uva 12307 - Smallest Enclosing Rectangle(旋转卡壳)

题目链接:uva 12307 - Smallest Enclosing Rectangle

两组踵对点围成长方形,枚举出所有可行长方形维护最小值。

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

using namespace std;
typedef pair<int,int> pii;
const double pi = 4 * atan(1);
const double eps = 1e-10;

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); }
inline double torad(double deg) { return deg / 180 * pi; }

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;
typedef vector<Point> Polygon;

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

struct DirLine {
	Point p;
	Vector v;
	double ang;
	DirLine () {}
	DirLine (Point p, Vector v): p(p), v(v) { ang = atan2(v.y, v.x); }
	bool operator < (const DirLine& u) const { return ang < u.ang; }
};

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 ComplexVector {
	typedef complex<double> Point;
	typedef Point Vector;

	double getDot(Vector a, Vector b) { return real(conj(a)*b); }
	double getCross(Vector a, Vector b) { return imag(conj(a)*b); }
	Vector rotate(Vector a, double rad) { return a*exp(Point(0, rad)); }
};

namespace Linear {
	using namespace Vectorial;

	Line getLine (double x1, double y1, double x2, double y2) { return Line(y2-y1, x1-x2, y1*x2-x1*y2); }
	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; }
	bool onLeft(DirLine l, Point p) { return dcmp(l.v * (p-l.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 = 0; i < n-1; i++)
			ret += (p[i]-p[0]) * (p[i+1]-p[0]);
		return 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;
	}

	/* 旋转卡壳 */
	void rotatingCalipers(Point *p, int n, vector<pii>& sol) {
		sol.clear();
		int j = 1; p[n] = p[0];
		for (int i = 0; i < n; i++) {
			while (getCross(p[j+1]-p[i+1], p[i]-p[i+1]) > getCross(p[j]-p[i+1], p[i]-p[i+1]))
				j = (j+1) % n;
			sol.push_back(make_pair(i, j));
			sol.push_back(make_pair(i + 1, j + 1));
		}
	}

	void rotatingCalipersGetRectangle (Point *p, int n, double& area, double& perimeter) {
		p[n] = p[0];
		int l = 1, r = 1, j = 1;
		area = perimeter = 1e20;

		for (int i = 0; i < n; i++) {
			Vector v = (p[i+1]-p[i]) / getLength(p[i+1]-p[i]);
			while (dcmp(getDot(v, p[r%n]-p[i]) - getDot(v, p[(r+1)%n]-p[i])) < 0) r++;
			while (j < r || dcmp(getCross(v, p[j%n]-p[i]) - getCross(v,p[(j+1)%n]-p[i])) < 0) j++;
			while (l < j || dcmp(getDot(v, p[l%n]-p[i]) - getDot(v, p[(l+1)%n]-p[i])) > 0) l++;
			double w = getDot(v, p[r%n]-p[i])-getDot(v, p[l%n]-p[i]);
			double h = getDistanceToLine (p[j%n], p[i], p[i+1]);
			area = min(area, w * h);
			perimeter = min(perimeter, 2 * w + 2 * h);
		}
	}

	/* 计算半平面相交可以用增量法,o(n^2),初始设置4条无穷大的半平面 */
	/* 用有向直线A->B切割多边形u,返回左侧。可能退化成单点或线段 */
	Polygon cutPolygon (Polygon u, Point a, Point b) {
		Polygon ret;
		int n = u.size();
		for (int i = 0; i < n; i++) {
			Point c = u[i], d = u[(i+1)%n];
			if (dcmp((b-a)*(c-a)) >= 0) ret.push_back(c);
			if (dcmp((b-a)*(c-d)) != 0) {
				Point t;
				getIntersection(a, b-a, c, d-c, t);
				if (onSegment(t, c, d))
					ret.push_back(t);
			}
		}
		return ret;
	}

	/* 半平面相交 */
	int halfPlaneIntersection(DirLine* li, int n, Point* poly) {
		sort(li, li + n);

		int first, last;
		Point* p = new Point[n];
		DirLine* q = new DirLine[n];
		q[first=last=0] = li[0];

		for (int i = 1; i < n; i++) {
			while (first < last && !onLeft(li[i], p[last-1])) last--;
			while (first < last && !onLeft(li[i], p[first])) first++;
			q[++last] = li[i];

			if (dcmp(q[last].v * q[last-1].v) == 0) {
				last--;
				if (onLeft(q[last], li[i].p)) q[last] = li[i];
			}

			if (first < last)
				getIntersection(q[last-1].p, q[last-1].v, q[last].p, q[last].v, p[last-1]);
		}

		while (first < last && !onLeft(q[first], p[last-1])) last--;
		if (last - first <= 1) { delete [] p; delete [] q; return 0; }
		getIntersection(q[last].p, q[last].v, q[first].p, q[first].v, p[last]);

		int m = 0;
		for (int i = first; i <= last; i++) poly[m++] = p[i];
		delete [] p; delete [] q;
		return m;
	}
};

