http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3726
题意：
新白书p267， 有说
给出三角形三点求外接圆，内接圆，给出一点求以及圆求过该点的切线， 给出一直线和一个点求过该点与直线想切的圆，圆半径给出。给出两条相交的直线求与这两条直线想切的圆， 给出两个相离的圆，求与这两个圆都想切的圆
思路：
其实没什么很难的就是模板运用，还要注意细节什么的，这题考了很多二维几何的模板，值得一做，话说这是做ACM 题目以来写的最长的题目。细心。。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue> #define CL(arr, val) memset(arr, val, sizeof(arr)) #define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define PI acos(-1.0)
#define L(x) (x) << 1
#define R(x) (x) << 1 | 1
#define MID(l, r) (l + r) >> 1
#define Min(x, y) (x) < (y) ? (x) : (y)
#define Max(x, y) (x) < (y) ? (y) : (x)
#define E(x) (1 << (x))
#define iabs(x) (x) < 0 ? -(x) : (x)
#define OUT(x) printf("%I64d\n", x)
#define lowbit(x) (x)&(-x)
#define Read() freopen("data.in", "r", stdin)
#define Write() freopen("d.out", "w", stdout)
#define ll unsigned long long #define M 100007
#define N 65736 using namespace std; const int inf = 0x7f7f7f7f;
const int mod = 1000000007;
const double eps = 1e-6; struct Point
{
double x,y;
Point(double tx = 0,double ty = 0) : x(tx),y(ty){}
};
typedef Point Vtor;
//向量的加减乘除
Vtor operator + (Vtor A,Vtor B) { return Vtor(A.x + B.x,A.y + B.y); }
Vtor operator - (Point A,Point B) { return Vtor(A.x - B.x,A.y - B.y); }
Vtor operator * (Vtor A,double p) { return Vtor(A.x*p,A.y*p); }
Vtor operator / (Vtor A,double p) { return Vtor(A.x/p,A.y/p); }
bool operator < (Point A,Point B) { return A.x < B.x || (A.x == B.x && A.y < B.y);}
int dcmp(double x){ if (fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; }
bool operator == (Point A,Point B) {return dcmp(A.x - B.x) == 0 && dcmp(A.y - B.y) == 0; }
//向量的点积，长度，夹角
double Dot(Vtor A,Vtor B) { return (A.x*B.x + A.y*B.y); }
double Length(Vtor A) { return sqrt(Dot(A,A)); }
double Angle(Vtor A,Vtor B) { return acos(Dot(A,B)/Length(A)/Length(B)); }
//叉积，三角形面积
double Cross(Vtor A,Vtor B) { return A.x*B.y - A.y*B.x; }
double Area2(Point A,Point B,Point C) { return Cross(B - A,C - A); }
//向量的旋转,求向量的单位法线（即左转90度，然后长度归一）
Vtor Rotate(Vtor A,double rad){ return Vtor(A.x*cos(rad) - A.y*sin(rad),A.x*sin(rad) + A.y*cos(rad)); }
Vtor Normal(Vtor A)
{
double L = Length(A);
return Vtor(-A.y/L, A.x/L);
}
//直线的交点
Point GetLineIntersection(Point P,Vtor v,Point Q,Vtor w)
{
Vtor u = P - Q;
double t = Cross(w,u)/Cross(v,w);
return P + v*t;
}
//点到直线的距离
double DistanceToLine(Point P,Point A,Point B)
{
Vtor v1 = B - A;
return fabs(Cross(P - A,v1))/Length(v1);
}
//点到线段的距离
double DistanceToSegment(Point P,Point A,Point B)
{
if (A == B) return Length(P - A);
Vtor v1 = B - A , v2 = P - A, v3 = P - B;
if (dcmp(Dot(v1,v2)) < 0) return Length(v2);
else if (dcmp(Dot(v1,v3)) > 0) return Length(v3);
else return fabs(Cross(v1,v2))/Length(v1);
}
//点到直线的映射
Point GetLineProjection(Point P,Point A,Point B)
{
Vtor v = B - A;
return A + v*Dot(v,P - A)/Dot(v,v);
} //判断线段是否规范相交
bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2)
{
double c1 = Cross(a2 - a1,b1 - a1), c2 = Cross(a2 - a1,b2 - a1),
c3 = Cross(b2 - b1,a1 - b1), c4 = Cross(b2 - b1,a2 - b1);
return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0;
}
//判断点是否在一条线段上
bool OnSegment(Point P,Point a1,Point a2)
{
return dcmp(Cross(a1 - P,a2 - P)) == 0 && dcmp(Dot(a1 - P,a2 - P)) < 0;
}
//多边形面积
double PolgonArea(Point *p,int n)
{
double area = 0;
for (int i = 1; i < n - 1; ++i)
area += Cross(p[i] - p[0],p[i + 1] - p[0]);
return area/2;
} struct Line
{
Point p,b;
Vtor v;
