Question

假设我有两个基于AABB的区域，每个区域由两个坐标mins{x, y}和maxs{x, y}定义，我想找到它们之间的中间连接点。

由于我的英语不好，我无法用言语解释所有，请参阅以下图片以便于理解： http://i.*.com/WokivEe.png

我需要找到的只是红点坐标。

所以如果我们将其转移到编程问题中，实际的数据结构将如下所示：

struct Vec2D {
    float x, y;
}

struct Rectangle {
    Vec2D min;
    Vec2D max;
}

Rectangle obj[2]

有人对算法有所了解吗？

Answer 1

沿X轴或Y轴，按顺序排列接触边的坐标。然后平均该列表中的第2和第3个以找到它们的中点。我希望这足以回答这个问题。

Answer 2

这是一个小算法，首先找到对象的哪一边最近，然后沿公共边使用4个点来制作一个列表，沿着公共轴排序。排序列表的2个中间点的平均值是答案。这适用于水平和垂直两侧。我将访问器函数添加到数据结构中，以便可以将它们编入索引;例如，对于Vec2D，坐标（0）是x值，坐标（1）是y值。

#include <math.h>
#include <iostream>
#include <limits>

struct Vec2D {
    float x, y;
    float coordinate(int axis)
    {
        return (axis & 1) ? y : x;
    }
};

struct Rectangle {
    Vec2D min;
    Vec2D max;
    Vec2D corner(int j)
    {
        return (j & 1) ? max : min;
    }

    // Get the other corner along the given axis
    Vec2D along(int j, int ax)
    {
        Vec2D p = corner(j);
        if (0 == ax)
        {
            p.x = corner(1-j).x;
        }
        else
        {
            p.y = corner(1-j).y;
        }
        return p;
    }
};

using namespace std;

inline Vec2D* vp(const void* p)
{
    return (Vec2D*) p;
}

static int compare_x(const void*a, const void*b)
{
    if (vp(a)->x < vp(b)->x)
    {
        return -1;
    }
    else
    if (vp(a)->x > vp(b)->x)
    {
        return 1;
    }
    return 0;
}

static int compare_y(const void*a, const void*b)
{
    if (vp(a)->y < vp(b)->y)
    {
        return -1;
    }
    else
    if (vp(a)->y > vp(b)->y)
    {
        return 1;
    }
    return 0;
}


int main(void) {
    int ax; // axis index
    int c0, c1;
    float gap = numeric_limits<float>::max();
    struct Rectangle obj[2] = {0,2,10,10,10,5,15,20};
    struct
    {
        int ax,c0,c1;
    } closest;

    // Find out which sides are the closest to each other
    for(ax = 0; 2 > ax; ++ax) // Look at x axis and y axis
    {
        for(c0 = 0; 2 > c0; ++c0) // Look at both corners of obj[0]
        {
            for(c1 = 0; 2 > c1; ++c1) // Look at both corners of obj[1]
            {
                float dist = fabs(obj[0].corner(c0).coordinate(ax) - obj[1].corner(c1).coordinate(ax));
                if (dist < gap)
                {
                    gap = dist;
                    closest.ax = ax;
                    closest.c0 = c0;
                    closest.c1 = c1;
                }
            }
        }
    }

    int other = 1 - closest.ax; // The other axis
    cout << "The closest gap is along the " << (closest.ax ? 'y' : 'x') << " axis\n";
    cout << "The common side is along the " << (other ? 'y' : 'x') << " direction\n";


    // Make a list of the 4 points along the common side
    Vec2D list[4];
    list[0] = obj[0].corner(closest.c0);
    list[1] = obj[0].along(closest.c0, other);
    list[2] = obj[1].corner(closest.c1);
    list[3] = obj[1].along(closest.c1, other);

    // Sort them into order along the common axis
    qsort(list, 4, sizeof(Vec2D), closest.ax ? compare_x : compare_y);

    // Get the average of the 2 middle points along the common axis.
    Vec2D answer = {
        (list[1].x + list[2].x) / 2,
        (list[1].y + list[2].y) / 2
    };

    cout << "(" << answer.x << "," << answer.y << ")\n";
}

找到两个AABB区域之间的边缘连接点

2 个答案: