做题UVA-12304——平面计算集合六合一

Posted 2020-11-08 cly_none

可真是道恶习题……
首先翻译一下6个任务：

1. 给出一个三角形，求它的外界圆。
2. 给出一个三角形，求它的内接圆。
3. 给出一个圆和一个点，求过这个点的切线的倾斜角\(\alpha \in [0,180)\)。（这个点可能在圆内或圆上）
4. 给出一条切线、圆上一点和圆的半径，求圆心位置。（此问题和后面的问题都可能无解或有多个解）
5. 给出两条切线和圆的半径，求圆心位置。
6. 给出两个外切圆和半径，求圆心位置。

若有多个解，从小到大输出答案。（点排序时以\(x\)为第一关键字）
可以发现，上面的问题经过简单的转换后，就是要求我们实现：

• 求点到直线的距离。
• 求点在直线上的垂足。
• 求直线的垂直平分线。
• 求角的平分线。
• 求两条直线的交点。
• 求圆和一条直线的交点。
• 求两个圆的交点。

然后，我们写\(O(1)\)个小时就能写出来了（大雾

#include <bits/stdc++.h>
using namespace std;
typedef double db;
const db eps = 1e-8, INF = 1.0 / 0.0, pi = acos(-1);
inline int judge(db x) {
  return x > -eps ? x > eps ? 1 : 0 : -1;
}
#define NULLP point(INF,INF)
struct point {
  db x,y;
  point(db x_=0,db y_=0): x(x_), y(y_) {}
  db abs() const {
    return sqrt(x * x + y * y);
  }
  db norm() const {
    return x * x + y * y;
  }
  bool operator < (const point& a) const {
    return judge(x - a.x) != 0 ? judge(x - a.x) < 0 : judge(y - a.y) < 0;
  }
  point operator - () const {
    return point(-x, -y);
  }
  point operator * (const db& a) const {
    return point(x * a, y * a);
  }
  point operator / (const db& a) const {
    return point(x / a, y / a);
  }
  point* operator *= (const db& a) {
    return *this = *this * a, this;
  }
  point* operator /= (const db& a) {
    return *this = *this / a, this;
  }
  point operator + (const point& a) const {
    return point(x + a.x, y + a.y);
  }
  point operator - (const point& a) const {
    return point(x - a.x, y - a.y);
  }
  point* operator += (const point& a) {
    return *this = *this + a, this;
  }
  point* operator -= (const point& a) {
    return *this = *this - a, this;
  }
  bool avail() {
    return (fabs(x) != INF) && (fabs(y) != INF);
  }
};
db dot(point a,point b) {
  return a.x * b.x + a.y * b.y;
}
db cross(point a,point b) {
  return a.x * b.y - a.y * b.x;
}
point unit(point a) {
  db t = a.abs();
  if (judge(t) == 0) return a;
  return a / t;
}
struct line {
  point u,v;
  line(point u_=point(),point v_=point()): u(u_) {
    v = unit(v_);
  }
};
db dist(point u,line l) {
  return fabs(cross(u - l.u,l.v));
}
point foot_point(point u,line l) {
  return l.u + (l.v * dot(u - l.u,l.v));
}
point perpen(point a) {
  return point(a.y, -a.x);
}
point crossover(line a,line b) {
  if (judge(cross(a.v,b.v)) == 0) return NULLP;
  point f1 = foot_point(b.u,a);
  point f2 = foot_point(f1,b);
  if (judge(dot(f2 - b.u,b.v)) < 0) b.v = -b.v;
  return b.u + b.v * ((f1 - b.u).norm() / (f2 - b.u).abs());
}
line perpen_bi(point a,point b) {
  return line((a + b) / 2, perpen(b - a));
}
struct angle {
  point o,u,v;
  angle(point o_=point(),point u_=point(),point v_=point()): o(o_) {
    u = unit(u_);
    v = unit(v_);
  }
  line bise() const {
    return line(o,u + v);
  }
};
struct circle {
  point o;
  db r;
  circle(point o_=point(),db r_=0): o(o_), r(r_) {}
};
pair<point,point> crossover(circle c,line l) {
  point f = foot_point(c.o,l);
  db d = dist(c.o,l);
  if (judge(d - c.r) > 0) return make_pair(NULLP,NULLP);
  if (judge(d - c.r) == 0) return make_pair(f,NULLP);
  db t = sqrt(c.r * c.r - d * d);
  return make_pair(f + (l.v * t), f - (l.v * t));
}
pair<point,point> crossover(circle c1,circle c2) {
  if (judge(c1.r - c2.r) < 0) swap(c1,c2);
  db d = (c1.o - c2.o).abs();
  if (judge(d - c1.r - c2.r) > 0) return make_pair(NULLP,NULLP);
  if (judge(c1.r - c2.r - d) > 0) return make_pair(NULLP,NULLP);
  db co = (d * d + c1.r * c1.r - c2.r * c2.r) / (2 * d * c1.r);
  line l = line(c1.o,c2.o - c1.o);
  point h = l.u + (l.v * c1.r * co);
  line t = line(h,perpen(l.v));
  return crossover(c1,t);
}
circle CircumscribedCircle(point a,point b,point c) {
  line l1 = perpen_bi(a,b);
  line l2 = perpen_bi(b,c);
  point o = crossover(l1,l2);
  return circle(o,(o - a).abs());
}
circle InscribedCircle(point a,point b,point c) {
  angle a1 = angle(a,b - a,c - a);
  angle a2 = angle(b,a - b,c - b);
