处理椭圆和矩形的面积相交问题
作者：互联网
首先将椭圆的圆心移动到原点处，同时矩形也进行相应的移动。
在次之后，使用仿射变换，选定任意一个轴变换使其变成圆，那么相应的矩形也同等比例的变换。
变换之后，来一个板子把他们整合起来，结束。
1 int dcmp(double x) {
2 if (fabs(x) < eps)return 0;
3 return x > 0 ? 1 : -1;
4 }
5 struct Point {
6 double x, y;
7 Point(double _x = 0, double _y = 0) {
8 x = _x;y = _y;
9 }
10 };
11 Point operator + (const Point& a, const Point& b) {
12 return Point(a.x + b.x, a.y + b.y);
13 }
14 Point operator - (const Point& a, const Point& b) {
15 return Point(a.x - b.x, a.y - b.y);
16 }
17 Point operator * (const Point& a, const double& p) {
18 return Point(a.x * p, a.y * p);
19 }
20 Point operator / (const Point& a, const double& p) {
21 return Point(a.x / p, a.y / p);
22 }
23 bool operator < (const Point& a, const Point& b) {
24 return a.x < b.x || (dcmp(a.x - b.x) == 0 && a.y < b.y);
25 }
26 bool operator == (const Point& a, const Point& b) {
27 return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
28 }
29 double Dot(Point a, Point b) {
30 return a.x * b.x + a.y * b.y;
31 }
32 double Length(Point a) {
33 return sqrt(Dot(a, a));
34 }
35 double Angle(Point a, Point b) {
36 return acos(Dot(a, b) / Length(a) / Length(b));
37 }
38 double angle(Point a) {
39 return atan2(a.y, a.x);
40 }
41 double Cross(Point a, Point b) {
42 return a.x * b.y - a.y * b.x;
43 }
44 Point vecunit(Point a) {
45 return a / Length(a);
46 }
47 Point Normal(Point a) {
48 return Point(-a.y, a.x) / Length(a);
49 }
50 Point Rotate(Point a, double rad) {
51 return Point(a.x * cos(rad) - a.y * sin(rad), a.x * sin(rad) + a.y * cos(rad));
52 }
53 double Area2(Point a, Point b, Point c) {
54 return Length(Cross(b - a, c - a));
55 }
56 bool OnSegment(Point p, Point a1, Point a2) {
57 return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) <= 0;
58 }
59 struct Line {
60 Point p, v;
61 double ang;
62 Line() {};
63 Line(Point p, Point v) :p(p), v(v) {
64 ang = atan2(v.y, v.x);
65 }
66 bool operator < (const Line& L) const {
67 return ang < L.ang;
68 }
69 Point point(double d) {
70 return p + (v * d);
71 }
72 };
73 bool OnLeft(const Line& L, const Point& p) {
74 return Cross(L.v, p - L.p) >= 0;
75 }
76 Point GetLineIntersection(Point p, Point v, Point q, Point w) {
77 Point u = p - q;
78 double t = Cross(w, u) / Cross(v, w);
79 return p + v * t;
80 }
81 Point GetLineIntersection(Line a, Line b) {
82 return GetLineIntersection(a.p, a.v, b.p, b.v);
83 }
84 double PolyArea(vector<Point> p) {
85 int n = p.size();
86 double ans = 0;
87 for (int i = 1;i < n - 1;i++)
88 ans += Cross(p[i] - p[0], p[i + 1] - p[0]);
89 return fabs(ans) / 2;
90 }
91 struct Circle
92 {
93 Point c;
94 double r;
95 Circle() {}
96 Circle(Point c, double r) :c(c), r(r) {}
97 Point point(double a) //根据圆心角求点坐标
98 {
99 return Point(c.x + cos(a) * r, c.y + sin(a) * r);
100 }
101 };
102
103 bool InCircle(Point x, Circle c) {
104 return dcmp(c.r - Length(c.c - x)) >= 0;
105 }
106 bool OnCircle(Point x, Circle c) {
107 return dcmp(c.r - Length(c.c - x)) == 0;
108 }
109 int getSegCircleIntersection(Line L, Circle C, Point* sol) {
110 Point nor = Normal(L.v);
111 Line p1 = Line(C.c, nor);
112 Point ip = GetLineIntersection(p1, L);
113 double dis = Length(ip - C.c);
114 if (dcmp(dis - C.r) > 0)return 0;
115 Point dxy = vecunit(L.v) * sqrt(C.r * C.r - dis * dis);
116 int ret = 0;
117 sol[ret] = ip + dxy;
118 if (OnSegment(sol[ret], L.p, L.point(1)))ret++;
119 sol[ret] = ip - dxy;
120 if (OnSegment(sol[ret], L.p, L.point(1)))ret++;
121 return ret;
122 }
123 double SegCircleArea(Circle C, Point a, Point b) {
124 double a1 = angle(a - C.c);
125 double a2 = angle(b - C.c);
126 double da = fabs(a1 - a2);
127 if (da > pi)da = pi * 2 - da;
128 return dcmp(Cross(b - C.c, a - C.c)) * da * C.r * C.r / 2.0;
129 }
130 double PolyCircleArea(Circle C, Point* p, int n) {
131 double ret = 0;
132 Point sol[2];
133 p[n] = p[0];
134 for (int i = 0;i < n;i++) {
135 double t1, t2;
136 int cnt = getSegCircleIntersection(Line(p[i], p[i + 1] - p[i]), C, sol); //判断线段与圆有几个交点，
137 // cout<<"cnt="<<cnt<<" "<<p[i].x<<" "<<p[i].y<<" "<<p[i+1].x<<" "<<p[i+1].y<<endl;
138 // cout<<"C: "<<C.c.x<<" "<<C.c.y<<" "<<C.r<<endl;
139 // cout<<"sol ";for(int j=0;j<cnt;j++)cout<<sol[j].x<<" "<<sol[j].y<<" ";cout<<endl;
140 //cout << cnt << endl;
141 if (cnt == 0) { //0个交点，判断线段在多边形内部还是外部。
142 if (!InCircle(p[i], C) || !InCircle(p[i + 1], C))ret += SegCircleArea(C, p[i], p[i + 1]); //外部直接计算圆弧面积
143 else ret += Cross(p[i + 1] - C.c, p[i] - C.c) / 2; //内部计算三角形面积。
144 }
145 if (cnt == 1) {
146 if (InCircle(p[i], C) && (!InCircle(p[i + 1], C) || OnCircle(p[i + 1], C)))ret += Cross(sol[0] - C.c, p[i] - C.c) / 2, ret += SegCircleArea(C, sol[0], p[i + 1]);//,cout<<"jj-1"<<endl;
147 else ret += SegCircleArea(C, p[i], sol[0]), ret += Cross(p[i + 1] - C.c, sol[0] - C.c) / 2;//,cout<<"jj-2"<<endl;
148 }
149 if (cnt == 2) { //两个交点
150 if ((p[i] < p[i + 1]) ^ (sol[0] < sol[1]))swap(sol[0], sol[1]);
151 ret += SegCircleArea(C, p[i], sol[0]);
152 ret += Cross(sol[1] - C.c, sol[0] - C.c) / 2;
153 ret += SegCircleArea(C, sol[1], p[i + 1]);
154 }
155 // cout<<ret<<endl;
156 }
157 return fabs(ret);
158 }
标签：椭圆,return,Point,double,相交,Length,dcmp,const,矩形 来源： https://www.cnblogs.com/xyisreallycuteeee/p/15081016.html