HDU 4766 模擬退火(最小圓覆蓋) + 二分


#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
const int maxn = 1E3 + 10;
using namespace std;
struct Point
{
	double x, y;
	Point(double x = 0, double y = 0): x(x), y(y) {}
};
typedef Point Vector;
typedef vector<Point> Polygon;
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);
}
const double eps = 1e-6;
int dcmp(double x)//
{
	if (fabs(x) < eps) return 0;
	else return x < 0 ? -1 : 1;
}
bool operator ==(const Point &a, const Point &b)//
{
	return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
double Dot(Vector A, Vector B)//
{
	return A.x * B.x + A.y * B.y;
}
double Length(Vector A)//
{
	return sqrt(Dot(A, A));
}
double Cross(Vector A, Vector B)//
{
	return A.x * B.y - A.y * B.x;
}
double Distance(Point A, Point B)
{
	return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}
double Distance2(Point A, Point B)
{
	return (A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y);
}
double DistanceToLine(Point p, Point A, Point B) //
{
	Vector v1 = B - A, v2 = p - A;
	return fabs(Cross(v1, v2)) / Length(v1);
}
Vector Normal(Vector A)//
{
	double L = Length(A);
	return Vector(-A.y / L, A.x / L);
}
double Angle(Vector A, Vector B)//
{
	return acos(Dot(A, B) / Length(A) / Length(B));
}
struct Line
{
	Point p;
	Vector v;
	double ang;
	Line() {};
	Line(Point P, Vector v): p(P), v(v) {ang = atan2(v.y, v.x);}
	bool operator < (const Line& L) const
	{
		return ang < L.ang;
	}
	Point point(double t)
	{
		return p + v * t;
	}
	Line move(double d)
	{
		return Line(p + Normal(v) * d, v);
	}
};
struct Circle
{
	Point c;
	double r;
	Circle(Point c, double r): c(c), r(r) {}
	Point point(double a)
	{
		return Point(c.x + cos(a) * r, c.y + sin(a) * r);
	}
};
int getLineCircleIntersection(Line L, Circle C, double& t1, double& t2, vector<Point>& sol)
{
	double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y - C.c.y;
	double e = a * a + c * c, f = 2 * (a * b + c * d), g = b * b + d * d - C.r * C.r;
	double delta = f * f - 4 * e * g; // 判別式
	if (dcmp(delta) < 0) return 0; // 相離
	if (dcmp(delta) == 0)  // 相切
	{
		t1 = t2 = -f / (2 * e); sol.push_back(L.point(t1));
		return 1;
	}
	// 相交
	t1 = (-f - sqrt(delta)) / (2 * e); sol.push_back(L.point(t1));
	t2 = (-f + sqrt(delta)) / (2 * e); sol.push_back(L.point(t2));
	return 2;
}
double angle(Vector v) {return atan2(v.y, v.x);}
int GetCircleCircleIntersection(Circle C1, Circle C2, Point &A, Point &B)
{
	double d = Length(C1.c - C2.c);
	if (dcmp(d) == 0)
	{
		if (dcmp(C1.r - C2.r) == 0) return -1; // 重合,無窮多交點
		return 0;
	}
	if (dcmp(C1.r + C2.r - d) < 0) return 0;
	if (dcmp(fabs(C1.r - C2.r) - d) > 0) return 0;

	double a = angle(C2.c - C1.c);
	double da = acos((C1.r * C1.r + d * d - C2.r * C2.r) / (2 * C1.r * d));
	Point p1 = C1.point(a - da), p2 = C1.point(a + da);

	A = p1;
	if (p1 == p2) return 1;
	B = p2;
	return 2;
}
int n;
Point P[maxn], p0;
double R[maxn], d;
bool Judge(double mid)
{
	double Left, Right;
	for (int i = 0; i <= n; i++)
	{
		if (i == 0)
		{
			Left = P[i].x - R[i];
			Right = P[i].x + R[i];
		}
		else
		{
			if (P[i].x - R[i] > Left) Left = P[i].x - R[i];
			if (P[i].x + R[i] < Right) Right = P[i].x + R[i];
		}
	}

	if (Left - Right > eps) return false;
	int step = 50;
	while (step--)
	{
		double mid = (Left + Right) * 0.5;
		double low, high, uy, dy;
		int low_id, high_id;
		for (int i = 0; i <= n; i++)
		{
			double d = sqrt(R[i] * R[i] - (P[i].x - mid) * (P[i].x - mid));
			uy = P[i].y + d;
			dy = P[i].y - d;
			if (i == 0)
			{
				low_id = high_id = 0;
				low = dy; high = uy;
			}
			else
			{
				if (uy < high) high = uy, high_id = i;
				if (dy > low) low = dy, low_id = i;
			}
		}

		if (high - low > -eps)
		{
			return 1;
		}
		Point a, b;
		if (GetCircleCircleIntersection(Circle(P[high_id], R[high_id]), Circle(P[low_id], R[low_id]), a, b))
		{
			if ((a.x + b.x) * 0.5 < mid)
			{
				Right = mid;
			}
			else Left = mid;
		}
		else return false;
	}
	return false;
}
void circle_center(Point p0, Point p1, Point p2, Point &cp)
{
	double a1 = p1.x - p0.x, b1 = p1.y - p0.y, c1 = (a1 * a1 + b1 * b1) / 2;
	double a2 = p2.x - p0.x, b2 = p2.y - p0.y, c2 = (a2 * a2 + b2 * b2) / 2;
	double d = a1 * b2 - a2 * b1;
	cp.x = p0.x + (c1 * b2 - c2 * b1) / d;
	cp.y = p0.y + (a1 * c2 - a2 * c1) / d;
}
void circle_center(Point p0, Point p1, Point &cp)
{
	cp.x = (p0.x + p1.x) / 2;
	cp.y = (p0.y + p1.y) / 2;
}
Point center;
double radius;
bool Point_in(const Point &p)
{
	return Distance(p, center) - radius < 0;
}
void min_circle_cover(Point a[], int n)
{
	radius = 0;
	center = a[0];
	for (int i = 1; i < n; i++)
		if (!Point_in(a[i]))
		{
			center = a[i];
			radius = 0;
			for (int j = 0; j < i; j++)
				if (!Point_in(a[j]))
				{
					circle_center(a[i], a[j], center);
					radius = Distance(a[j], center);
					for (int k = 0; k < j; k++)
						if (!Point_in(a[k]))
						{
							circle_center(a[i], a[j], a[k], center);
							radius = Distance(a[k], center);
						}
				}
		}
}
int main(int argc, char const *argv[])
{
	while (~scanf("%lf%lf%lf", &p0.x, &p0.y, &d))
	{
		scanf("%d",  &n);
		for (int i = 0; i < n; i++)
		{
			R[i] = d;
			scanf("%lf%lf", &P[i].x, &P[i].y);
		}
		P[n] = p0; min_circle_cover(P, n);
		double Left = 0, Right = Distance(center, p0) + eps, mid;
		int ok = 0;
		while (dcmp(Right - Left) > 0)
		{
			mid = (Left + Right) / 2;
			R[n] = mid;
			if (Judge(mid)) Right = mid, ok = 1;
			else Left = mid;
		}
		if (ok) printf("%.2lf\n", mid);
		else cout << "X" << endl;
	}
	return 0;
}


一看題目,顯然是二分,二分枚舉到房子的距離也就是路由器到房子的距離,確定最大值應該用模擬退火,退火時間復雜度O(n)。


比網上的O(n3)的暴力高得不知道到哪去了,暴力普遍需要2K毫秒,這個只要78毫秒就可以了。哈哈哈。


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