uva 1463 - Largest Empty Circle on a Segment（二分+三分+几何）

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题目链接：uva 1463 - Largest Empty Circle on a Segment

二分半径，对于每个半径，用三分求出线段到线段的最短距离，根据最短距离可以确定当前R下每条线段在[0,L]上的可行区间，存在一个点被可行区间覆盖n次。

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

using namespace std;
const double pi = 4 * atan(1);
const double eps = 1e-9;

inline int dcmp (double x) { if (fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; }
inline double getDistance (double x, double y) { return sqrt(x * x + y * y); }

struct Point {
	double x, y;
	Point (double x = 0, double y = 0): x(x), y(y) {}
	void read () { scanf("%lf%lf", &x, &y); }
	void write () { printf("%lf %lf", x, y); }

	bool operator == (const Point& u) const { return dcmp(x - u.x) == 0 && dcmp(y - u.y) == 0; }
	bool operator != (const Point& u) const { return !(*this == u); }
	bool operator < (const Point& u) const { return x < u.x || (x == u.x && y < u.y); }
	bool operator > (const Point& u) const { return u < *this; }
	bool operator <= (const Point& u) const { return *this < u || *this == u; }
	bool operator >= (const Point& u) const { return *this > u || *this == u; }
	Point operator + (const Point& u) { return Point(x + u.x, y + u.y); }
	Point operator - (const Point& u) { return Point(x - u.x, y - u.y); }
	Point operator * (const double u) { return Point(x * u, y * u); }
	Point operator / (const double u) { return Point(x / u, y / u); }
	double operator * (const Point& u) { return x*u.y - y*u.x; }
};

typedef Point Vector;

struct Line {
	double a, b, c;
	Line (double a = 0, double b = 0, double c = 0): a(a), b(b), c(c) {}
};

struct Circle {
	Point o;
	double r;
	Circle () {}
	Circle (Point o, double r = 0): o(o), r(r) {}
	void read () { o.read(), scanf("%lf", &r); }
	Point point(double rad) { return Point(o.x + cos(rad)*r, o.y + sin(rad)*r); }
	double getArea (double rad) { return rad * r * r / 2; }
};


namespace Punctual {
	double getDistance (Point a, Point b) { double x=a.x-b.x, y=a.y-b.y; return sqrt(x*x + y*y); }
};

namespace Vectorial {
	/* 点积: 两向量长度的乘积再乘上它们夹角的余弦, 夹角大于90度时点积为负 */
	double getDot (Vector a, Vector b) { return a.x * b.x + a.y * b.y; }

	/* 叉积: 叉积等于两向量组成的三角形有向面积的两倍, cross(v, w) = -cross(w, v) */
	double getCross (Vector a, Vector b) { return a.x * b.y - a.y * b.x; }

	double getLength (Vector a) { return sqrt(getDot(a, a)); }
	double getPLength (Vector a) { return getDot(a, a); }
	double getAngle (Vector u) { return atan2(u.y, u.x); }
	double getAngle (Vector a, Vector b) { return acos(getDot(a, b) / getLength(a) / getLength(b)); }
	Vector rotate (Vector a, double rad) { return Vector(a.x*cos(rad)-a.y*sin(rad), a.x*sin(rad)+a.y*cos(rad)); }
	/* 单位法线 */
	Vector getNormal (Vector a) { double l = getLength(a); return Vector(-a.y/l, a.x/l); }
};

namespace Linear {
	using namespace Vectorial;

	Line getLine (double x1, double y1, double x2, double y2) { return Line(y2-y1, x1-x2, y1*(x2-x1)-x1*(y2-y1)); }
	Line getLine (double a, double b, Point u) { return Line(a, -b, u.y * b - u.x * a); }

	bool getIntersection (Line p, Line q, Point& o) {
		if (fabs(p.a * q.b - q.a * p.b) < eps)
			return false;
		o.x = (q.c * p.b - p.c * q.b) / (p.a * q.b - q.a * p.b);
		o.y = (q.c * p.a - p.c * q.a) / (p.b * q.a - q.b * p.a);
		return true;
	}

	/* 直线pv和直线qw的交点 */
	bool getIntersection (Point p, Vector v, Point q, Vector w, Point& o) {
		if (dcmp(getCross(v, w)) == 0) return false;
		Vector u = p - q;
		double k = getCross(w, u) / getCross(v, w);
		o = p + v * k;
		return true;
	}

