HDU 4766 模擬退火(最小圓覆蓋) + 二分
阿新 • • 發佈:2018-12-24
#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,二分列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,退火時間複雜度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。