10652.cpp

// Steven Kester Yuwono - UVa 10652
// Computational Geometry - Convex Hull

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <cmath>
#include <stack>
#include <vector>
using namespace std;

#define EPS 1e-9
#define PI acos(-1.0)

double DEG_to_RAD(double d) { return d * PI / 180.0; }

double RAD_to_DEG(double r) { return r * 180.0 / PI; }

struct point { double x, y;   // only used if more precision is needed
	point() { x = y = 0.0; }                      // default constructor
	point(double _x, double _y) : x(_x), y(_y) {}        // user-defined
	bool operator == (point other) const {
	 return (fabs(x - other.x) < EPS && (fabs(y - other.y) < EPS)); } };

struct vec { double x, y;  // name: `vec' is different from STL vector
	vec(double _x, double _y) : x(_x), y(_y) {} };

vec toVec(point a, point b) {       // convert 2 points to vector a->b
	return vec(b.x - a.x, b.y - a.y); }

double dist(point p1, point p2) {                // Euclidean distance
	return hypot(p1.x - p2.x, p1.y - p2.y); }           // return double

// returns the perimeter, which is the sum of Euclidian distances
// of consecutive line segments (polygon edges)
double perimeter(const vector<point> &P) {
	double result = 0.0;
	for (int i = 0; i < (int)P.size()-1; i++)  // remember that P[0] = P[n-1]
		result += dist(P[i], P[i+1]);
	return result; }

// returns the area, which is half the determinant
double area(const vector<point> &P) {
	double result = 0.0, x1, y1, x2, y2;
	for (int i = 0; i < (int)P.size()-1; i++) {
		x1 = P[i].x; x2 = P[i+1].x;
		y1 = P[i].y; y2 = P[i+1].y;
		result += (x1 * y2 - x2 * y1);
	}
	return fabs(result) / 2.0; }

double dot(vec a, vec b) { return (a.x * b.x + a.y * b.y); }

double norm_sq(vec v) { return v.x * v.x + v.y * v.y; }

double angle(point a, point o, point b) {  // returns angle aob in rad
	vec oa = toVec(o, a), ob = toVec(o, b);
	return acos(dot(oa, ob) / sqrt(norm_sq(oa) * norm_sq(ob))); }

double cross(vec a, vec b) { return a.x * b.y - a.y * b.x; }

// note: to accept collinear points, we have to change the `> 0'
// returns true if point r is on the left side of line pq
bool ccw(point p, point q, point r) {
	return cross(toVec(p, q), toVec(p, r)) > 0; }

// returns true if point r is on the same line as the line pq
bool collinear(point p, point q, point r) {
	return fabs(cross(toVec(p, q), toVec(p, r))) < EPS; }

// returns true if we always make the same turn while examining
// all the edges of the polygon one by one
bool isConvex(const vector<point> &P) {
	int sz = (int)P.size();
	if (sz <= 3) return false;   // a point/sz=2 or a line/sz=3 is not convex
	bool isLeft = ccw(P[0], P[1], P[2]);               // remember one result
	for (int i = 1; i < sz-1; i++)            // then compare with the others
		if (ccw(P[i], P[i+1], P[(i+2) == sz ? 1 : i+2]) != isLeft)
			return false;            // different sign -> this polygon is concave
	return true; }                                  // this polygon is convex

// returns true if point p is in either convex/concave polygon P
bool inPolygon(point pt, const vector<point> &P) {
	if ((int)P.size() == 0) return false;
	double sum = 0;    // assume the first vertex is equal to the last vertex
	for (int i = 0; i < (int)P.size()-1; i++) {
		if (ccw(pt, P[i], P[i+1]))
				 sum += angle(P[i], pt, P[i+1]);                   // left turn/ccw
		else sum -= angle(P[i], pt, P[i+1]); }                 // right turn/cw
	return fabs(fabs(sum) - 2*PI) < EPS; }	

// line segment p-q intersect with line A-B.
point lineIntersectSeg(point p, point q, point A, point B) {
	double a = B.y - A.y;
	double b = A.x - B.x;
	double c = B.x * A.y - A.x * B.y;
	double u = fabs(a * p.x + b * p.y + c);
	double v = fabs(a * q.x + b * q.y + c);
	return point((p.x * v + q.x * u) / (u+v), (p.y * v + q.y * u) / (u+v)); }