Line(){}
Line(Point a,Point b,Vtor v) : p(a),b(b),v(v) {}
Line(Point p,Vtor v) : p(p),v(v){}
Point point(double t) { return p + v*t; }
};
struct Circle
{
Point c;
double r;
Circle(Point tc,double tr) : c(tc),r(tr){}
Point point(double a)
{
return Point(c.x + cos(a)*r,c.y + sin(a)*r);
}
};
//判断圆与直线是否相交以及求出交点
int getLineCircleIntersection(Line L,Circle C,double &t1,double &t2,vector<Point> &sol)
{
// printf(">>>>>>>>>>>>>>>>>>>>>>>>\n");
//注意sol没有清空哦
double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y - C.c.y;
double e = a*a + c*c , f = 2*(a*b + c*d), g = b*b + d*d - C.r*C.r;
double delta = f*f - 4.0*e*g;
if (dcmp(delta) < 0) return 0;
else if (dcmp(delta) == 0)
{
t1 = t2 = -f/(2.0*e);
sol.push_back(L.point(t1));
return 1;
}
t1 = (-f - sqrt(delta))/(2.0 * e); sol.push_back(L.point(t1));
t2 = (-f + sqrt(delta))/(2.0 * e); sol.push_back(L.point(t2));
return 2;
}
//判断并求出两圆的交点
double angle(Vtor v) { return atan2(v.y, v.x); }
int getCircleIntersection(Circle C1,Circle C2,vector<Point> &sol)
{
double d = Length(C1.c - C2.c);
// 圆心重合
if (dcmp(d) == 0)
{
if (dcmp(C1.r - C2.r) == 0) return -1; // 两圆重合
return 0; // 包含
} // 圆心不重合
if (dcmp(C1.r + C2.r - d) < 0) return 0; // 相离
if (dcmp(fabs(C1.r - C2.r) - d) > 0) return 0; // 包含 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;
}
//求点到圆的切线
int getTangents(Point p,Circle C,Vtor* v)
{
Vtor u = C.c - p;
double dis = Length(u);
if (dis < C.r) return 0;
else if (dcmp(dis - C.r) == 0)
{
v[0] = Rotate(u,PI/2.0);
return 1;
}
else
{
double ang = asin(C.r / dis);
v[0] = Rotate(u, -ang);
v[1] = Rotate(u, +ang);
return 2;
}
}
//求两圆的切线
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
if (dcmp(d2 - rsum*rsum) == 0)
{
a[cnt] = A.point(base); b[cnt] = B.point(PI + base); cnt++;
}//两条内切线
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;
} //**********************************
Circle CircumscribedCircle(Point A,Point B,Point C)
{
Point tmp1 = Point((B.x + C.x) / 2.0,(B.y + C.y) / 2.0);
Vtor u = C - tmp1;
u = Rotate(u,PI/2.0);
Point tmp2 = Point((A.x + C.x) / 2.0,(A.y + C.y) / 2.0);
Vtor v = C - tmp2;
v = Rotate(v,-PI/2.0);
Point c = GetLineIntersection(tmp1,u,tmp2,v);
double r = Length(C - c);
return Circle(c,r);
}
//得到法向量就得到了这个方向上的向量了
//Circle work1(Point p1, Point p2, Point p3)
// {
// Vtor nor1 = Normal(p1 - p2);
// Vtor nor2 = Normal(p2 - p3);
// Point mid1 = (p1 + p2) / 2.0;
// Point mid2 = (p2 + p3) / 2.0;
// Point O = GetLineIntersection(mid1, nor1, mid2, nor2);
// double r = Length(O - p1);
// return Circle(O, r);
//} //不知道为什么我按常规的求法就是不对
//Circle InscribedCircle(Point A,Point B,Point C)
//{
// Vtor u = A - B;
// Vtor v = C - B;
// double ang = Angle(u,v);
// Vtor vv= Rotate(v,ang / 2.0);
// u = A - C;
// v = B - C;
// ang = Angle(u,v);
// Vtor uu = Rotate(u,ang / 2.0);
// Point c = GetLineIntersection(B,vv,C,uu);
// double r = DistanceToLine(c,A,C);
// return Circle(c,r);
//}
Circle work2(Point p1, Point p2, Point p3) {
Vtor v11 = p2 - p1;
Vtor v12 = p3 - p1;
Vtor v21 = p1 - p2;
Vtor v22 = p3 - p2;
double ang1 = (angle(v11) + angle(v12)) / 2.0;
double ang2 = (angle(v21) + angle(v22)) / 2.0;
Vtor vec1 = Vtor(cos(ang1), sin(ang1));
Vtor vec2 = Vtor(cos(ang2), sin(ang2));
Point O = GetLineIntersection(p1, vec1, p2, vec2);
double r = DistanceToLine(O, p1, p2);
return Circle(O, r);
}
vector<Point> solve4(Point A,Point B,double r,Point C)
{
Vtor normal = Normal(B - A);
normal = normal / Length(normal) * r;