	/* 点p到直线ab的距离 */
	double getDistanceToLine (Point p, Point a, Point b) { return fabs(getCross(b-a, p-a) / getLength(b-a)); }
	double getDistanceToSegment (Point p, Point a, Point b) {
		if (a == b) return getLength(p-a);
		Vector v1 = b - a, v2 = p - a, v3 = p - b;
		if (dcmp(getDot(v1, v2)) < 0) return getLength(v2);
		else if (dcmp(getDot(v1, v3)) > 0) return getLength(v3);
		else return fabs(getCross(v1, v2) / getLength(v1));
	}

	/* 点p在直线ab上的投影 */
	Point getPointToLine (Point p, Point a, Point b) { Vector v = b-a; return a+v*(getDot(v, p-a) / getDot(v,v)); }

	/* 判断线段是否存在交点 */
	bool haveIntersection (Point a1, Point a2, Point b1, Point b2) {
		double c1=getCross(a2-a1, b1-a1), c2=getCross(a2-a1, b2-a1), c3=getCross(b2-b1, a1-b1), c4=getCross(b2-b1,a2-b1);
		return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0;
	}

	/* 判断点是否在线段上 */
	bool onSegment (Point p, Point a, Point b) { return dcmp(getCross(a-p, b-p)) == 0 && dcmp(getDot(a-p, b-p)) < 0; }
}

namespace Triangular {
	using namespace Vectorial;

	double getAngle (double a, double b, double c) { return acos((a*a+b*b-c*c) / (2*a*b)); }
	double getArea (double a, double b, double c) { double s =(a+b+c)/2; return sqrt(s*(s-a)*(s-b)*(s-c)); }
	double getArea (double a, double h) { return a * h / 2; }
	double getArea (Point a, Point b, Point c) { return fabs(getCross(b - a, c - a)) / 2; }
	double getDirArea (Point a, Point b, Point c) { return getCross(b - a, c - a) / 2; }
};

namespace Polygonal {
	using namespace Vectorial;
	using namespace Linear;

	double getArea (Point* p, int n) {
		double ret = 0;
		for (int i = 1; i < n-1; i++)
			ret += getCross(p[i]-p[0], p[i+1]-p[0]);
		return fabs(ret)/2;
	}

	/* 凸包 */
	int getConvexHull (Point* p, int n, Point* ch) {
		sort(p, p + n);
		int m = 0;
		for (int i = 0; i < n; i++) {
			/* 可共线 */
			//while (m > 1 && dcmp(getCross(ch[m-1]-ch[m-2], p[i]-ch[m-1])) < 0) m--;
			while (m > 1 && dcmp(getCross(ch[m-1]-ch[m-2], p[i]-ch[m-1])) <= 0) m--;
			ch[m++] = p[i];
		}
		int k = m;
		for (int i = n-2; i >= 0; i--) {
			/* 可共线 */
			//while (m > k && dcmp(getCross(ch[m-1]-ch[m-2], p[i]-ch[m-2])) < 0) m--;
			while (m > k && dcmp(getCross(ch[m-1]-ch[m-2], p[i]-ch[m-2])) <= 0) m--;
			ch[m++] = p[i];
		}
		if (n > 1) m--;
		return m;
	}

	int isPointInPolygon (Point o, Point* p, int n) {
		int wn = 0;
		for (int i = 0; i < n; i++) {
			int j = (i + 1) % n;
			if (onSegment(o, p[i], p[j])) return 0; // 边界上
			int k = dcmp(getCross(p[j] - p[i], o-p[i]));
			int d1 = dcmp(p[i].y - o.y);
			int d2 = dcmp(p[j].y - o.y);
			if (k > 0 && d1 <= 0 && d2 > 0) wn++;
			if (k < 0 && d2 <= 0 && d1 > 0) wn--;
		}
		return wn ? -1 : 1;
	}
};

namespace Circular {
	using namespace Linear;
	using namespace Vectorial;
	using namespace Triangular;

	/* 直线和原的交点 */
	int getLineCircleIntersection (Point p, Point q, Circle O, double& t1, double& t2, vector<Point>& sol) {
		Vector v = q - p;
		/* 使用前需清空sol */
		//sol.clear();
		double a = v.x, b = p.x - O.o.x, c = v.y, d = p.y - O.o.y;
		double e = a*a+c*c, f = 2*(a*b+c*d), g = b*b+d*d-O.r*O.r;
		double delta = f*f - 4*e*g;
		if (dcmp(delta) < 0) return 0;
		if (dcmp(delta) == 0) {
			t1 = t2 = -f / (2 * e);
			sol.push_back(p + v * t1);
			return 1;
		}

		t1 = (-f - sqrt(delta)) / (2 * e); sol.push_back(p + v * t1);
		t2 = (-f + sqrt(delta)) / (2 * e); sol.push_back(p + v * t2);
		return 2;
	}

