UVA 12304 - 2D Geometry 110 in 1!(计算几何)

这题真的是恶心到爆炸啊

通过这题整理了下圆相关的计算几何模板（基本都是参考别人的)

代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;

struct Point {
double x, y;
Point() {}
Point(double x, double y) {
this->x = x;
this->y = y;
}
void read() {
scanf("%lf%lf", &x, &y);
}
};

typedef Point Vector;

Vector operator + (Vector A, Vector B) {
return Vector(A.x + B.x, A.y + B.y);
}

Vector operator - (Vector A, Vector B) {
return Vector(A.x - B.x, A.y - B.y);
}

Vector operator * (Vector A, double p) {
return Vector(A.x * p, A.y * p);
}

Vector operator / (Vector A, double p) {
return Vector(A.x / p, A.y / p);
}

bool operator < (const Point& a, const Point& b) {
return a.x < b.x || (a.x == b.x && a.y < b.y);
}

const double eps = 1e-8;
const double PI = acos(-1.0);

int dcmp(double x) {
if (fabs(x) < eps) return 0;
else return x < 0 ? -1 : 1;
}

bool operator == (const Point& a, const Point& b) {
return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

double Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y;} //点积
double Length(Vector A) {return sqrt(Dot(A, A));} //向量的模
double Angle(Vector A, Vector B) {return acos(Dot(A, B) / Length(A) / Length(B));} //向量夹角
double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;} //叉积
double Area2(Point A, Point B, Point C) {return Cross(B - A, C - A);} //有向面积
double angle(Vector v) {return atan2(v.y, v.x);}

Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {
Vector u = P - Q;
double t = Cross(w, u) / Cross(v, w);
return P + v * t;
}

Vector Rotate(Vector A, double rad) {
return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}

double DistanceToLine(Point P, Point A, Point B) {
Vector v1 = B - A, v2 = P - A;
return fabs(Cross(v1, v2)) / Length(v1);
}

Vector AngleBisector(Point p, Vector v1, Vector v2){//给定两个向量，求角平分线
double rad = Angle(v1, v2);
return Rotate(v1, dcmp(Cross(v1, v2)) * 0.5 * rad);
}

//求线与x轴的真实角(0<=X<180)
double RealAngleWithX(Vector a){
Vector b(1, 0);
if (dcmp(Cross(a, b)) == 0) return 0.0;
else if (dcmp(Dot(a, b) == 0)) return 90.0;
double rad = Angle(a, b);
rad = (rad / PI) * 180.0;
if (dcmp(a.y) < 0) rad = 180.0 - rad;
return rad;
}

struct Circle {
Point c;
double r;
Circle(Point c, double r) {
this->c = c;
this->r = r;
}
Point point(double a) {
return Point(c.x + cos(a) * r, c.y + sin(a) * r);
}
};

//求直线与圆的交点
int getLineCircleIntersection(Point p, Vector v, Circle c, vector<Point> &sol) {
double a1 = v.x, b1 = p.x - c.c.x, c1 = v.y, d1 = p.y - c.c.y;
double e1 = a1 * a1 + c1 * c1, f1 = 2 * (a1 * b1 + c1 * d1), g1 = b1 * b1 + d1 * d1 - c.r * c.r;
double delta = f1 * f1 - 4 * e1 * g1, t;
if(dcmp(delta) < 0) return 0;
else if(dcmp(delta) == 0){
t = (-f1) / (2 * e1);
sol.push_back(p + v * t);
return 1;
} else{
t = (-f1 + sqrt(delta)) / (2 * e1); sol.push_back(p + v * t);
t = (-f1 - sqrt(delta)) / (2 * e1); sol.push_back(p + v * t);
return 2;
}
}

//两圆相交
int getCircleCircleIntersection(Circle C1, Circle C2, vector<Point> &sol) {
double d = Length(C1.c - C2.c);
if (dcmp(d) == 0) {
if (dcmp(C1.r - C2.r) == 0) return -1; // 重合
return 0;
}
if (dcmp(C1.r + C2.r - d) < 0) return 0;
if (dcmp(fabs(C1.r - C2.r) - d) > 0) return 0;
double a = angle(C2.c - C1.c);
double da = acos((C1.r * C1.r + d * d - C2.r * C2.r) / (2 * C1.r * d));
Point p1 = C1.point(a - da), p2 = C1.point(a + da);
sol.push_back(p1);
if(p1 == p2) return 1;
sol.push_back(p2);
return 2;

