（beginer） 凸包+点在多边形内 UVA 11072 Points

Points 

Time limit: 10s

In this problem you will be given a set of points in the Euclidian plane. The number of points in the set will never exceed100000. The coordinates of these points will be integer coordinates and will have an absolute value smaller than10000. There will be no identical points in the first set. Then you will be given a second set of points. For each point in the second set you will have to determine whether it lies in a triangle spanned by three points in the first set. A point lying on the edge of a triangle is considered to be "inside" the triangle.

In the following example the points p₁,p₂,p₃,p₄ belong to the first set. The pointsr and s belong to the second set. The point r isn't contained in any triangle spanned by three points of the first set. The points is contained in two triangles. For example, the triangle spanned by p₂,p₃,p₄.

Input

You will be given several testcases. A testcases consists of the number of pointsp, 3 ≤ p ≤ 100000 in the first set. It is followed by p pairs of numbers, each describing a point of the first set, the first number of a pair denoting thex-coordinate of the point, the second the y-coordinate. Each pair is on a seperate line. There may be colinear points in the first set. The next number in the input gives you the number of pointsr in the second set. It is followed by r pairs of numbers, each describing a point, each on a separate line. The first number of a pair being thex-coordinate, the second number being the y-coordinate of the point. All coordinates in the input will be integer coordinates.

Output

For each point in the second set, output if the point lies in a triangle spanned by three points of the first set. If the point lies inside a triangle outputinside otherwise output outside.

Sample input

40 04 40 44 062 24 41 10 20 1010 0

Sample output

insideinsideinsideinsideoutsideoutside

---

FAU Local Summer Contest 2005
Author: Tilmann Spiegelhauer 

题意：给出一些列的点的集合，然后询问某个点是否在这些点能形成的三角形里面。在边上算在里面。

思路：我们先对集合求一个凸包。然后就判断点是否在这个凸包里面就行了。显然在这个凸包里的任何点都能在一个三角形里面，在凸包上面总能选出三个点，形成的三角形能包围这个点，而在凸包外面的点，一定画不出三角形把它围起来。

