题意:求圆和圆的切线,并将每根直线按在第一个圆的切点排序。
分析:排序可以使用冒泡,记住比较大小的时候要处理精度问题。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-8;
const double PI = acos(-1.0);
struct Point
{
double x, y;
Point(double x = 0, double y = 0) : x(x), y(y){}
};
typedef Point Vector;
Vector operator+(Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator-(Point A, Point B) { return Vector(A.x - B.x, A.y - B.x); }
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); }
double Dot(Vector A, Vector B)
{
return A.x * B.x + A.y + B.y;
}
double Length(Vector A)
{
return sqrt(Dot(A, A));
}
Vector Normal(Vector A)
{
double L = Length(A);
return Vector(-A.y / L, A.x / L);
}
double dist(Point a, Point b)
{
return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
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.x - b.x) == 0 && dcmp(a.y - b.y) < 0;
}
struct Circle
{
Point c;
double r;
Circle(Point c, double r):c(c), r(r){}
Point point(double a)
{
return Point(c.x + cos(a) * r, c.y + sin(a) * r);
}
};
struct Line
{
Point p;
Vector v;
Line(Point p, Vector v):p(p),v(v){}
Point point(double t)
{
return p + v * t;
}
Line move(double d)
{
return Line(p + Normal(v) * d, v);
}
};
int getTangents(Circle A, Circle B, Point* a, Point* b)
{
//存储切点
int cnt = 0;
if (dcmp(A.r - B.r) < 0) { swap(A, B); swap(a, b); }
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;
if (dcmp(d2 - rdiff * rdiff) < 0) return 0;
//两个圆心之间的向量的极角
double base = atan2(B.c.y - A.c.y, B.c.x - A.c.x);
//无限多条切线
if (dcmp(d2) == 0 && dcmp(A.r - B.r) == 0) return -1;
//内切,一条切线
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;
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(PI + base + ang); ++cnt;
a[cnt] = A.point(base - ang); b[cnt] = B.point(PI + base - ang); ++cnt;
}
return cnt;
}
bool check(Circle A)
{
if (A.c.x == 0 && A.c.y == 0 && A.r == 0)
return 1;
return 0;
}
int main()
{
double x1, y1, r1, x2, y2, r2;
while (scanf("%lf%lf%lf%lf%lf%lf", &x1, &y1, &r1, &x2, &y2, &r2) != EOF)
{
Circle A(Point(x1, y1), r1);
Circle B(Point(x2, y2), r2);
if (check(A) && check(B)) break;
Point a[10], b[10];
int res = getTangents(A, B, a, b);
if (res == -1)
{
puts("-1");
continue;
}
printf("%d
", res);
//冒泡
for (int i = 0; i < res; ++i)
{
for (int j = res - 1; j > i; --j)
{
if (a[j] < a[j - 1])
{
swap(a[j - 1], a[j]);
swap(b[j - 1], b[j]);
}
}
}
for (int i = 0; i < res; ++i)
{
printf("%.5lf %.5lf %.5lf %.5lf %.5lf
", a[i].x, a[i].y, b[i].x, b[i].y, dist(a[i], b[i]));
}
}
return 0;
}