}

//点到圆的切线
int getTangents(Point p, Circle C, Vector *v) {
Vector u = C.c - p;
double dist = Length(u);
if (dist < C.r) return 0;
else if (dcmp(dist - C.r) == 0) {
v[0] = Rotate(u, PI / 2);
return 1;
} else {
double ang = asin(C.r / dist);
v[0] = Rotate(u, -ang);
v[1] = Rotate(u, +ang);
return 2;
}
}

//两圆公切线
//a[i], b[i]分别是第i条切线在圆A和圆B上的切点
int getCircleTangents(Circle A, Circle B, Point *a, Point *b) {
int cnt = 0;
if (A.r < B.r) { swap(A, B); swap(a, b); }
//圆心距的平方
double d2 = (A.c.x - B.c.x) * (A.c.x - B.c.x) + (A.c.y - B.c.y) * (A.c.y - B.c.y);
double rdiff = A.r - B.r;
double rsum = A.r + B.r;
double base = angle(B.c - A.c);
//重合有无限多条
if (d2 == 0 && dcmp(A.r - B.r) == 0) return -1;
//内切
if (dcmp(d2 - rdiff * rdiff) == 0) {
a[cnt] = A.point(base);
b[cnt] = B.point(base);
cnt++;
return 1;
}
//有外公切线
double ang = acos((A.r - B.r) / sqrt(d2));
a[cnt] = A.point(base + ang); b[cnt] = B.point(base + ang); cnt++;
a[cnt] = A.point(base - ang); b[cnt] = B.point(base - ang); cnt++;

//一条内切线，两条内切线
if (dcmp(d2 - rsum*rsum) == 0) {
a[cnt] = A.point(base); b[cnt] = B.point(PI + base); cnt++;
} else if (dcmp(d2 - rsum*rsum) > 0) {
double ang = acos((A.r + B.r) / sqrt(d2));
a[cnt] = A.point(base + ang); b[cnt] = B.point(base + ang); cnt++;
a[cnt] = A.point(base - ang); b[cnt] = B.point(base - ang); cnt++;
}
return cnt;
}

//三角形外切圆
Circle CircumscribedCircle(Point p1, Point p2, Point p3) {
double Bx = p2.x - p1.x, By = p2.y - p1.y;
double Cx = p3.x - p1.x, Cy = p3.y - p1.y;
double D = 2 * (Bx * Cy - By * Cx);
double cx = (Cy * (Bx * Bx + By * By) - By * (Cx * Cx + Cy * Cy)) / D + p1.x;
double cy = (Bx * (Cx * Cx + Cy * Cy) - Cx * (Bx * Bx + By * By)) / D + p1.y;
Point p = Point(cx, cy);
return Circle(p, Length(p1 - p));
}

//三角形内切圆
Circle InscribedCircle(Point p1, Point p2, Point p3) {
double a = Length(p2 - p3);
double b = Length(p3 - p1);
double c = Length(p1 - p2);
Point p = (p1 * a + p2 * b + p3 * c) / (a + b + c);
return Circle(p, DistanceToLine(p, p1, p2));
}

//求经过点p1,与直线(p2, w)相切,半径为r的一组圆
int CircleThroughAPointAndTangentToALineWithRadius(Point p1, Point p2, Vector w, double r, vector<Point> &sol) {
Circle c1 = Circle(p1, r);
double t = r / Length(w);
Vector u = Vector(-w.y, w.x);
Point p4 = p2 + u * t;
int tot = getLineCircleIntersection(p4, w, c1, sol);
u = Vector(w.y, -w.x);
p4 = p2 + u * t;
tot += getLineCircleIntersection(p4, w, c1, sol);
return tot;
}

//给定两个向量，求两向量方向内夹着的圆的圆心。圆与两线均相切，圆的半径已给定
Point Centre_CircleTangentTwoNonParallelLineWithRadius(Point p1, Vector v1, Point p2, Vector v2, double r){
Point p0 = GetLineIntersection(p1, v1, p2, v2);
Vector u = AngleBisector(p0, v1, v2);
double rad = 0.5 * Angle(v1, v2);
double l = r / sin(rad);
double t = l / Length(u);
return p0 + u * t;
}

//求与两条不平行的直线都相切的4个圆，圆的半径已给定
int CircleThroughAPointAndTangentALineWithRadius(Point p1, Vector v1, Point p2, Vector v2, double r, Point *sol) {
int ans = 0;
sol[ans++] = Centre_CircleTangentTwoNonParallelLineWithRadius(p1, v1, p2, v2, r);
sol[ans++] = Centre_CircleTangentTwoNonParallelLineWithRadius(p1, v1 * -1, p2, v2, r);
sol[ans++] = Centre_CircleTangentTwoNonParallelLineWithRadius(p1, v1, p2, v2 * -1, r);
sol[ans++] = Centre_CircleTangentTwoNonParallelLineWithRadius(p1, v1 * -1, p2, v2 * -1, r);
return ans;
}

//求与两个相离的圆均外切的一组圆，三种情况
int CircleTangentToTwoDisjointCirclesWithRadius(Circle c1, Circle c2, double r, Point *sol){
double dis1 = c1.r + r + r + c2.r;
double dis2= Length(c1.c - c2.c);
if(dcmp(dis1 - dis2) < 0) return 0;
Vector u = c2.c - c1.c;
double t = (r + c1.r) / Length(u);
if(dcmp(dis1 - dis2)==0){
Point p0 = c1.c + u * t;
sol[0] = p0;
return 1;
}
double aa = Length(c1.c - c2.c);
double bb = r + c1.r, cc = r + c2.r;
double rad = acos((aa * aa + bb * bb - cc * cc) / (2 * aa * bb));
Vector w = Rotate(u, rad);
Point p0 = c1.c + w * t;
sol[0] = p0;
w = Rotate(u, -rad);
p0 = c1.c + w * t;
sol[1] = p0;
return 2;
}

char op[25];
Point p[4];
double r[3];

int main() {
while (~scanf("%s", op)) {
if (strcmp(op, "CircumscribedCircle") == 0) {
for (int i = 0; i < 3; i++) p[i].read();
Circle ans = CircumscribedCircle(p[0], p[1], p[2]);
printf("(%.6f,%.6f,%.6f)\n", ans.c.x, ans.c.y, ans.r);
} else if (strcmp(op, "InscribedCircle") == 0) {
for (int i = 0; i < 3; i++) p[i].read();
Circle ans = InscribedCircle(p[0], p[1], p[2]);
printf("(%.6f,%.6f,%.6f)\n", ans.c.x, ans.c.y, ans.r);
} else if (strcmp(op, "TangentLineThroughPoint") == 0) {
p[0].read();
scanf("%lf", &r[0]);
p[1].read();
Vector v[3];
int tot = getTangents(p[1], Circle(p[0], r[0]), v);
double ans[3];
for (int i = 0; i < tot; i++)
ans[i] = RealAngleWithX(v[i]);
sort(ans, ans + tot);
printf("[");
for (int i = 0; i < tot; i++) {
printf("%.6f", ans[i]);
if (i != tot - 1) printf(",");
}
printf("]\n");
} else if (strcmp(op, "CircleThroughAPointAndTangentToALineWithRadius") == 0) {
for (int i = 0; i < 3; i++) p[i].read();
scanf("%lf", &r[0]);
vector<Point> ans;
int tot = CircleThroughAPointAndTangentToALineWithRadius(p[0], p[1], p[2] - p[1], r[0], ans);
sort(ans.begin(), ans.end());
printf("[");
for (int i = 0; i < tot; i++) {
printf("(%.6f,%.6f)", ans[i].x, ans[i].y);
if (i != tot - 1) printf(",");
}
printf("]\n");
} else if (strcmp(op, "CircleTangentToTwoLinesWithRadius") == 0) {
Point ans[4];
for (int i = 0; i < 4; i++) p[i].read();
scanf("%lf", &r[0]);
int tot = CircleThroughAPointAndTangentALineWithRadius(p[0], p[1] - p[0], p[3], p[3] - p[2], r[0], ans);
sort(ans, ans + tot);
printf("[");
for (int i = 0; i < tot; i++) {
printf("(%.6f,%.6f)", ans[i].x, ans[i].y);
if (i != tot - 1) printf(",");
}
printf("]\n");
} else {
p[0].read(); scanf("%lf", &r[0]);
p[1].read(); scanf("%lf", &r[1]);
scanf("%lf", &r[2]);
Point ans[4];
int tot = CircleTangentToTwoDisjointCirclesWithRadius(Circle(p[0], r[0]), Circle(p[1], r[1]), r[2], ans);
sort(ans, ans + tot);
printf("[");
for (int i = 0; i < tot; i++) {
printf("(%.6f,%.6f)", ans[i].x, ans[i].y);
if (i != tot - 1) printf(",");
}
printf("]\n");
}
}
return 0;
}