in r2
System.out.print("r1 is inside r2");
} else if (x2max <= x1max && x2min >= x1min && y2max <= y1max
&& y2min >= y1min) {
// Check if r2 is in r1
System.out.print("r2 is inside r1");
} else if (x1max < x2min || x1min > x2max || y1max < y2min
|| y2min > y1max) {
// Check if the two overlap
System.out.print("r2 does not overlaps r1");
} else {
System.out.print("r2 overlaps r1");
}
}
}
bool Square::IsOverlappig(Square &other)
{
bool result1 = other.x >= x && other.y >= y && other.x <= (x + width) && other.y <= (y + height); // other's top left falls within this area
bool result2 = other.x >= x && other.y <= y && other.x <= (x + width) && (other.y + other.height) <= (y + height); // other's bottom left falls within this area
bool result3 = other.x <= x && other.y >= y && (other.x + other.width) <= (x + width) && other.y <= (y + height); // other's top right falls within this area
bool result4 = other.x <= x && other.y <= y && (other.x + other.width) >= x && (other.y + other.height) >= y; // other's bottom right falls within this area
return result1 | result2 | result3 | result4;
}
for ( int i = 0; i < n; i++ ) {
for ( int j = 0; j < n; j++ ) {
if ( i != j ) {
Rectangle rectangle1 = rectangles.get(i);
Rectangle rectangle2 = rectangles.get(j);
int l1 = rectangle1.l; //left
int r1 = rectangle1.r; //right
int b1 = rectangle1.b; //bottom
int t1 = rectangle1.t; //top
int l2 = rectangle2.l;
int r2 = rectangle2.r;
int b2 = rectangle2.b;
int t2 = rectangle2.t;
boolean topOnBottom = t2 == b1;
boolean bottomOnTop = b2 == t1;
boolean topOrBottomContact = topOnBottom || bottomOnTop;
boolean rightOnLeft = r2 == l1;
boolean leftOnRight = l2 == r1;
boolean rightOrLeftContact = leftOnRight || rightOnLeft;
boolean leftPoll = l2 <= l1 && r2 >= l1;
boolean rightPoll = l2 <= r1 && r2 >= r1;
boolean leftRightInside = l2 >= l1 && r2 <= r1;
boolean leftRightPossiblePlaces = leftPoll || rightPoll || leftRightInside;
boolean bottomPoll = t2 >= b1 && b2 <= b1;
boolean topPoll = b2 <= b1 && t2 >= b1;
boolean topBottomInside = b2 >= b1 && t2 <= t1;
boolean topBottomPossiblePlaces = bottomPoll || topPoll || topBottomInside;
boolean topInBetween = t2 > b1 && t2 < t1;
boolean bottomInBetween = b2 > b1 && b2 < t1;
boolean topBottomInBetween = topInBetween || bottomInBetween;
boolean leftInBetween = l2 > l1 && l2 < r1;
boolean rightInBetween = r2 > l1 && r2 < r1;
boolean leftRightInBetween = leftInBetween || rightInBetween;
if ( (topOrBottomContact && leftRightPossiblePlaces) || (rightOrLeftContact && topBottomPossiblePlaces) ) {
path[i][j] = true;
}
}
}
}
#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));
}
struct point { int x, y; };
struct rect { point tl, br; }; // top left and bottom right points
// return true if rectangles overlap
bool overlap(const rect &a, const rect &b)
{
return a.tl.x <= b.br.x && a.br.x >= b.tl.x &&
a.tl.y >= b.br.y && a.br.y <= b.tl.y;
}