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?
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;
}
A和B是两个矩形。C是它们的覆盖矩形。
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)
它考虑到所有可能的情况。
这是来自《Java编程入门-综合版》中的练习3.28。该代码测试两个矩形是否缩进，一个矩形是否在另一个矩形内，一个矩形是否在另一个矩形外。如果这些条件都不满足，则两者重叠。
**3.28(几何:两个矩形)编写一个程序，提示用户进入
中心x, y坐标，宽度和高度的两个矩形，并确定
第二个矩形是在第一个矩形的内部还是与第一个矩形重叠，如图所示
如图3.9所示。测试您的程序以覆盖所有情况。
下面是示例运行:
输入r1的中心x坐标，y坐标，宽度和高度:2.5 4 2.5 43
输入r2的中心x坐标，y坐标，宽度和高度:1.5 5 0.5 3
R2在r1里面
输入r1的中心x坐标，y坐标，宽度和高度:1 2 3 5.5
输入r2的中心x坐标，y坐标，宽度和高度:3 4 4.5 5
R2和r1重叠
输入r1的中心x坐标，y坐标，宽度和高度:1 2 3 3
输入r2的中心x坐标，y坐标，宽度和高度:40 45 3 2
R2不与r1重叠
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");
}
}
}
最简单的方法是
/**
* 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);
}
首先要记住在计算机中坐标系统是颠倒的。x轴与数学中的相同，但y轴向下增大，向上减小。
如果矩形是从中心画的。
如果x1坐标大于x2加上它的一半宽。然后这意味着他们会互相接触。用同样的方法向下+一半高。它会碰撞的。
