Question

假设我有两个圆圈，其中（x ₁，y ₁，r ₁）和 （x ₂，y ₂，r ₂）分别为其坐标和半径。这样，它们之间的最大和最小距离是多少。到目前为止，我所做的事情：

double centreDistance = Math.sqrt(Math.pow((x[i] - x[j]), 2) + Math.pow((x[i] - y[j]), 2));
if (centreDistance >= r[i] + r[j]) {
    double maxDistance = centreDistance + (r[i] + r[j]);
    double minDistance = centreDistance - (r[i] + r[j]);
} else if (centreDistance == 0) {
    double minDistance = Math.abs(r[i] - r[j]);
    double maxDistance = Math.abs(r[i] - r[j]);
} else {
    int rmax = Math.max(r[i], r[j]);
    int rmin = Math.min(r[i], r[j]);
    double minDistance = rmax - (centreDistance + rmin);
    double maxDistance = rmax + (centreDistance - rmin);
}

这里x，y，r是分别具有每个圆和半径的中心的x，y坐标的数组。这给了我错误的答案。

Answer 1

else路径错误。

此：

double maxDistance = rmax + (centreDistance-rmin);

应为：

double maxDistance = rmax + centreDistance + rmin;

您还缺少圆圈相交的情况。

centreDistance < r1+r2 && centredistance + rmin >= rmax

在这种情况下

minDistance = 0 
maxDistance = rmax + centreDistance + rmin

centreDistance == 0也是错误的，它应该是maxDistance = r1 + r2（因为您应该比较中心相对侧的点。无论如何，这只是else路径的特例，可以被忽略。

实际情况是：

centreDistance>=r[i]+r[j]       (centres are so far apart the circles don't overlap, or touch in one point)
centreDistance < r1+r2 && centredistance + rmin >= rmax        (circles intersect)
centreDistance < r1+r2 && centredistance + rmin < rmax        (one circle inside the other)

Answer 2

在圆圈接触的情况下，您必须进行一些调整

if (centreDistance >= r[i] + r[j]) {
        // the circle dont touch, min and max distance are the closest and most far apart parts of the circles
        // code is correct
        double maxDistance = centreDistance + (r[i] + r[j]);
        double minDistance = centreDistance - (r[i] + r[j]);
    } else if (centreDistance == 0) {
        // the circle are on top of each other
        // check if they have the same radius -> max min dist = 0
        // if not max and min distance = Math.abs(r[i] - r[j]);
        // code is correct
        double minDistance = Math.abs(r[i] - r[j]);
        double maxDistance = Math.abs(r[i] - r[j]);
    } else {
        // the circles touch 
        // max distance is the same as in case where they dont touch
        // and min distance is 0 because they touch
        // code has to be adjusted
        int rmax = Math.max(r[i], r[j]);
        int rmin = Math.min(r[i], r[j]);
        double minDistance = rmax - (centreDistance + rmin);
        double maxDistance = rmax + (centreDistance - rmin);
    }

Answer 3

问题是Array ( [0] => 1428 [1] => 1428 ) Array ( [0] => 217 )和rmax是整数。返回max和min的结果时，它会四舍五入为rmin值。

不同圆上两点之间的最大和最小距离是多少？

3 个答案: