题意：要求解答6个关于圆的问题。
1.给出三角形坐标求外接圆
2.给出三角形坐标求内切圆
3.给出一个圆心和半径已知的圆，求过点(x,y)的所有和这个圆相切的直线
4.求所有和已知直线相切的过定点(x,y)的已知半径的圆的圆心
5.给出两个不平行的直线，求所有半径为r的同时和这两个直线相切的圆
6.给定两个相离的圆，求出所有和这两个圆外切的半径为r的圆。
题解：花了一天做这个，就当整理模板吧。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const double PI = acos(-1);
const int N = 100;
char str[N];
struct Point {
double x, y;
Point(double x = 0, double y = 0): x(x), y(y) {}
};
struct Circle {
Point c;
double r;
Circle() {}
Circle(Point c, double r = 0): c(c), r(r) {}
Point point(double a) {
return Point(c.x + cos(a) * r, c.y + sin(a) * c.y);
}
};
//重载运算符
Point operator + (Point A, Point B) {
return Point(A.x + B.x, A.y + B.y);
}
Point operator - (Point A, Point B) {
return Point(A.x - B.x, A.y - B.y);
}
Point operator * (Point A, double p) {
return Point(A.x * p, A.y * p);
}
Point operator / (Point A, double p) {
return Point(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);
}
bool operator == (const Point& a, const Point& b) {
return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
//计算点积的正负 负值夹角为钝角
int dcmp(double x) {
if (fabs(x) < 1e-9)
return 0;
return x < 0 ? -1 : 1;
}
//计算点积 正负和夹角有关 |a||b|cosC
double Dot(Point A, Point B) {
return A.x * B.x + A.y * B.y;
}
//计算叉积，也就是数量积 向量A和B组成的三角形的有向面积的两倍
//叉积等于0 三角形成为线段
double Cross(Point A, Point B) {
return A.x * B.y - A.y * B.x;
}
//计算向量长度
double Length(Point A) {
return sqrt(Dot(A, A));
}
//向量A旋转rad弧度，rad负值为顺时针旋转
Point Rotate(Point A, double rad) {
return Point(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}
//得到两直线交点
Point GetLineIntersection(Point P, Point v, Point Q, Point w) {
Point u = P - Q;
double t = Cross(w, u) / Cross(v, w);
return P + v * t;
}
//点p到线段AB的距离
double DistanceToSegment(Point p, Point A, Point B) {
if (A == B)
return Length(p - A);
Point AB = B - A, AP = p - A, BP = p - B;
if (dcmp(Dot(AB, AP)) < 0)
return Length(AP);
else if (dcmp(Dot(AB, BP)) > 0)
return Length(BP);
else
return fabs(Cross(AB, AP)) / Length(AB);
}
//判断两个线段是否有交点(不包括端点)
bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {
double c1 = Cross(a2 - a1, b1 - a1);
double c2 = Cross(a2 - a1, b2 - a1);
double c3 = Cross(b2 - b1, a1 - b1);
double c4 = Cross(b2 - b1, a2 - b1);
return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}
//判断点p是否在线段a1--a2上(不包括端点)
bool OnSegment(Point p, Point a1, Point a2) {
return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
}
//三角形外接圆
Circle CircumscribedCircle(Point a, Point b, Point c) {
Point mid1 = Point((a.x + b.x) / 2, (a.y + b.y) / 2);
Point mid2 = Point((b.x + c.x) / 2, (b.y + c.y) / 2);
Point t1 = b - a;
Point t2 = b - c;
Point cc = GetLineIntersection(mid1, Point(t1.y, -t1.x), mid2, Point(t2.y, -t2.x));
return Circle(cc, Length(cc - a));
}
//三角形内切圆
Circle InscribedCircle(Point a, Point b, Point c) {
double l1 = Length(b - c);
double l2 = Length(c - a);
double l3 = Length(a - b);
Point cc = (a * l1 + b * l2 + c * l3) / (l1 + l2 + l3);
return Circle(cc, DistanceToSegment(cc, a, b));
}
//将角度规范化到0 <= x < 180
double solve(double x) {
if (x >= 180)
x -= 180;
else if (x < 0)
x = 180 + x;
return x;
}
//过定点的圆的切线
int getTangents(Point p, Circle C, double* v) {
Point temp;
Point u = C.c - p;
double dist = Length(u);
if (dist < C.r)
return 0;
else if (dcmp(dist - C.r) == 0) {
temp = Rotate(u, PI / 2);
v[0] = atan2(temp.y, temp.x) * 180 / PI;
v[0] = solve(v[0]);
return 1;
}
else {
double ang = asin(C.r / dist);
temp = Rotate(u, -ang);
v[0] = atan2(temp.y, temp.x) * 180 / PI;
temp = Rotate(u, ang);
v[1] = atan2(temp.y, temp.x) * 180 / PI;
v[0] = solve(v[0]);
v[1] = solve(v[1]);
if (v[0] > v[1])
swap(v[0], v[1]);
return 2;
}
}
//求经过点(x,y)的所有和直线(x1,y1)-(x2,y2)相切的半径为r的圆
int CircleThroughPointAndTangentToLineWithRadius(Point p, Point a, Point b, double rr, Point* v) {
Point pcl = Point(p.x + a.y - b.y, p.y + b.x - a.x);
Point pcz = GetLineIntersection(p, p - pcl, a, a - b);
if (!dcmp(Cross(a - p, b - p))) {
v[0] = p + (pcl - p) * (rr / Length(p - pcl));
v[1] = p - (pcl - p) * (rr / Length(p - pcl));
if (v[0].x > v[1].x || (fabs(v[0].x - v[1].x) < 1e-9 && v[0].y > v[1].y))
swap(v[0], v[1]);
return 2;
}
else if (!dcmp(rr - Length(p - pcz) * 0.5)) {
v[0] = (p + pcz) * 0.5;
return 1;
}
else if (dcmp(rr - Length(p - pcz) * 0.5) > 0) {
double rcp = sqrt(rr * rr - (rr - Length(p - pcz)) * (rr - Length(p - pcz)));
v[0] = pcz + (a - b) * (rcp / Length(a - b)) + (p - pcz) * (rr / Length(p - pcz));
v[1] = pcz - (a - b) * (rcp / Length(a - b)) + (p - pcz) * (rr / Length(p - pcz));
if (v[0].x > v[1].x || (fabs(v[0].x - v[1].x) < 1e-9 && v[0].y > v[1].y))
swap(v[0], v[1]);
return 2;
}
else
return 0;
}
//求和两条不平行直线相切的半径为r的圆
void CircleTangentToTwoWithRadius(Point a, Point b, Point c, Point d, double rr, Point* v) {
Point t1 = Point(a.y - b.y, b.x - a.x);
t1 = t1 * (rr / Length(t1));
Point t2 = Point(c.y - d.y, d.x - c.x);
t2 = t2 * (rr / Length(t2));
v[0] = GetLineIntersection(a + t1, a - b, c + t2, c - d);
v[1] = GetLineIntersection(a + t1, a - b, c - t2, c - d);
v[2] = GetLineIntersection(a - t1, a - b, c + t2, c - d);
v[3] = GetLineIntersection(a - t1, a - b, c - t2, c - d);
sort(v, v + 4);
}
//求两个相离的圆的所有半径为r的外切圆
int CircleTangentToTwoDisjointCirclesWithRadius(Circle c1, Circle c2, double rr, Point* v) {
if (!dcmp(rr * 2 + c1.r + c2.r - Length(c1.c - c2.c))) {
v[0] = c1.c + (c2.c - c1.c) * ((rr + c1.r) / (rr * 2 + c1.r + c2.r));
return 1;
}
else if (dcmp(rr * 2 + c1.r + c2.r - Length(c1.c - c2.c)) < 0)
return 0;
else {
double p = (rr + c1.r + rr + c2.r + Length(c1.c - c2.c)) * 0.5;
double S = sqrt(p * (p - rr - c1.r) * (p - rr - c2.r) * (p - Length(c1.c - c2.c))); //海伦公式
double h = S * 2.0 / Length(c1.c - c2.c);
double temp = sqrt((rr + c1.r) * (rr + c1.r) - h * h);
Point mid = c1.c + (c2.c - c1.c) * (temp / (Length(c2.c - c1.c))); //对称圆心的中点
double dis = sqrt((rr + c1.r) * (rr + c1.r) - Length(mid - c1.c) * Length(mid - c1.c)); //中点到所求圆心距离
Point t = Point(c1.c.y - c2.c.y, c2.c.x - c1.c.x); //中点到所求圆心的向量
v[0] = mid + t * (dis / Length(t));
v[1] = mid - t * (dis / Length(t));
sort(v, v + 2);
return 2;
}
}
int main() {
while (scanf("%s", str) == 1) {
if (str[4] == 'u') {
Point a, b, c;
scanf("%lf%lf%lf%lf%lf%lf", &a.x, &a.y, &b.x, &b.y, &c.x, &c.y);
Circle res = CircumscribedCircle(a, b, c);
printf("(%lf,%lf,%lf)\n", res.c.x, res.c.y, res.r);
}
else if (str[0] == 'I') {
Point a, b, c;
scanf("%lf%lf%lf%lf%lf%lf", &a.x, &a.y, &b.x, &b.y, &c.x, &c.y);
Circle res = InscribedCircle(a, b, c);
printf("(%lf,%lf,%lf)\n", res.c.x, res.c.y, res.r);
}
else if (str[0] == 'T') {
Circle cc;
Point p;
scanf("%lf%lf%lf%lf%lf", &cc.c.x, &cc.c.y, &cc.r, &p.x, &p.y);
double v[5];
int cnt = getTangents(p, cc, v);
printf("[");
for (int i = 0; i < cnt; i++) {
if (i == 0)
printf("%lf", v[0]);
else
printf(",%lf", v[i]);
}
printf("]\n");
}
else if (str[7] == 'h') {
double rr;
Point p, a, b;
scanf("%lf%lf%lf%lf%lf%lf%lf", &p.x, &p.y, &a.x, &a.y, &b.x, &b.y, &rr);
Point v[5];
int cnt = CircleThroughPointAndTangentToLineWithRadius(p, a, b, rr, v);
printf("[");
for (int i = 0; i < cnt; i++)
if (i == 0)
printf("(%lf,%lf)", v[0].x, v[0].y);
else
printf(",(%lf,%lf)", v[i].x, v[i].y);
printf("]\n");
}
else if (str[18] == 'L') {
double rr;
Point a, b, c, d;
scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf", &a.x, &a.y, &b.x, &b.y, &c.x, &c.y, &d.x, &d.y, &rr);
Point v[5];
CircleTangentToTwoWithRadius(a, b, c, d, rr, v);
printf("[(%lf,%lf),(%lf,%lf),(%lf,%lf),(%lf,%lf)]\n", v[0].x, v[0].y, v[1].x, v[1].y, v[2].x, v[2].y, v[3].x, v[3].y);
}
else {
Circle c1, c2;
double rr;
Point v[5];
scanf("%lf%lf%lf%lf%lf%lf%lf", &c1.c.x, &c1.c.y, &c1.r, &c2.c.x, &c2.c.y, &c2.r, &rr);
int cnt = CircleTangentToTwoDisjointCirclesWithRadius(c1, c2, rr, v);
printf("[");
for (int i = 0; i < cnt; i++)
if (i == 0)
printf("(%lf,%lf)", v[0].x, v[0].y);
else
printf(",(%lf,%lf)", v[i].x, v[i].y);
printf("]\n");
}
}
return 0;
}