代码：

#include<iostream>#include<cstdio>#include<string.h>#include<math.h>#include<string>#include<set>#include<cstring>#include<map>#include<algorithm>#include<vector>#include<queue>using namespace std;#define mp make_pair#define eps 1e-10const double inf = 1e16;const double PI = acos(-1.0);struct Point{Point(const Point&p) { x = p.x , y = p.y; }Point (double xx=0,double yy=0) : x(xx) , y(yy) { }double x;double y;};typedef Point Vector;Vector operator+(Vector  v1,Vector  v2) { return Vector(v1.x+v2.x,v1.y+v2.y); }Vector operator-(Vector  v1,Vector  v2) { return Vector(v1.x-v2.x,v1.y-v2.y); }Vector operator*(Vector  v, double p) { return Vector(v.x*p,v.y*p); }Vector operator/(Vector  v,double p) { return Vector(v.x/p,v.y/p); }int dcmp(double x) {if (fabs(x) < eps) return 0;return x < 0 ? -1 : 1; }bool operator < (Point  a,Point  b) { return dcmp(a.x-b.x)<0 || dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)<0; }bool operator==(const Point & a,const Point & b) {return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;}inline double toRad(double x) { return x * PI/180; }inline double toDegreed(double rad) { return rad*180/PI; }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 Angle(Vector A,Vector B) { double cosine = Dot(A,B)/Length(A)/Length(B);if (dcmp(cosine-1.0)==0) return 0;return acos(cosine); }double Cross(Vector A,Vector B) { return A.x*B.y-A.y*B.x; }double Area2(Point a,Point b,Point c) {  return Cross(b-a,c-a); }double PolyArea(Point *poly, int n){double ret = 0;for (int i = 1 ; i < n-1 ; ++i) ret += Cross(poly[i]-poly[0],poly[i+1]-poly[0]);return ret/2;}//旋转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 Normal(Vector A) { double L = Length(A); return Vector(-A.y/L,A.x/L); }//点和直线Point GetLineIntersection(Point P,Vector v,Point Q,Vector w){Vector u = P-Q;double t = Cross(w,u) / Cross(v,w);return P+v*t;}double DistanceToLine(Point P,Point A,Point B) {Vector v1 = B-A , v2 = P-A;return fabs(Cross(v1,v2))/Length(v1);}double DistanceToSegment(Point P,Point A,Point B){if (A==B) return Length(P-A);Vector v1 = B-A , v2 = P-A , v3 = P-B;if (dcmp(Dot(v1,v2)) < 0) return Length(v2);else if (dcmp(Dot(v1,v3)) > 0) return Length(v3);else return fabs(Cross(v1,v2))/Length(v1);}//点在直线上的投影Point GetLineProjection(Point P,Point A,Point B) {Vector v = B-A;return A+v*(Dot(v,P-A)/Dot(v,v));}//线段相交(不包括端点）bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2) {double c1 = Cross(a2-a1,b1-a1) , c2 = Cross(a2-a1,b2-a1) ,     c3 = Cross(b2-b1,a1-b1) , c4 = Cross(b2-b1,a2-b1);return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4)<0;}//点在线段上（不包括端点）bool OnSegment(Point p,Point a,Point b){return dcmp(Cross(a-p,b-p))==0 && dcmp(Dot(a-p,b-p)) < 0;}//点在线上bool OnLine(Point p,Point a,Point b){if (p==a || p==b) return true;return dcmp(Cross(a-p,b-p))==0;}//线段相交（包括端点）bool SegmentIntersection(Point a1,Point a2,Point b1,Point b2) {if (SegmentProperIntersection(a1,a2,b1,b2)) return true;if (OnSegment(a1,b1,b2) || OnSegment(a2,b1,b2)) return true;if (OnSegment(b1,a1,a2) || OnSegment(b2,a1,a2)) return true;if (a1==b1 || a1==b2 || a2==b1 || a2==b2) return true;return false;}//点在多边形内：0:外部 1：内部 -1：线上 -2:在顶点int isPointInPolygon(Point a,Point *p,int n){int wn = 0;for (int i = 0 ; i < n ; ++i) {if (a==p[i]) return -2; //点重合if (OnSegment(a,p[i],p[(i+1)%n])) return -1;int k = dcmp(Cross(p[(i+1)%n]-p[i],a-p[i]));int d1 = dcmp(p[i].y-a.y);int d2 = dcmp(p[(i+1)%n].y-a.y);if (k > 0 && d1<=0 && d2>0) ++wn;if (k<0 && d2<=0 && d1>0) --wn;}if (wn!