vector<Point> ans;
double t1 = 0,t2 = 0;
Vtor tA = A + normal,tB = B + normal;
getLineCircleIntersection(Line(tA,tB,tB - tA),Circle(C, r),t1,t2,ans);
tA = A - normal,tB = B - normal;
getLineCircleIntersection(Line(tA,tB,tB - tA),Circle(C, r),t1,t2,ans);
return ans;
}
vector<Point> solve5(Point A,Point B,Point C,Point D,double r)
{
Line lines[5];
Vtor normal = Normal(B - A) * r;
Point ta,tb,tc,td;
ta = A + normal,tb = B + normal;
lines[0] = Line(ta,tb,tb - ta);
ta = A - normal,tb = B - normal;
lines[1] = Line(ta,tb,tb - ta); normal = Normal(D - C) * r;
tc = C + normal,td = D + normal;
lines[2] = Line(tc,td,td - tc);
tc = C - normal,td = D - normal;
lines[3] = Line(tc,td,td - tc);
vector<Point> ans;
ans.push_back(GetLineIntersection(lines[0].p,lines[0].v,lines[2].p,lines[2].v));
ans.push_back(GetLineIntersection(lines[0].p,lines[0].v,lines[3].p,lines[3].v));
ans.push_back(GetLineIntersection(lines[1].p,lines[1].v,lines[2].p,lines[2].v));
ans.push_back(GetLineIntersection(lines[1].p,lines[1].v,lines[3].p,lines[3].v));
return ans;
}
vector<Point> solve6(Circle C1,Circle C2,double r)
{
vector<Point> vc;
getCircleIntersection(Circle(C1.c, C1.r + r),Circle(C2.c, C2.r + r),vc);
return vc;
} string op;
double x[10]; int main()
{
// Read(); while (cin>>op)
{
if (op == "CircumscribedCircle")
{
for (int i = 0; i < 6; ++i) cin>>x[i];
Circle ans = CircumscribedCircle(Point(x[0],x[1]),Point(x[2],x[3]),Point(x[4],x[5]));
// Circle ans = work1(Point(x[0],x[1]),Point(x[2],x[3]),Point(x[4],x[5]));
printf("(%.6lf,%.6lf,%.6lf)\n",ans.c.x,ans.c.y,ans.r);
}
else if (op == "InscribedCircle")
{
for (int i = 0; i < 6; ++i) cin>>x[i];
// Circle ans = InscribedCircle(Point(x[0],x[1]),Point(x[2],x[3]),Point(x[4],x[5]));
Circle ans = work2(Point(x[0],x[1]),Point(x[2],x[3]),Point(x[4],x[5]));
printf("(%.6lf,%.6lf,%.6lf)\n",ans.c.x,ans.c.y,ans.r);
}
else if (op == "TangentLineThroughPoint")
{
for (int i = 0; i < 5; ++i) cin>>x[i];
Vtor vc[5];
int len = getTangents(Point(x[3],x[4]),Circle( Point(x[0],x[1]), x[2] ),vc);
double tmp[5];
for (int i = 0; i < len; ++i)
{
double ang = angle(vc[i]);
if (ang < 0) ang += PI;
ang = fmod(ang,PI);
tmp[i] = ang*180/PI;
}
sort(tmp,tmp + len);
printf("[");
for (int i = 0; i < len; ++i)
{
printf("%.6lf",tmp[i]);
if (i != len - 1) printf(",");
}
printf("]\n");
}
else if (op == "CircleThroughAPointAndTangentToALineWithRadius")
{
for (int i = 0; i < 7; ++i) cin>>x[i];
vector<Point> vc = solve4(Point(x[2],x[3]),Point(x[4],x[5]),x[6],Point(x[0],x[1]));
sort(vc.begin(),vc.end());
printf("[");
for (size_t i = 0; i < vc.size(); ++i)
{
printf("(%.6lf,%.6lf)",vc[i].x,vc[i].y);
if (i != vc.size() - 1) printf(",");
}
printf("]\n");
}
else if (op == "CircleTangentToTwoLinesWithRadius")
{
for (int i = 0; i < 9; ++i) cin>>x[i];
vector<Point> vc = solve5(Point(x[0],x[1]),Point(x[2],x[3]),Point(x[4],x[5]),Point(x[6],x[7]),x[8]);
sort(vc.begin(),vc.end());
printf("[");
for (size_t i = 0; i < vc.size(); ++i)
{
printf("(%.6lf,%.6lf)",vc[i].x,vc[i].y);
if (i != vc.size() - 1) printf(",");
}
printf("]\n");
}
else
{
for (int i = 0; i < 7; ++i) cin>>x[i];
vector<Point> vc = solve6(Circle(Point(x[0],x[1]),x[2]),Circle(Point(x[3],x[4]),x[5]),x[6]);
sort(vc.begin(),vc.end());
printf("[");
for (size_t i = 0; i < vc.size(); ++i)
{
printf("(%.6lf,%.6lf)",vc[i].x,vc[i].y);
if (i != vc.size() - 1) printf(",");
}
printf("]\n");
}
}
return 0;
}