题目链接:uva 11978 - Fukushima Nuclear Blast
二分,圆和多边形面积交。将多边形拆成n份三角形(每两点与圆心构成),计算有向面积。每个三角形和圆的面积分四类讨论。
#include <cstdio> #include <cstring> #include <cmath> #include <vector> #include <algorithm> using namespace std; 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); } 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; struct Line { double a, b, c; Line (double a = 0, double b = 0, double c = 0): a(a), b(b), c(c) {} }; 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 Linear { using namespace Vectorial; Line getLine (double x1, double y1, double x2, double y2) { return Line(y2-y1, x1-x2, y1*(x2-x1)-x1*(y2-y1)); } 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; } } 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 = 1; i < n-1; i++) ret += getCross(p[i]-p[0], p[i+1]-p[0]); return fabs(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; } }; 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 Vectorial; using namespace Circular; using namespace Polygonal; const int maxn = 5005; const double inf = 1e4; int N; Point P[maxn], C; int main () { int cas; scanf("%d", &cas); for (int kcas = 1; kcas <= cas; kcas++) { double x, y, k; scanf("%d", &N); for (int i = 0; i < N; i++) P[i].read(); scanf("%lf%lf%lf", &x, &y, &k); double S = getArea(P, N) * k / 100; double l = 0, r = inf; while (dcmp(r - l) > 0) { double mid = (l + r) / 2; double area = getPublicAreaToPolygon(Circle(Point(x, y), mid), P, N); if (dcmp(area-S) >= 0) r = mid; else l = mid; } double ans = (l + r) / 2; printf("Case %d: %.0lf\n", kcas, ans); } return 0; }
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时间: 2024-08-24 01:32:08