=0) return 1;   //内部return 0;                 //外部}//--------------------------------------------------------------------------------------------//直线和直线struct Line{Point P;//直线上任意一点Vector v;// 方向向量。它的左边就是对应的半平面double ang;//极角Line() { }Line(Point P,Vector v) { this->P = P ; this->v = v; ang = atan2(v.y,v.x); }bool operator < (const Line& L) const { return ang < L.ang; } //排序用的比较运算符Point point(double t) { return v*t+P; }};//点p在有向直线L的左边（线上不算）bool OnLeft(Line L , Point p) { return Cross(L.v,p-L.P) > 0; }//二直线交点。假定交点唯一存在Point GetIntersection(Line a,Line b) {Vector u = a.P-b.P;double t = Cross(b.v,u) / Cross(a.v,b.v);return a.P+a.v*t;}//--------------------------------------------//与圆相关struct Circle {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;//判别式if (dcmp(delta) < 0) return 0;//相离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,vector<Point>& sol){double d = Length(C1.c-C2.c);if (dcmp(d)==0) {if (dcmp(C1.r-C2.r)==0) return -1;//两圆重合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));//向量C1C2的极角//C1C2到C1P1的角Point p1 = C1.point(a-da) , p2 = C1.point(a+da);sol.push_back(p1);if (p1==p2) return 1;sol.push_back(p2);return 2;}//国电p到圆C的切线。v[i]是第i条切线的向量。返回切线条数int getTangents(Point p,Circle C,Vector* v){Vector u= C.c-p;double dist = Length(u);if (dist < C.r) return 0;else if (dcmp(dist-C.r)==0) {v[0] = Rotate(u,PI/2);return 1;} else {double ang = asin(C.r/dist);v[0] = Rotate(u,-ang);v[1] = Rotate(u,+ang);return 2;}}int getTangents(Circle A,Circle B,Point* a, Point* b){int cnt = 0;if (A.r < B.r) { swap(A,B); swap(a,b); }double d2 = Dot(A.c-B.c,A.c-B.c);double rdiff = A.r-B.r;double rsum = A.r+B.r;if (dcmp(d2-rdiff*rdiff) < 0) return 0;//内含double base = atan2(B.c.y-A.c.y,B.c.x-A.c.x);if (d2==0 && dcmp(A.r-B.r)==0) return -1;//无限多条切线if (dcmp(d2-rdiff*rdiff)==0) {                       //内切，1条切线a[cnt] = A.point(base);b[cnt] = B.point(base);++cnt;return 1;}//有外切共线double ang = acos((A.r-B.r)/sqrt(d2));a[cnt] = A.point(base+ang); b[cnt] = B.point(base+ang); ++cnt;a[cnt] = A.point(base-ang); b[cnt] = B.point(base-ang); ++cnt;if (dcmp(d2-rsum*rsum)==0) {                  //一条内公切线a[cnt] = A.point(base);b[cnt] = B.point(PI+base);++cnt;} else if (dcmp(d2-rsum*rsum)>0) {           //两条公切线double ang = acos((A.r+B.r)/sqrt(d2));a[cnt] = A.point(base+ang); b[cnt] = B.point(PI+base+ang); ++cnt;a[cnt] = A.point(base-ang); b[cnt] = B.point(PI+base-ang); ++cnt;}return cnt;}//点p和圆的关系: 0:在圆上 1：在圆外 -1：在圆内int PointCircleRelation(Point p,Circle c) {return dcmp(Dot(p-c.c,p-c.c)-c.r*c.r);}//A在B内bool InCircle(Circle A,Circle B){if (dcmp(A.r-B.r)>0) return false;double d2 = Dot(A.c-B.c,A.c-B.c);double rdiff = A.r-B.r;double rsum = A.r+B.r;if (dcmp(d2-rdiff*rdiff) <= 0) return true;//内含或内切或重合return false;}//----------------------------------------------------------------------------//与球相关的转化//角度转换为弧度double torad(double deg) { return deg/180*PI; }//经纬度（角度）转化为空间坐标void get_coord(double R,double lat,double lng,double& x,double& y,double& z){lat = torad(lat);lng = torad(lng);x = R*cos(lat)*cos(lng);y = R*cos(lat)*sin(lng);z = R*sin(lat);}//-----------------------------------------------------------------------//几何算法：//凸包:O(nlogn) 得到的凸包逆时针int ConvexHull(Point* p , int n, Point* ch) {sort(p,p+n);int m = 0;for (int i = 0 ; i < n ; ++i) {while (m>1 && dcmp(Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2]))<=0) --m;ch[m++] = p[i];}int k = m;for (int i = n-2 ; i >= 0 ; --i) {while (m>k && dcmp(Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2]))<=0) --m;ch[m++] = p[i];}if (n > 1) --m;return m;}//半平面交 int HalfplaneIntersection(Line* L,int n,Point* poly){sort(L,L+n);     //按极角排序int first , last;    //双端队列的第一个元素和最后一个元素的下表Point *p = new Point[n];    //p[i]为q[i]和q[i+1]的交点Line *q = new Line[n];   //双端队列q[first=last=0] = L[0];  //双端队列初始化为只有一个半平面L[0]for (int i = 1 ; i < n ; ++i) {while (first < last && !OnLeft(L[i],p[last-1])) --last;while (first < last && !OnLeft(L[i],p[first])) ++first;q[++last] = L[i];if (dcmp(Cross(q[last].v,q[last-1].v))==0 ) {//两向量平行且同向，取内侧的一个--last;if (OnLeft(q[last],L[i].P)) q[last] = L[i];}if (first < last) p[last-1] = GetIntersection(q[last-1],q[last]);}while (first < last && !OnLeft(q[first],p[last-1])) --last;//删除无用平面（*）if (last - first <=1 ) { delete [] p; delete [] q; return 0; }                   //空集(**)p[last] = GetIntersection(q[last],q[first]);//计算首尾两个半平面的交点//从deque复制到输出中int m = 0;for (int i = first ; i <= last ; ++i) poly[m++] = p[i];delete [] p; delete[] q;return m ;}bool OnLine(Point p,Line L){Vector v = p-L.P;return dcmp(Cross(v,L.v))==0;}//旋转卡壳：pair<double,double> RotateCalipers(Point* hull,int n){hull[n] = hull[0];Line L1, L2, L3, L4;int q=0, w=0, e=0, r=0;for (int i = 0 ; i < n ; ++i) {if (dcmp(hull[i].y-hull[q].y)<0) q = i;if (dcmp(hull[i].x-hull[w].x)<0) w = i;if (dcmp(hull[i].y-hull[e].y)>0) e = i;if (dcmp(hull[i].x-hull[r].x)>0) r = i;}L1 = Line(hull[q],Vector(1,0));L2 = Line(hull[w],Vector(0,-1));L3 = Line(hull[e],Vector(-1,0));L4 = Line(hull[r],Vector(0,1));double sum = 0;double ansA = inf , ansC = inf;while (dcmp(sum-PI/2)<0) {double ang = Angle(hull[q+1]-hull[q],L1.v);ang = min(ang,Angle(hull[w+1]-hull[w],L2.v));ang = min(ang,Angle(hull[e+1]-hull[e],L3.v));ang = min(ang,Angle(hull[r+1]-hull[r],L4.v));sum += ang;L1.v = Rotate(L1.v,ang);L2.v = Rotate(L2.v,ang);L3.v = Rotate(L3.v,ang);L4.v = Rotate(L4.v,ang);if (dcmp(ang-Angle(hull[q+1]-hull[q],L1.v))==0) { L1.P = hull[q+1]; q = (q+1)%n; }if (dcmp(ang-Angle(hull[w+1]-hull[w],L2.v))==0) { L2.P = hull[w+1]; w = (w+1)%n; }if (dcmp(ang-Angle(hull[e+1]-hull[e],L3.v))==0) { L3.P = hull[e+1]; e = (e+1)%n; }if (dcmp(ang-Angle(hull[r+1]-hull[r],L4.v))==0) { L4.P = hull[r+1]; r = (r+1)%n; }double A = DistanceToLine(L1.P,L3.P,L3.P+L3.v*Length(L1.P-L3.P));double B = DistanceToLine(L2.P,L4.P,L4.P+L4.v*Length(L2.P-L4.P));ansA = min(ansA,A*B);ansC = min(ansC,A+B);}ansC *= 2;return mp(ansA,ansC);}//-----------------------------------------------------------------------const int maxn = 100000+5;Point p[maxn] , hull[maxn];int n , r , m;int main(){while (scanf("%d",&n)==1) {for (int i = 0 ; i  < n ; ++i) scanf("%lf%lf",&p[i].x,&p[i].y);m = ConvexHull(p,n,hull);scanf("%d",&r);while (r--) {Point a; scanf("%lf%lf",&a.x,&a.y);if (isPointInPolygon(a,hull,m)==0) printf("outside\n");else printf("inside\n");}}}