Uva - 12304 - 2D Geometry 110 in 1!

题意：在二维平面上求解6个子问题：

1.三角形的外接圆；

2.三角形的内切圆；

3.点到圆的切线；

4.过定点且与定直线相切的半径为r的圆；

5.同时与两相交直线相切的半径为r的圆；

6.同时与两相离圆相切的半径为r的圆。

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=24008

——>>做到第4小问的时候，出现了一个精度问题：本来用dcmp(delta) == 0来判断直线与圆相切，可以在此处硬是不正确。。。书中有言：“圆和直线相切的判定很容易受到误差影响，因为这里的直线是计算出来的，而不是题目中输入的。”。。。所以——“特殊判断圆心到输入直线的距离”（“当然也可以通过调整eps的数值来允许一定的误差，但不推荐这样做。调节eps只是掩盖了问题，并没有消除“）。

代码虽长，但1A啦。。。

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

const int maxn = 50;
const double eps = 1e-10;        //调到1e-6以上第4问就可以用delta判断切线,但《训练指南》建议，尽量不要调eps
const double pi = acos(-1);

char type[maxn];

int dcmp(double x){
    return fabs(x) < eps ? 0 : (x > 0 ? 1 : -1);
}

struct Point{
    double x;
    double y;

    Point(double x = 0, double y = 0):x(x), y(y){}

    bool operator < (const Point& e) const{
        return dcmp(x - e.x) < 0 || (dcmp(x - e.x) == 0 && dcmp(y - e.y) < 0);
    }

    bool operator == (const Point& e) const{
        return dcmp(x - e.x) == 0 && dcmp(y - e.y) == 0;
    }

    int read(){
        return scanf("%lf%lf", &x, &y);
    }
}p[3];

typedef Point Vector;

Vector operator + (Point A, Point 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.y);
}

Vector operator * (Point A, double p){
    return Vector(A.x * p, A.y * p);
}

Vector operator / (Point A, double p){
    return Vector(A.x / p, A.y / p);
}

struct Line{
    Point p;
    Point v;

    Line(){}
    Line(Point p, Point v):p(p), v(v){}

    int read(){
        return scanf("%lf%lf%lf%lf", &p.x, &p.y, &v.x, &v.y);
    }

    Point point(double t){
        return p + v * t;
    }
};

struct Circle{
    Point c;
    double r;

    Circle(){}
    Circle(Point c, double r):c(c), r(r){}

    int read(){
        return scanf("%lf%lf%lf", &c.x, &c.y, &r);
    }

    Point point(double a){
        return Point(c.x + r * cos(a), c.y + r * sin(a));
    }
};

double Dot(Vector A, Vector B){
    return A.x * B.x + A.y * B.y;
}

double Cross(Vector A, Vector B){
    return A.x * B.y - B.x * A.y;
}

double Length(Vector A){
    return sqrt(Dot(A, A));
}

Vector Rotate(Vector A, double rad){
    return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}

Vector Normal(Vector A){
    double L = Length(A);
    return Vector(-A.y / L, A.x / L);
}

double DistanceToLine(Point P, Point A, Point B){       //点到直线的距离
    Vector v1 = B - A;
    Vector v2 = P - A;
    return fabs(Cross(v1, v2) / Length(v1));
}

double angle(Vector v){     //求向量的极角
    return atan2(v.y, v.x);
}

Point GetLineIntersection(Line l1, Line l2){        //求两直线的交点（前提：相交）
    Vector u = l1.p - l2.p;
    double t = Cross(l2.v, u) / Cross(l1.v, l2.v);
    return l1.point(t);
}

int getLineCircleIntersection(Line l, Circle C, double& t1, double& t2, vector<Point>& sol){        //求直线与圆的交点
    double a = l.v.x;
    double b = l.p.x - C.c.x;
    double c = l.v.y;
    double d = l.p.y - C.c.y;
    double e = a * a + c * c;
    double f = 2 * (a * b + c * d);
    double g = b * b + d * d - C.r * C.r;
    double delta = f * f - 4 * e * g;
    double dist = DistanceToLine(C.c, l.p, l.p+l.v);
    if(dcmp(dist - C.r) == 0){      //相切，此处需特殊判断，不能用delta
        t1 = t2 = -f / (2 * e);
        sol.push_back(l.point(t1));
        return 1;
    }
    if(dcmp(delta) < 0) return 0;       //相离
    else{       //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<Point>& sol){      //求圆与圆的交点
    double d = Length(C1.c - C2.c);
    if(dcmp(d) == 0){
        if(dcmp(C1.r - C2.r) == 0) return -1;       //两圆重合
        return 0;       //同心圆但不重合
    }
    if(dcmp(C1.r + C2.r - d) < 0) return 0;     //外离
    if(dcmp(fabs(C1.r - C2.r) - d) > 0) return 0;       //内含
    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);
    Point p2 = C1.point(a - da);
    sol.push_back(p1);
    if(p1 == p2) return 1;      //外切
    sol.push_back(p2);
    return 2;
}

Circle CircumscribedCircle(Point p1, Point p2, Point p3){       //求三角形的外心
    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(cx, cy);
    return Circle(p, Length(p1-p));
}

