// uva12304 2d geometry 110 in 1!
// rujia liu
#include#include#include#include#include#include#includeusing namespace std;
const double eps = 1e-6;
int dcmp(double x)
const double pi = acos(-1);
struct point
};typedef point vector;
vector operator + (vector a, vector b)
vector operator - (point a, point b)
vector operator * (vector a, double p)
vector operator / (vector a, double p)
bool operator < (const point& a, const point& b)
bool operator == (const point& a, const point &b)
double dot(vector a, vector b)
double length(vector a)
double angle(vector a, vector b)
double cross(vector a, vector b)
vector rotate(vector a, double rad)
vector normal(vector a)
point getlineintersection(point p, point v, point q, point w)
point getlineprojection(point p, point a, point b)
double distancetoline(point p, point a, point b)
struct line
point point(double t)
line move(double d)
};struct circle
point point(double a)
};point getlineintersection(line a, line b)
double angle(vector v)
int getlinecircleintersection(line l, circle c, double& t1, double& t2, vector& sol)
// 相交
t1 = (-f - sqrt(delta)) / (2 * e); sol.push_back(l.point(t1));
t2 = (-f + sqrt(delta)) / (2 * e); sol.push_back(l.point(t2));
return 2;
}int getcirclecircleintersection(circle c1, circle c2, vector& sol)
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;
}/******************* problem 1 **********************/
circle circumscribedcircle(point p1, point p2, point p3)
/******************* problem 2 **********************/
circle inscribedcircle(point p1, point p2, point p3)
/******************* problem 3 **********************/
// 過點p到圓c的切線。v[i]是第i條切線的向量。返回切線條數
int gettangents(point p, circle c, vector* v) else
}/******************* problem 4 **********************/
vectorcirclethroughpointtangenttolinegivenradius(point p, line l, double r)
/******************* problem 5 **********************/
vectorcircletangenttolinesgivenradius(line a, line b, double r)
/******************* problem 6 **********************/
vectorcircletangenttotwodisjointcircleswithradius(circle c1, circle c2, double r)
// formatting
double lineangledegree(vector v)
void format(circle c)
void format(const vector& ans)
printf("]\n");
}void format(const vector& ans)
printf("]\n");
}line getline(double x1, double y1, double x2, double y2)
int main()
if(cmd == "inscribedcircle")
if(cmd == "tangentlinethroughpoint")
if(cmd == "circlethroughapointandtangenttoalinewithradius")
if(cmd == "circletangenttotwolineswithradius")
if(cmd == "circletangenttotwodisjointcircleswithradius")
} return 0;
}
劉汝佳的**，看了這麼多模板，還是最可靠的；
要熟練運用定比分點，直接算距離誤差大。
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include include include include include include define eps 1e 9 include define pi acos 1 using namespace std const double pi acos 1.0 int dcmp double ...