计算几何： 堆几何模版就能够了。
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Description
Problem E2D Geometry 110 in 1!This is a collection of 110 (in binary) 2D geometry problems. CircumscribedCircle x1 y1 x2 y2 x3 y3 Find out the circumscribed circle of triangle (x1,y1)-(x2,y2)-(x3,y3). These three points are guaranteed to be non-collinear. The circle is formatted as (x,y,r) where (x,y) is the center of circle, r is the InscribedCircle x1 y1 x2 y2 x3 y3 Find out the inscribed circle of triangle (x1,y1)-(x2,y2)-(x3,y3). These three points are guaranteed to be non-collinear. The circle is formatted as (x,y,r) where (x,y) is the center of circle, r is the radius. TangentLineThroughPoint xc yc r xp yp Find out the list of tangent lines of circle centered (xc,yc) with radius r that pass through point (xp,yp). Each tangent line is formatted as a single real number "angle" (in degrees), the angle of the line
Find out the list of circles passing through point (xp, yp) that is tangent to a line (x1,y1)-(x2,y2) with radius r. Each circle is formatted as (x,y), since the radius is already given. Note that the answer
Find out the list of circles tangent to two non-parallel lines (x1,y1)-(x2,y2) and (x3,y3)-(x4,y4), having radius r. Each circle is formatted as (x,y), since the radius is already given. Note that the answer
Find out the list of circles externally tangent to two disjoint circles (x1,y1,r1) and (x2,y2,r2), having radius r. By "externally" we mean it should not enclose the two given circles. Each circle is formatted
For each line described above, the two endpoints will not be equal. When formatting a list of real numbers, the numbers should be sorted in increasing order; when formatting a list of (x,y) pairs, the pairs InputThere will be at most 1000 sub-problems, one in each line, formatted as above. The coordinates will be integers with absolute value not greater than 1000. The input is terminated by end of file (EOF). OutputFor each input line, print out your answer formatted as stated in the problem description. Each number in the output should be rounded to six digits after the decimal point. Note that the list should be enclosed Sample InputCircumscribedCircle 0 0 20 1 8 17 Output for the Sample Input(9.734940,5.801205,11.332389) Rujia Liu’s Present 4: A Contest Dedicated to Geometry and CG Lovers Special Thanks: Di Tang and Yi Chen Source Root :: Prominent Problemsetters :: Rujia Liu Root :: Rujia Liu’s Presents :: Present 4: Dedicated to Geometry and CG Lovers Root :: AOAPC I: Beginning Algorithm Contests — Training Guide (Rujia Liu) :: Chapter 4. Geometry :: Geometric Computations in 2D :: Submit Status |
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#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>using namespace std;const double eps=1e-6;int dcmp(double x){if(fabs(x)<eps) return 0; return (x<0)?-1:1;}struct Point
{
double x,y;
Point(double _x=0,double _y=0):x(_x),y(_y){};
};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;}double Dot(Point A,Point B) {return A.x*B.x+A.y*B.y;}
double Length(Point A) {return sqrt(Dot(A,A));}
double Angle(Point A,Point B) {return acos(Dot(A,B)/Length(A)/Length(B));}
double Angle(Point v) {return atan2(v.y,v.x);}
double Cross(Point A,Point B) {return A.x*B.y-A.y*B.x;}/**Cross
P*Q > 0 P在Q的顺时针方向