	/* 圆和圆的交点 */
	int getCircleCircleIntersection (Circle o1, Circle o2, vector<Point>& sol) {
		double d = getLength(o1.o - o2.o);

		if (dcmp(d) == 0) {
			if (dcmp(o1.r - o2.r) == 0) return -1;
			return 0;
		}

		if (dcmp(o1.r + o2.r - d) < 0) return 0;
		if (dcmp(fabs(o1.r-o2.r) - d) > 0) return 0;

		double a = getAngle(o2.o - o1.o);
		double da = acos((o1.r*o1.r + d*d - o2.r*o2.r) / (2*o1.r*d));

		Point p1 = o1.point(a-da), p2 = o1.point(a+da);

		sol.push_back(p1);
		if (p1 == p2) return 1;
		sol.push_back(p2);
		return 2;
	}

	/* 过定点作圆的切线 */
	int getTangents (Point p, Circle o, Vector* v) {
		Vector u = o.o - p;
		double d = getLength(u);
		if (d < o.r) return 0;
		else if (dcmp(d - o.r) == 0) {
			v[0] = rotate(u, pi / 2);
			return 1;
		} else {
			double ang = asin(o.r / d);
			v[0] = rotate(u, -ang);
			v[1] = rotate(u, ang);
			return 2;
		}
	}

	/* a[i] 和 b[i] 分别是第i条切线在O1和O2上的切点 */
	int getTangents (Circle o1, Circle o2, Point* a, Point* b) {
		int cnt = 0;
		if (o1.r < o2.r) { swap(o1, o2); swap(a, b); }
		double d2 = getLength(o1.o - o2.o); d2 = d2 * d2;
		double rdif = o1.r - o2.r, rsum = o1.r + o2.r;
		if (d2 < rdif * rdif) return 0;
		if (dcmp(d2) == 0 && dcmp(o1.r - o2.r) == 0) return -1;

		double base = getAngle(o2.o - o1.o);
		if (dcmp(d2 - rdif * rdif) == 0) {
			a[cnt] = o1.point(base); b[cnt] = o2.point(base); cnt++;
			return cnt;
		}

		double ang = acos( (o1.r - o2.r) / sqrt(d2) );
		a[cnt] = o1.point(base+ang); b[cnt] = o2.point(base+ang); cnt++;
		a[cnt] = o1.point(base-ang); b[cnt] = o2.point(base-ang); cnt++;

		if (dcmp(d2 - rsum * rsum) == 0) {
			a[cnt] = o1.point(base); b[cnt] = o2.point(base); cnt++;
		} else if (d2 > rsum * rsum) {
			double ang = acos( (o1.r + o2.r) / sqrt(d2) );
			a[cnt] = o1.point(base+ang); b[cnt] = o2.point(base+ang); cnt++;
			a[cnt] = o1.point(base-ang); b[cnt] = o2.point(base-ang); cnt++;
		}
		return cnt;
	}

	/* 三点确定外切圆 */
	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, getLength(p1 - p));  
	}

	/* 三点确定内切圆 */
	Circle InscribedCircle(Point p1, Point p2, Point p3) {  
		double a = getLength(p2 - p3);  
		double b = getLength(p3 - p1);  
		double c = getLength(p1 - p2);  
		Point p = (p1 * a + p2 * b + p3 * c) / (a + b + c);  
		return Circle(p, getDistanceToLine(p, p1, p2));  
	} 

	/* 三角形一顶点为圆心 */
	double getPublicAreaToTriangle (Circle O, Point a, Point b) {
		if (dcmp((a-O.o)*(b-O.o)) == 0) return 0;
		int sig = 1;
		double da = getPLength(O.o-a), db = getPLength(O.o-b);
		if (dcmp(da-db) > 0) {
			swap(da, db);
			swap(a, b);
			sig = -1;
		}

		double t1, t2;
		vector<Point> sol;
		int n = getLineCircleIntersection(a, b, O, t1, t2, sol);

		if (dcmp(da-O.r*O.r) <= 0) {
			if (dcmp(db-O.r*O.r) <= 0)	return getDirArea(O.o, a, b) * sig;

			int k = 0;
			if (getPLength(sol[0]-b) > getPLength(sol[1]-b)) k = 1;

