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题目链接:uva 10335 - Ray Inside a Polygon
恶心题,注意精度和输出等问题,代码中有标识,后面有一些数据。
#include <cstdio> #include <cstring> #include <cmath> #include <vector> #include <algorithm> using namespace std; const double pi = 4 * atan(1); const double eps = 1e-9; 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); } }; 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) {} Point point(double rad) { return Point(o.x + cos(rad)*r, o.y + sin(rad)*r); } void read() { scanf("%lf%lf%lf", &o.x, &o.y, &r); } }; 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 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; } }; namespace Polygonal { using namespace Vectorial; 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; } }; namespace Circular { using namespace Vectorial; using namespace Linear; /* 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)); } }; using namespace Vectorial; using namespace Linear; typedef long long ll; int V, N; double rad; Point S, P[15]; bool cmp(Point a, Point b) { ll x = (ll)(a.x * 100 + 0.5) - (ll)(b.x * 100 + 0.5); ll y = (ll)(a.y * 100 + 0.5) - (ll)(b.y * 100 + 0.5); return x == 0 && y == 0; } bool move () { Point o; double len; int rec = -1; for (int i = 0; i < V; i++) { int u = (i + 1) % V; if (getIntersection(P[i], P[u]-P[i], S, Vector(cos(rad), sin(rad)), o)) { if (cmp(S, P[i]) || cmp(S, P[u])) continue; double k = getAngle(o-S); if (dcmp(k-rad) != 0 && dcmp(k+2*pi-rad) != 0 && dcmp(k-2*pi-rad) != 0) continue; // 在射线的反向延长线上 if (!onSegment(o, P[i], P[u])) continue; // 交点不在线段上 if (cmp(o, P[i]) || cmp(o, P[u])) continue; //printf("%d %lf %lf\n", i, k, rad); if (rec == -1 || dcmp(len - getLength(o-S)) >= 0) { // 维护最早向交的线段 len = getLength(o-S); rec = i; } } else { // 平行 if (onSegment(S, P[i], P[u]) || cmp(S, P[i]) || cmp(S, P[u])) {// 与线段重合 return false; } } } if (rec == -1) return false; // 找不到交点, 一开始就远离多边形 int u = (rec + 1) % V; Vector v = P[u] - P[rec]; getIntersection(P[rec], v, S, Vector(cos(rad), sin(rad)), o); double tmp = getAngle(v, o - S); if (dcmp(getCross(v, o-S)) > 0) tmp = -tmp; S = o; rad = getAngle(rotate(v, tmp)); return true; } int main () { while (scanf("%d%d", &V, &N) == 2 && V + N) { S.read(), scanf("%lf", &rad); rad = rad * pi / 180; for (int i = 0; i < V; i++) P[i].read(); bool flag = true; for (int i = 0; i <= N && flag; i++) { flag = move(); //S.write(), printf(" %d %lf\n", i, rad); } if (fabs(S.x) + 0.005 < 1e-2) S.x = 0; if (fabs(S.y) + 0.005 < 1e-2) S.y = 0; if (flag) printf("%.2lf %.2lf\n", S.x, S.y); else printf("lost forever...\n"); } return 0; } /* 4 2 2345.67 0.00 45.01 0.00 0.00 4000.00 0.00 4000.00 4000.00 -0.01 4000.00 4 6 1345.67 1.00 45.01 0.00 0.00 4000.00 0.00 4000.00 4000.00 -0.01 4000.00 */
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uva 10335 - Ray Inside a Polygon(几何)
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原文地址:http://blog.csdn.net/keshuai19940722/article/details/47732435