uva 10652 凸包

#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
struct Point
{
	double x, y;
	Point(double x = 0, double y = 0): x(x), y(y) {}
};
typedef Point Vector;
Vector operator +(Vector A, Vector 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 *(Vector A, double p)//
{
	return Vector(A.x * p, A.y * p);
}
Vector operator /(Vector A, double p)//
{
	return Vector(A.x / p, A.y / p);
}
bool operator <(const Point &a, const Point &b)//
{
	return a.x < b.x || (a.x == b.x && a.y < b.y);
}
double Cross(Vector A, Vector B)//
{
	return A.x * B.y - A.y * B.x;
}
double PolygonArea(Point *p, int n) //
{
	double area = 0;
	for (int i = 1; i < n - 1; i++)
		area += Cross(p[i] - p[0], p[i + 1] - p[0]);
	return area / 2;
}
Vector Rotate(Vector A, double rad) //
{
	return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}
double torad(double deg) { return deg / 180 * PI; }
int ConvexHull(Point*p, int n, Point *ch)
{
	sort(p, p + n);
	int m = 0;
	for (int i = 0; i < n; i++)
	{
		while (m > 1 && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0) m--;
		ch[m++] = p[i];
	}
	int k = m;
	for (int i = n - 2; i >= 0; i--)
	{
		while (m > k && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0) m--;
		ch[m++] = p[i];
	}
	if (n > 1) m--;
	return m;
}
int main(int argc, char const *argv[])
{
	int T;
	Point P[2500], ch[2500];
	scanf("%d", &T);
	while (T--)
	{
		int n, pc = 0;
		double areal = 0;
		scanf("%d", &n);
		for (int i = 0; i < n; i++)
		{
			double x, y, w, h, j, ang;
			scanf("%lf%lf%lf%lf%lf", &x, &y, &w, &h, &j);
			Point o(x, y);
			ang = -torad(j);
			P[pc++] = o + Rotate(Vector(-w / 2, -h / 2), ang);
			P[pc++] = o + Rotate(Vector(w / 2, -h / 2), ang);
			P[pc++] = o + Rotate(Vector(-w / 2, h / 2), ang);
			P[pc++] = o + Rotate(Vector(w / 2, h / 2), ang);
			areal += w * h;
		}
		int m = ConvexHull(P, pc, ch);
		double area2 = PolygonArea(ch, m);
		printf("%.1lf %%\n", areal * 100 / area2);
	}
	return 0;
}

大意：有n块矩形木板，你的任务是用一个面积尽量小的凸多边形把它们包起来，并计算出木板站整个包装面积的百分比。


思路：按照题意将所有矩形顶点坐标存起来，旋转时先旋转从中心出发的向量，求得各个坐标之后，求凸包即可。