			double ret = getArea(O.o, a, sol[k]) + O.getArea(getAngle(sol[k]-O.o, b-O.o));
			double tmp = (a-O.o)*(b-O.o);
			return ret * sig * dcmp(tmp);
		}

		double d = getDistanceToSegment(O.o, a, b);
		if (dcmp(d-O.r) >= 0) {
			double ret = O.getArea(getAngle(a-O.o, b-O.o));
			double tmp = (a-O.o)*(b-O.o);
			return ret * sig * dcmp(tmp);
		}


		double k1 = O.r / getLength(a - O.o), k2 = O.r / getLength(b - O.o);
		Point p = O.o + (a - O.o) * k1, q = O.o + (b - O.o) * k2;
		double ret1 = O.getArea(getAngle(p-O.o, q-O.o));
		double ret2 = O.getArea(getAngle(sol[0]-O.o, sol[1]-O.o)) - getArea(O.o, sol[0], sol[1]);
		double ret = (ret1 - ret2), tmp = (a-O.o)*(b-O.o);
		return ret * sig * dcmp(tmp);
	}

	double getPublicAreaToPolygon (Circle O, Point* p, int n) {
		if (dcmp(O.r) == 0) return 0;
		double area = 0;
		for (int i = 0; i < n; i++) {
			int u = (i + 1) % n;
			area += getPublicAreaToTriangle(O, p[i], p[u]);
		}
		return fabs(area);
	}
};
//double getDistanceToSegment (Point p, Point a, Point b);

using namespace Linear;

const int maxn = 2005;
const double inf = 100000;

int N;
double L;
Point S[maxn], E[maxn];

struct Seg {
	int d;
	double x;
	Seg (double x = 0, int d = 0): x(x), d(d) {}
	bool operator < (const Seg& u) const { return dcmp(x- u.x) < 0 || (dcmp(x - u.x) == 0 && d < u.d); }
};

double getDistance (int idx, double l, double r) {
	//while (dcmp(r - l) > 0) {
	while (r - l >= eps) {
		double ll = (l + r) / 2;
		double rr = (ll + r) / 2;
		double d1 = getDistanceToSegment(Point(ll, 0), S[idx], E[idx]);
		double d2 = getDistanceToSegment(Point(rr, 0), S[idx], E[idx]);
		if (d1 < d2)
			r = rr - eps;
		else
			l = ll + eps;
	}
	return l;
}

double handle1 (int idx, double l, double r, double d) {
	while (dcmp(r - l) > 0) {
		double mid = (l + r) / 2;
		if (getDistanceToSegment(Point(mid, 0), S[idx], E[idx]) > d)
			l = mid;
		else
			r = mid;
	}
	return (l + r) / 2;
}

double handle2 (int idx, double l, double r, double d) {
	while (dcmp(r - l) > 0) {
		double mid = (l + r) / 2;
		if (getDistanceToSegment(Point(mid, 0), S[idx], E[idx]) > d)
			r = mid;
		else
			l = mid;
	}
	return (l + r) / 2;
}

bool judge (double r) {
	vector<Seg> g;

	for (int i = 0; i < N; i++) {
		double s = getDistance(i, 0, L);
		if (dcmp(getDistanceToSegment(Point(s, 0), S[i], E[i]) - r) >= 0) {
			g.push_back(Seg(0, 1));
			g.push_back(Seg(L, -1));
			continue;
		}

		if (dcmp(getDistanceToSegment(Point(0, 0), S[i], E[i]) - r) >= 0) {
			double d = handle1 (i, 0, s, r);
			g.push_back(Seg(0, 1));
			g.push_back(Seg(d, -1));
		}

		if (dcmp(getDistanceToSegment(Point(L, 0), S[i], E[i]) - r) >= 0) {
			double d = handle2(i, s, L, r);
			g.push_back(Seg(d, 1));
			g.push_back(Seg(L, -1));
		}
	}

	sort(g.begin(), g.end());
	int c = 0;
	for (int i = 0; i < g.size(); i++) {
		c += g[i].d;
		if (c >= N)
			return true;
	}
	return false;
}

int main () {
	int cas;
	scanf("%d", &cas);
	while (cas--) {
		scanf("%d%lf", &N, &L);
		for (int i = 0; i < N; i++)
			S[i].read(), E[i].read();

		double l = 0, r = inf;
		while (dcmp(r-l) > 0) {
			double mid = (l + r) / 2;
			if (judge(mid)) l = mid;
			else r = mid;
		}
		printf("%.3lf\n", (l + r) / 2);
	}
	return 0;
}

uva 1463 - Largest Empty Circle on a Segment（二分+三分+几何）

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/47791659