  line l1 = a1.bise();
  line l2 = a2.bise();
  point o = crossover(l1,l2);
  return circle(o,dist(o,line(a,b-a)));
}
void TangentLineThroughPoint(circle c,point p) {
  db d = (p - c.o).abs();
  if (judge(d - c.r) < 0) return (void) (puts("[]"));
  if (judge(d - c.r) == 0) {
    point t = perpen(p - c.o);
    db ret = atan(t.y  / t.x) / pi * 180;
    if (judge(ret) < 0) ret = ret + 180;
    return (void) (printf("[%.6lf]\n",ret));
  }
  db r = sqrt(d * d - c.r * c.r);
  pair<point,point> tmp = crossover(circle(p,r),c);
  point t;
  db ret1, ret2;
  t = tmp.first - p;
  ret1 = atan(t.y / t.x) / pi * 180;
  if (judge(ret1) < 0) ret1 += 180;
  t = tmp.second - p;
  ret2 = atan(t.y / t.x) / pi * 180;
  if (judge(ret2) < 0) ret2 += 180;
  if (judge(ret1 - ret2) > 0) swap(ret1,ret2);
  printf("[%.6lf,%.6lf]\n",ret1,ret2);
}
void output(vector<point>& ret) {
  sort(ret.begin(),ret.end());
  if (ret.size() == 0) return (void) (puts("[]"));
  printf("[(%.6lf,%.6lf)",ret[0].x,ret[0].y);
  for (int i = 1 ; i < (int)ret.size() ; ++ i)
    printf(",(%.6lf,%.6lf)",ret[i].x,ret[i].y);
  puts("]");
}
void CircleThroughAPointAndTangentToALineWithRadius(point p,line l,db r) {
  static vector<point> ret;
  ret.clear();
  point t = perpen(l.v);
  line l1 = line(l.u + (t * r),l.v);
  line l2 = line(l.u - (t * r),l.v);
  circle c = circle(p,r);
  pair<point,point> tmp;
  tmp = crossover(c,l1);
  if (tmp.first.avail()) ret.push_back(tmp.first);
  if (tmp.second.avail()) ret.push_back(tmp.second);
  tmp = crossover(c,l2);
  if (tmp.first.avail()) ret.push_back(tmp.first);
  if (tmp.second.avail()) ret.push_back(tmp.second);
  output(ret);
}
void CircleTangentToTwoLinesWithRadius(line a,line b,db r) {
  static vector<point> ret;
  ret.clear();
  point t;
  t = perpen(a.v);
  line a1 = line(a.u + (t * r),a.v);
  line a2 = line(a.u - (t * r),a.v);
  t = perpen(b.v);
  line b1 = line(b.u + (t * r),b.v);
  line b2 = line(b.u - (t * r),b.v);
  t = crossover(a1,b1);
  if (t.avail()) ret.push_back(t);
  t = crossover(a1,b2);
  if (t.avail()) ret.push_back(t);
  t = crossover(a2,b1);
  if (t.avail()) ret.push_back(t);
  t = crossover(a2,b2);
  if (t.avail()) ret.push_back(t);
  output(ret);
}
void CircleTangentToTwoDisjointCirclesWithRadius(circle c1,circle c2,db r) {
  static vector<point> ret;
  ret.clear();
  c1.r += r;
  c2.r += r;
  pair<point,point> tmp;
  tmp = crossover(c1,c2);
  if (tmp.first.avail()) ret.push_back(tmp.first);
  if (tmp.second.avail()) ret.push_back(tmp.second);
  output(ret);
}
int main() {
  string tmp;
  point a,b,c,d;
  circle cir;
  db t,t1,t2;
  while (cin >> tmp) {
    if (tmp == "CircumscribedCircle") {
      scanf("%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y);
      cir = CircumscribedCircle(a,b,c);
      printf("(%.6lf,%.6lf,%.6lf)\n",cir.o.x,cir.o.y,cir.r);
    }
    if (tmp == "InscribedCircle") {
      scanf("%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y);
      cir = InscribedCircle(a,b,c);
      printf("(%.6lf,%.6lf,%.6lf)\n",cir.o.x,cir.o.y,cir.r);
    }
    if (tmp == "TangentLineThroughPoint") {
      scanf("%lf%lf%lf%lf%lf",&cir.o.x,&cir.o.y,&cir.r,&a.x,&a.y);
      TangentLineThroughPoint(cir,a);
    }
    if (tmp == "CircleThroughAPointAndTangentToALineWithRadius") {
      scanf("%lf%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y,&t);
      CircleThroughAPointAndTangentToALineWithRadius(a,line(b,c - b),t);
    }
    if (tmp == "CircleTangentToTwoLinesWithRadius") {
      scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y,&d.x,&d.y,&t);
      CircleTangentToTwoLinesWithRadius(line(a,b-a),line(c,d-c),t);
    }
    if (tmp == "CircleTangentToTwoDisjointCirclesWithRadius") {
      scanf("%lf%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&t1,&b.x,&b.y,&t2,&t);
      CircleTangentToTwoDisjointCirclesWithRadius(circle(a,t1),circle(b,t2),t);
    }
  }
  return 0;
}

小结：我写得太久了。敲键盘时还总是有迟疑。