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
{
   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;
}

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;
};

bool
overlap(const Rect &r1, const Rect &r2)
{
    // The rectangles don't overlap if
    // one rectangle's minimum in some dimension 
    // is greater than the other's maximum in
    // that dimension.

    bool noOverlap = r1.x1 > r2.x2 ||
                     r2.x1 > r1.x2 ||
                     r1.y1 > r2.y2 ||
                     r2.y1 > r1.y2;

    return !noOverlap;
}

下面是如何在Java API中完成的:

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));
}
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;
}

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");
    }

}
}

更容易检查一个矩形是否完全在另一个矩形之外,如果它是其中之一

在左边……

(r1.x + r1.width < r2.x)

或者在右边…

(r1.x > r2.x + r2.width)

或者在上面…

(r1.y + r1.height < r2.y)

或者在底部…

(r1.y > r2.y + r2.height)

对于第二个矩形,它不可能与它碰撞。因此,要有一个返回布尔值的函数,表示矩形是否碰撞,我们只需通过逻辑or组合这些条件,并对结果求反:

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);
}

如果只在触摸时得到阳性结果,我们可以通过“<=”和“>=”来更改“<”和“>”。