// cuts polygon Q along the line formed by point a -> point b
// (note: the last point must be the same as the first point)
vector<point> cutPolygon(point a, point b, const vector<point> &Q) {
	vector<point> P;
	for (int i = 0; i < (int)Q.size(); i++) {
		double left1 = cross(toVec(a, b), toVec(a, Q[i])), left2 = 0;
		if (i != (int)Q.size()-1) left2 = cross(toVec(a, b), toVec(a, Q[i+1]));
		if (left1 > -EPS) P.push_back(Q[i]);       // Q[i] is on the left of ab
		if (left1 * left2 < -EPS)        // edge (Q[i], Q[i+1]) crosses line ab
			P.push_back(lineIntersectSeg(Q[i], Q[i+1], a, b));
	}
	if (!P.empty() && !(P.back() == P.front()))
		P.push_back(P.front());        // make P's first point = P's last point
	return P; }

point pivot;
bool angleCmp(point a, point b) {                 // angle-sorting function
	if (collinear(pivot, a, b))                               // special case
		return dist(pivot, a) < dist(pivot, b);    // check which one is closer
	double d1x = a.x - pivot.x, d1y = a.y - pivot.y;
	double d2x = b.x - pivot.x, d2y = b.y - pivot.y;
	return (atan2(d1y, d1x) - atan2(d2y, d2x)) < 0; }   // compare two angles

vector<point> CH(vector<point> P) {   // the content of P may be reshuffled
	int i, j, n = (int)P.size();
	if (n <= 3) {
		if (!(P[0] == P[n-1])) P.push_back(P[0]); // safeguard from corner case
		return P;                           // special case, the CH is P itself
	}

	// first, find P0 = point with lowest Y and if tie: rightmost X
	int P0 = 0;
	for (i = 1; i < n; i++)
		if (P[i].y < P[P0].y || (P[i].y == P[P0].y && P[i].x > P[P0].x))
			P0 = i;

	point temp = P[0]; P[0] = P[P0]; P[P0] = temp;    // swap P[P0] with P[0]

	// second, sort points by angle w.r.t. pivot P0
	pivot = P[0];                    // use this global variable as reference
	sort(++P.begin(), P.end(), angleCmp);              // we do not sort P[0]

	// third, the ccw tests
	vector<point> S;
	S.push_back(P[n-1]); S.push_back(P[0]); S.push_back(P[1]);   // initial S
	i = 2;                                         // then, we check the rest
	while (i < n) {           // note: N must be >= 3 for this method to work
		j = (int)S.size()-1;
		if (ccw(S[j-1], S[j], P[i])) S.push_back(P[i++]);  // left turn, accept
		else S.pop_back(); }   // or pop the top of S until we have a left turn
	return S; }                                          // return the result


int main() {
	int tc; scanf("%d",&tc);
	while(tc--){
		int n;
		scanf("%d",&n);

		double boxArea = 0;
		vector<point> p;
		for(int i=0;i<n;i++){
			double x,y,w,h,j;
			scanf("%lf %lf %lf %lf %lf",&x,&y,&w,&h,&j);
			boxArea += (w*h);

			double angle = DEG_to_RAD(-j);

			double upperLeftX = x + (w/2.0) * cos(angle) - (h/2.0) * sin(angle);
			double upperLeftY = y + (h/2.0) * cos(angle) + (w/2.0) * sin(angle);
			double upperRightX =  x - (w/2.0) * cos(angle) - (h/2.0) * sin(angle);
			double upperRightY =  y + (h/2.0) * cos(angle) - (w/2.0) * sin(angle);
			double bottomLeftX =  x + (w/2.0) * cos(angle) + (h/2.0) * sin(angle);
			double bottomLeftY =  y - (h/2.0) * cos(angle) + (w/2.0) * sin(angle);
			double bottomRightX =  x - (w/2.0) * cos(angle) + (h/2.0) * sin(angle);
			double bottomRightY =  y - (h/2.0) * cos(angle) - (w/2.0) * sin(angle);

			p.push_back(point(upperLeftX,upperLeftY));
			p.push_back(point(upperRightX,upperRightY));
			p.push_back(point(bottomLeftX,bottomLeftY));
			p.push_back(point(bottomRightX,bottomRightY));
		}
		p = CH(p);
		printf("%.1f \%\n", boxArea/area(p)*100.0);
	}
	return 0;
}