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);
	}
};

using namespace Polygonal;

//void rotatingCalipersGetRectangle (Point *p, int n, double& area, double& perimeter) {

const int maxn = 1e5 + 5;

int N;
Point P[maxn], ch[maxn];

int main () {
	while (scanf("%d", &N) == 1 && N) {
		for (int i = 0; i < N; i++) P[i].read();
		N = getConvexHull(P, N, ch);
		double area, perimeter;
		rotatingCalipersGetRectangle(ch, N, area, perimeter);
		printf("%.2lf %.2lf\n", area, perimeter);
	}
	return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

时间: 2024-11-05 15:51:11

uva 12307 - Smallest Enclosing Rectangle(旋转卡壳)的相关文章

【最小矩形面积覆盖:凸包+旋转卡壳】UVA 10173 Smallest Bounding Rectangle

[最小矩形面积覆盖:凸包+旋转卡壳]UVA 10173 Smallest Bounding Rectangle 题目链接:UVA 10173 Smallest Bounding Rectangle 题目大意 给你n个点,求能够覆盖所有点集的最小矩形面积. 笔者的第2道凸包题目,凸包 + 旋转卡壳,实现点集的最小矩形面积覆盖问题 ">=0"写成"<=0"坑了我一下午!QAQ 说一下思路 ①Graham's Scan法构建凸包,时间复杂度O(nlogn) ②

此坑待填 离散化思想和凸包 UVA - 10173 Smallest Bounding Rectangle

Smallest Bounding Rectangle Given the Cartesian coordinates of n(>0)2-dimensional points, write a program that computes the area of their smallest bounding rectangle (smallest rectangle containing all the given points). Input The input le may contain

UVA 4728 Squares(凸包+旋转卡壳)

题目链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=17267 [思路] 凸包+旋转卡壳 求出凸包,用旋转卡壳算出凸包的直径即可. [代码] 1 #include<cstdio> 2 #include<vector> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 struct Pt { 8

BZOJ 1185 HNOI 2007 最小矩形覆盖 旋转卡壳

题目大意:给出平面上的一些点,问面积最小的矩形满足覆盖所有的点. 思路:覆盖问题和不是凸包上的点没关系,先做凸包.根据贪心的思想,这个覆盖了所有点的矩形肯定至少有一条边与凸包上的边重合,那么我们枚举凸包上的每一条边,对于这个已经确定了一条边的矩形,不难确定其他三个边.注意到已知当前直线的向量,就可以求出两侧和对面的向量,而这三个向量随着枚举的边的移动是单调的,所以就可以用旋转卡壳来卡住剩下的三条边. 但是旋转卡壳时的初值会出问题,如果按照逆时针的顺序求出剩下的三条边的时候,要想通过向量直接卡第三

NYOJ_253:LK的旅行(旋转卡壳入门)

题目链接 求平面最大点对. 找凸包 -> 根据凸包运用旋转卡壳算法求最大点对(套用kuang巨模板) 关于旋转卡壳算法 #include<bits/stdc++.h> using namespace std; struct point { int x,y; point operator -(const point& rhs)const { point ret; ret.x=x-rhs.x; ret.y=y-rhs.y; return ret; } int operator *(c

POJ 2079 Triangle [旋转卡壳]

Triangle Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 9525   Accepted: 2845 Description Given n distinct points on a plane, your task is to find the triangle that have the maximum area, whose vertices are from the given points. Input

【旋转卡壳】poj3608 Bridge Across Islands

给你俩凸包,问你它们的最短距离. 咋做就不讲了,经典题,网上一片题解. 把凸包上的点逆时针排序.可以取它们的平均点,然后作极角排序. 旋转卡壳其实是个很模板化的东西-- 先初始化分别在凸包P和Q上取哪个点,一般在P上取纵坐标最小的点,在Q上取纵坐标最大的点 for i=1 to n(n是凸包P上的点数) { while(两条边的叉积>0) { 凸包Q上的点++ } 计算当前两条边之间的答案(或者说每条边的2个端点到另一条边的答案) 凸包P上的点++ } #include<cstdio>

HDU 5251 矩形面积(二维凸包旋转卡壳最小矩形覆盖问题) --2015百度之星题目

B - 矩形面积 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description 小度熊有一个桌面,小度熊剪了很多矩形放在桌面上,小度熊想知道能把这些矩形包围起来的面积最小的矩形的面积是多少. Input 第一行一个正整数 T,代表测试数据组数(),接下来 T 组测试数据. 每组测试数据占若干行,第一行一个正整数 ,代表矩形的数量.接下来 N 行,每行 8

poj 2079 Triangle(旋转卡壳)

Triangle Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 8917   Accepted: 2650 Description Given n distinct points on a plane, your task is to find the triangle that have the maximum area, whose vertices are from the given points. Input