P*Q < 0 P在Q的逆时针方向
P*Q = 0 PQ共线
*/Point Horunit(Point x) {return x/Length(x);}///单位向量
Point Verunit(Point x) {return Point(-x.y,x.x)/Length(x);}///单位法向量Point Rotate(Point A,double rad)///逆时针旋转
{
return Point(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}double Area2(const Point A,const Point B,const Point C)
{
return Cross(B-A,C-A);
}/// 过两点p1, p2的直线一般方程ax+by+c=0 (x2-x1)(y-y1) = (y2-y1)(x-x1)
void getLineGeneralEquation(const Point& p1, const Point& p2, double& a, double&b, double &c)
{
a = p2.y-p1.y;
b = p1.x-p2.x;
c = -a*p1.x - b*p1.y;
}///P+t*v Q+w*t的焦点
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;
}///点到直线距离
double DistanceToLine(Point P,Point A,Point B)
{
Point v1=B-A,v2=P-A;
return fabs(Cross(v1,v2))/Length(v1);
}///点到线段距离
double DistanceToSegment(Point P,Point A,Point B)
{
if(A==B) return Length(P-A);
Point 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);
}///点到直线投影
Point GetLineProjection(Point P,Point A,Point B)
{
Point v=B-A;
return A+v*(Dot(v,P-A)/Dot(v,v));
}///推断规范相交
bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2)
{
double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1);
double c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1); return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}///一个点是否在直线端点上
bool OnSegment(Point p,Point a1,Point a2)
{
return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}///多边形有向面积
double PolygonArea(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;
}///有向直线
struct Line
{
Point p;
Point v;
double ang;
Line(Point _p,Point _v):p(_p),v(_v){ang=atan2(v.y,v.x);}
Point point(double a) {return p+(v*a);}
bool operator<(const Line& L)const
{
return ang<L.ang;
}
};///直线平移距离d
Line LineTransHor(Line l,int d)
{
Point vl=Verunit(l.v);
Point p1=l.p+vl*d,p2=l.p-vl*d;
Line ll=Line(p1,l.v);
return ll;
}///直线交点（如果存在）
Point GetLineIntersection(Line a,Line b)
{
return GetLineIntersection(a.p,a.v,b.p,b.v);
}///点p在有向直线的左边
bool OnLeft(const Line& L,const Point& p)
{
return Cross(L.v,p-L.p)>=0;
}///圆
const double pi=acos(-1.0);struct Circle
{
Point c;
double r;
Circle(Point _c=0,double _r=0):c(_c),r(_r){}
Point point(double a)///依据圆心角算圆上的点
{
return Point(c.x+cos(a)*r,c.y+sin(a)*r);
}
};///a点到b点（逆时针）在圆上的圆弧长度
double D(Point a,Point b,Circle C)
{
double ang1,ang2;
Point v1,v2;
v1=a-C.c; v2=b-C.c;
ang1=atan2(v1.y,v1.x);
ang2=atan2(v2.y,v2.x);
if(ang2<ang1) ang2+=2*pi;
return C.r*(ang2-ang1);
}///直线与圆交点 返回交点个数
int getLineCircleIntersection(Line L,Circle C,double& t1,double& t2,vector<Point>& sol)
{
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.*e*g;
if(dcmp(delta)<0) return 0;//相离
if(dcmp(delta)==0)//相切
{
t1=t2=-f/(2.*e); sol.push_back(L.point(t1));
return 1;
}
//相切
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;
}///圆与圆交点 返回交点个数
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;
}///P到圆的切线 v[] 储存切线向量
int getTangents(Point p,Circle C,Point* v)
{
Point u=C.c-p;
double dist=Length(u);
if(dist<C.r) return 0;
else if(dcmp(dist-C.r)==0)
{
///p在圆上仅仅有一条切线
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,b 公切线再 圆 A B 上的切点
int getTengents(Circle A,Circle B,Point* a,Point* b)
{
int cnt=0;
if(A.r<B.r) { swap(A,B); swap(a,b); }
int 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);
int rdiff=A.r-B.r;
int rsum=A.r+B.r;
if(d2<rdiff*rdiff) return 0;///内含 double base=atan2(B.c.y-A.c.y,B.c.x-A.c.x);
