javascript 找到给定点集的简单闭合路径（Find Simple Closed Path for a given set of points）

本文链接：https://blog.youkuaiyun.com/hefeng_aspnet/article/details/141715931

给定一组点，将这些点连接起来而不相交

例子：

输入：points[] = {(0, 3), (1, 1), (2, 2), (4, 4),
(0, 0), (1, 2), (3, 1}, {3, 3}};

输出：按以下顺序连接点将
不造成任何交叉
{(0, 0), (3, 1), (1, 1), (2, 2), (3, 3),
（4，4），（1，2），（0，3）}

我们强烈建议您最小化浏览器并先自己尝试一下。

这个想法是使用排序。

通过比较所有点的 y 坐标来找到最底部的点。如果有两个点的 y 值相同，则考虑 x 坐标值较小的点。将最底部的点放在第一个位置。

考虑剩余的 n-1 个点，并围绕 points[0] 按照极角逆时针顺序排列它们。如果两个点的极角相同，则将最近的点放在最前面。

遍历排序数组（按角度升序排序）产生简单的闭合路径。

如何计算角度？
一种解决方案是使用三角函数。
观察：我们不关心角度的实际值。我们只想按角度排序。
想法：使用方向来比较角度，而无需实际计算它们！

以下是上述想法的实现:

// A javascript program to find simple closed path for n points
// for explanation of orientation()

// A global point needed for sorting points with reference
// to the first point. Used in compare function of qsort()
let p0;

// A utility function to return square of distance between
// p1 and p2
function dist(p1, p2)
{
return (p1[0] - p2[0])*(p1[0] - p2[0]) +
(p1[1] - p2[1])*(p1[1] - p2[1]);
}

// To find orientation of ordered triplet (p, q, r).
// The function returns following values
// 0 --> p, q and r are collinear
// 1 --> Clockwise
// 2 --> Counterclockwise
function orientation(p, q, r)
{
let val = (q[1] - p[1]) * (r[0] - q[0]) - (q[0] - p[0]) * (r[1] - q[1]);

if (val == 0) return 0; // collinear
return (val > 0)? 1: 2; // clockwise or counterclock wise
}

// A function used by library function qsort() to sort
// an array of points with respect to the first point
function compare(vp1, vp2)
{
let p1 = vp1;
let p2 = vp2;

// Find orientation
let o = orientation(p0, p1, p2);
if (o == 0)
return (dist(p0, p2) >= dist(p0, p1))? -1 : 1;

return (o == 2)? -1: 1;
}

// Prints simple closed path for a set of n points.
function printClosedPath(points, n)
{
// Find the bottommost point
let ymin = points[0][1];
let min = 0;
for (let i = 1; i < n; i++)
{
let y = points[i][1];

// Pick the bottom-most. In case of tie, choose the
// left most point
if ((y < ymin) || (ymin == y && points[i][0] < points[min][0])){
ymin = points[i][1];
min = i;
}

}

// Place the bottom-most point at first position
let temp = points[0];
points[0] = points[min];
points[min] = temp;

// Sort n-1 points with respect to the first point.
// A point p1 comes before p2 in sorted output if p2
// has larger polar angle (in counterclockwise
// direction) than p1
p0 = points[0];
points.sort(compare);

// Now stack has the output points, print contents
// of stack
for (let i=0; i<n; i++)
console.log("(" + points[i][0] + "," + points[i][1] + "), ");
}

// Driver program to test above functions
let points = [[0, 3], [1, 1], [2, 2], [4, 4], [0, 0], [1, 2], [3, 1], [3, 3]];
let n = points.length;

printClosedPath(points, n);

// The code is contributed by Nidhi goel.