Circle InscribedCircle(Point p1, Point p2, Point p3){       //求三角形的内切圆
    double a = Length(p3 - p2);
    double b = Length(p3 - p1);
    double c = Length(p2 - p1);
    Point p = (p1 * a + p2 * b + p3 * c) / (a + b + c);
    return Circle(p, DistanceToLine(p, p2, p3));
}

int TangentLineThroughPoint(Point p, Circle C, Vector *v){      //求点到圆的直线
    Vector u = C.c - p;
    double dist = Length(u);
    if(dcmp(dist - C.r) < 0) return 0;
    else if(dcmp(dist - C.r) < eps){
        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;
    }
}

void CircleThroughAPointAndTangentToALineWithRadius(Point p, Point p1, Point p2, double r){
    Vector AB = p2 - p1;
    Vector change1 = Rotate(AB, pi / 2) / Length(AB) * r;
    Vector change2 = Rotate(AB, -pi / 2) / Length(AB) * r;
    Line l1(p1 + change1, AB);
    Line l2(p1 + change2, AB);
    vector<Point> sol;
    sol.clear();
    double t1, t2;
    int cnt1 = getLineCircleIntersection(l1, Circle(p, r), t1, t2, sol);
    int cnt2 = getLineCircleIntersection(l2, Circle(p, r), t1, t2, sol);
    int cnt = cnt1 + cnt2;
    if(cnt) sort(sol.begin(), sol.end());
    printf("[");
    for(int i = 0; i < cnt; i++){
        printf("(%.6f,%.6f)", sol[i].x, sol[i].y);
        if(cnt == 2 && !i) printf(",");
    }
    puts("]");
}

void CircleTangentToTwoLinesWithRadius(Point A, Point B, Point C, Point D, double r){
    Vector AB = B - A;
    Vector change = Normal(AB) * r;
    Point newA1 = A + change;
    Point newA2 = A - change;
    Vector CD = D - C;
    Vector update = Normal(CD) * r;
    Point newC1 = C + update;
    Point newC2 = C - update;
    Point p[5];
    p[0] = GetLineIntersection(Line(newA1, AB), Line(newC1, CD));
    p[1] = GetLineIntersection(Line(newA1, AB), Line(newC2, CD));
    p[2] = GetLineIntersection(Line(newA2, AB), Line(newC1, CD));
    p[3] = GetLineIntersection(Line(newA2, AB), Line(newC2, CD));
    sort(p, p + 4);
    printf("[");
    printf("(%.6f,%.6f)", p[0].x, p[0].y);
    for(int i = 1; i < 4; i++){
        printf(",(%.6f,%.6f)", p[i].x, p[i].y);
    }
    puts("]");
}

void CircleTangentToTwoDisjointCirclesWithRadius(Circle C1, Circle C2, double r){
    Vector CC = C2.c - C1.c;
    double rdist = Length(CC);
    if(dcmp(2 * r - rdist + C1.r + C2.r) < 0) puts("[]");
    else if(dcmp(2 * r - rdist + C1.r + C2.r) == 0){
        double ang = angle(CC);
        Point A = C1.point(ang);
        Point B = C2.point(ang + pi);
        Point ret = (A + B) / 2;
        printf("[(%.6f,%.6f)]\n", ret.x, ret.y);
    }
    else{
        Circle A = Circle(C1.c, C1.r + r);
        Circle B = Circle(C2.c, C2.r + r);
        vector<Point> sol;
        sol.clear();
        GetCircleCircleIntersection(A, B, sol);
        sort(sol.begin(), sol.end());
        printf("[(%.6f,%.6f),(%.6f,%.6f)]\n", sol[0].x, sol[0].y, sol[1].x, sol[1].y);
    }
}

int main()
{
    while(scanf("%s", type) == 1){
        if(strcmp(type, "CircumscribedCircle") == 0){
            Point p1, p2, p3;
            p1.read();
            p2.read();
            p3.read();
            Circle ret = CircumscribedCircle(p1, p2, p3);
            printf("(%f,%f,%f)\n", ret.c.x, ret.c.y, ret.r);
        }
        else if(strcmp(type, "InscribedCircle") == 0){
            Point p1, p2, p3;
            p1.read();
            p2.read();
            p3.read();
            Circle ret = InscribedCircle(p1, p2, p3);
            printf("(%f,%f,%f)\n", ret.c.x, ret.c.y, ret.r);
        }
        else if(strcmp(type, "TangentLineThroughPoint") == 0){
            Circle C;
            Point p;
            C.read();
            p.read();
            Vector v[3];
            int cnt = TangentLineThroughPoint(p, C, v);
            double ret[3];
            for(int i = 0; i < cnt; i++){
                ret[i] = angle(v[i]);
                if(dcmp(ret[i] - pi) == 0) ret[i] = 0;
                if(dcmp(ret[i]) < 0) ret[i] += pi;
            }
            sort(ret, ret + cnt);
            printf("[");
            for(int i = 0; i < cnt; i++){
                printf("%.6f", ret[i] / pi * 180);
                if(cnt == 2 && !i) printf(",");
            }
            puts("]");
        }
        else if(strcmp(type, "CircleThroughAPointAndTangentToALineWithRadius") == 0){
            Point p, p1, p2;
            double r;
            p.read();
            p1.read();
            p2.read();
            scanf("%lf", &r);
            CircleThroughAPointAndTangentToALineWithRadius(p, p1, p2, r);
        }
        else if(strcmp(type, "CircleTangentToTwoLinesWithRadius") == 0){
            Point A, B, C, D;
            double r;
            A.read();
            B.read();
            C.read();
            D.read();
            scanf("%lf", &r);
            CircleTangentToTwoLinesWithRadius(A, B, C, D, r);
        }
        else{
            Circle C1, C2;
            double r;
            C1.read();
            C2.read();
            scanf("%lf", &r);
            CircleTangentToTwoDisjointCirclesWithRadius(C1, C2, r);
        }
    }
    return 0;
}