if(d2==0&&A.r==B.r) return -1; ///无穷多
if(d2==rdiff*rdiff)//内切 1条
{
a[cnt]=A.point(base); b[cnt]=B.point(base); cnt++;
return 1;
}
///外切
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(d2==rsum*rsum)// one
{
a[cnt]=A.point(base); b[cnt]=B.point(pi+base); cnt++;
}
else if(d2>rsum*rsum)// two
{
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;
}///三角形外接圆
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));
}double RtoDegree(double x) {return x/pi*180.;}char op[200];
double a[10];
Point v[10];
double degree[10];
vector<Point> sol;int main()
{
while(scanf("%s",op)!=EOF)
{
if(strcmp(op,"CircumscribedCircle")==0)
{
for(int i=0;i<6;i++) scanf("%lf",a+i);
Circle C=CircumscribedCircle(Point(a[0],a[1]),Point(a[2],a[3]),Point(a[4],a[5]));
printf("(%.6lf,%.6lf,%.6lf)\n",C.c.x,C.c.y,C.r);
}
else if(strcmp(op,"InscribedCircle")==0)
{
for(int i=0;i<6;i++) scanf("%lf",a+i);
Circle C=InscribedCircle(Point(a[0],a[1]),Point(a[2],a[3]),Point(a[4],a[5]));
printf("(%.6lf,%.6lf,%.6lf)\n",C.c.x,C.c.y,C.r);
}
else if(strcmp(op,"TangentLineThroughPoint")==0)
{
for(int i=0;i<5;i++) scanf("%lf",a+i);
int sz=getTangents(Point(a[3],a[4]),Circle(Point(a[0],a[1]),a[2]),v);
for(int i=0;i<sz;i++)
{
double de=RtoDegree(Angle(v[i]));
if(dcmp(de)<0) de=180.+de;
else while(dcmp(de-180.)>=0) de-=180.;
degree[i]=de;
}
sort(degree,degree+sz);
putchar('[');if(sz==0) putchar(']');
for(int i=0;i<sz;i++) printf("%.6lf%c",degree[i],(i!=sz-1)?',':']');
putchar(10);
}
else if(strcmp(op,"CircleThroughAPointAndTangentToALineWithRadius")==0)
{
for(int i=0;i<7;i++) scanf("%lf",a+i);
Point A=Point(a[2],a[3]),B=Point(a[4],a[5]);
Circle C(Point(a[0],a[1]),a[6]); Point normal=Verunit(B-A);
normal=normal/Length(normal)*a[6]; Point ta=A+normal,tb=B+normal;
Line l1=Line(ta,tb-ta);
ta=A-normal,tb=B-normal;
Line l2=Line(ta,tb-ta); sol.clear();
double t1,t2;
int aa=getLineCircleIntersection(l1,C,t1,t2,sol);
int bb=getLineCircleIntersection(l2,C,t1,t2,sol);
sort(sol.begin(),sol.end()); putchar('[');
for(int i=0,sz=sol.size();i<sz;i++)
{
if(i) putchar(',');
printf("(%.6lf,%.6lf)",sol[i].x,sol[i].y);
}
putchar(']'); putchar(10);
}
else if(strcmp(op,"CircleTangentToTwoLinesWithRadius")==0)
{
for(int i=0;i<9;i++) scanf("%lf",a+i);
Line LA=Line(Point(a[0],a[1]),Point(a[2],a[3])-Point(a[0],a[1]));
Line LB=Line(Point(a[4],a[5]),Point(a[6],a[7])-Point(a[4],a[5]));
Line la1=LineTransHor(LA,a[8]),la2=LineTransHor(LA,-a[8]);
Line lb1=LineTransHor(LB,a[8]),lb2=LineTransHor(LB,-a[8]); sol.clear();
sol.push_back(GetLineIntersection(la1,lb1));
sol.push_back(GetLineIntersection(la1,lb2));
sol.push_back(GetLineIntersection(la2,lb1));
sol.push_back(GetLineIntersection(la2,lb2));
sort(sol.begin(),sol.end()); putchar('[');
for(int i=0,sz=sol.size();i<sz;i++)
{
if(i) putchar(',');
printf("(%.6lf,%.6lf)",sol[i].x,sol[i].y);
}
putchar(']'); putchar(10); }
else if(strcmp(op,"CircleTangentToTwoDisjointCirclesWithRadius")==0)
{
for(int i=0;i<7;i++) scanf("%lf",a+i);
Circle C1=Circle(Point(a[0],a[1]),a[2]+a[6]);
Circle C2=Circle(Point(a[3],a[4]),a[5]+a[6]);
sol.clear();
getCircleCircleIntersection(C1,C2,sol);
sort(sol.begin(),sol.end());
putchar('[');
for(int i=0,sz=sol.size();i<sz;i++)
{
if(i) putchar(',');
printf("(%.6lf,%.6lf)",sol[i].x,sol[i].y);
}
putchar(']'); putchar(10);
}
}
return 0;
}
