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https://codeberg.org/forgejo/forgejo.git
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70 lines
2.1 KiB
Go
70 lines
2.1 KiB
Go
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// Copyright 2015 by caixw, All rights reserved.
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// Use of this source code is governed by a MIT
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// license that can be found in the LICENSE file.
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package identicon
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var (
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// 4个元素分别表示cos(0),cos(90),cos(180),cos(270)
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cos = []float64{1, 0, -1, 0}
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// 4个元素分别表示sin(0),sin(90),sin(180),sin(270)
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sin = []float64{0, 1, 0, -1}
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)
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// 将points中的所有点,以x,y为原点旋转angle个角度。
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// angle取值只能是[0,1,2,3],分别表示[0,90,180,270]
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func rotate(points []float64, x, y float64, angle int) {
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if angle < 0 || angle > 3 {
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panic("rotate:参数angle必须0,1,2,3三值之一")
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}
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for i := 0; i < len(points); i += 2 {
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px := points[i] - x
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py := points[i+1] - y
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points[i] = px*cos[angle] - py*sin[angle] + x
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points[i+1] = px*sin[angle] + py*cos[angle] + y
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}
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}
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// 判断某个点是否在多边形之内,不包含构成多边形的线和点
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// x,y 需要判断的点坐标
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// points 组成多边形的所顶点,每两个元素表示一点顶点,其中最后一个顶点必须与第一个顶点相同。
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func pointInPolygon(x float64, y float64, points []float64) bool {
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if len(points) < 8 { // 只有2个以上的点,才能组成闭合多边形
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return false
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}
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// 大致算法如下:
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// 把整个平面以给定的测试点为原点分两部分:
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// - y>0,包含(x>0 && y==0)
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// - y<0,包含(x<0 && y==0)
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// 依次扫描每一个点,当该点与前一个点处于不同部分时(即一个在y>0区,一个在y<0区),
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// 则判断从前一点到当前点是顺时针还是逆时针(以给定的测试点为原点),如果是顺时针r++,否则r--。
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// 结果为:2==abs(r)。
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r := 0
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x1, y1 := points[0], points[1]
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prev := (y1 > y) || ((x1 > x) && (y1 == y))
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for i := 2; i < len(points); i += 2 {
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x2, y2 := points[i], points[i+1]
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curr := (y2 > y) || ((x2 > x) && (y2 == y))
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if curr == prev {
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x1, y1 = x2, y2
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continue
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}
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mul := (x1-x)*(y2-y) - (x2-x)*(y1-y)
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if mul > 0 {
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r++
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} else if mul < 0 {
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r--
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}
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x1, y1 = x2, y2
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prev = curr
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}
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return r == 2 || r == -2
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}
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