uva 361 - Cops and Robbers（凸包）

题目链接：uva 361 - Cops and Robbers

点P在点集S中某三个点构成的三角形内的充要条件是点P在点集S的凸包内。

#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); }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 (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); }};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;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 onPolygonal (Point a, Point* p, int n) {for (int i = 0; i < n; i++) if (a == p[i]) return 0;if (n >= 2) {for (int i = 0; i < n; i++) {int v = (i + 1) % n;if (onSegment(a, p[i], p[v])) return 0;}}if (n <= 2) return 1;int sig = dcmp((a - p[0]) * (p[1] - p[0]));for (int i = 0; i < n; i++) {int v = (i + 1) % n;int tsig = dcmp((a - p[i]) * (p[v] - p[i]));if (tsig != sig)return 1;}return -1;}};namespace Circular {using namespace Linear;using namespace Vectorial;/* 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 Polygonal;const int maxn = 205;int N, M, Q, n, m;Point A[maxn], B[maxn], a[maxn], b[maxn];void init () {for (int i = 0; i < N; i++) A[i].read();for (int i = 0; i < M; i++) B[i].read();sort(A, A + N);n = unique(A, A + N) - A;sort(B, B + N);m = unique(B, B + M) - B;n = getConvexHull(A, n, a);m = getConvexHull(B, m, b);}int main () {int cas = 1;while (scanf("%d%d%d", &N, &M, &Q) == 3 && N + M + Q) {init ();printf("Data set %d:\n", cas++);int x, y;for (int i = 0; i < Q; i++) {scanf("%d%d", &x, &y);if (N > 2 && onPolygonal(Point(x, y), a, n) <= 0)printf("     Citizen at (%d,%d) is safe.\n", x, y);else if (M > 2 && onPolygonal(Point(x, y), b, m) <= 0)printf("     Citizen at (%d,%d) is robbed.\n", x, y);elseprintf("     Citizen at (%d,%d) is neither.\n", x, y);}printf("\n");}return 0;}