好贷网好贷款

(beginer) 半平面交 UVA 1304 Art Gallery

发布时间:2016-12-3 8:18:28 编辑:www.fx114.net 分享查询网我要评论
本篇文章主要介绍了"(beginer) 半平面交 UVA 1304 Art Gallery",主要涉及到(beginer) 半平面交 UVA 1304 Art Gallery方面的内容,对于(beginer) 半平面交 UVA 1304 Art Gallery感兴趣的同学可以参考一下。

The art galleries of the new and very futuristic building of the Center for Balkan Cooperation have the form of polygons (not necessarily convex). When a big exhibition is organized, watching over all of the pictures is a big security concern. Your task is that for a given gallery to write a program which finds the surface of the area of the floor, from which each point on the walls of the gallery is visible. On the first figure a map of a gallery is given in some co-ordinate system. The area wanted is shaded on the second figure. Input  The number of tasks T that your program have to solve will be on the first row of the input file. Input data for each task start with an integer N, 5N1500. Each of the next N rows of the input will contain the co-ordinates of a vertex of the polygon two integers that fit in 16-bit integer type, separated by a single space. Following the row with the co-ordinates of the last vertex for the task comes the line with the number of vertices for the next test and so on. Output  For each test you must write on one line the required surface - a number with exactly two digits after the decimal point (the number should be rounded to the second digit after the decimal point). Sample Input  1 7 0 0 4 4 4 7 9 7 13 -1 8 -6 4 -4 Sample Output  80.00 题意:给出一个多边形,在里面能放一个录像头,录像头能看到任意不被挡住的地方,求出放摄像头后多边形内任意一个地方都能看到的这些所有的放置位置的面积。 思路:求一个半平面交就行了,套个模板就过了。。。怎么想到的呢?其实我们把所有的线延长后你就会发现有些区域是不能看到所有地方的。这个时候会因为延长而联想到半平面交。 代码: #include<iostream> #include<cstdio> #include<string.h> #include<math.h> #include<string> #include<cstring> #include<map> #include<algorithm> #include<vector> using namespace std; #define eps 1e-10 const 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); } bool operator < (Point a,Point b) { return a.x < b.x || (a.x==b.x && a.y > b.y); } int dcmp(double x) { if (fabs(x) < eps) return 0; 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; } 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) { return acos(Dot(A,B)/Length(A)/Length(B)); } 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; } //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; } //-------------------------------------------------------------------------------------------- //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) 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 ; } //----------------------------------------------------------------------- const int maxn = 1500+5; int n; Point p[maxn] , poly[maxn]; Line L[maxn]; void input() { scanf("%d",&n); for (int i = 0 ; i < n ; ++i) scanf("%lf%lf",&p[i].x,&p[i].y); for (int i = 0 ; i < n ; ++i) L[i] = Line(p[(i+1)%n],p[i]-p[(i+1)%n]); } void solve() { int m = HalfplaneIntersection(L,n,poly); printf("%.2lf\n",PolyArea(poly,m)+eps); } int main() { int T; cin>>T; while (T--) { input(); solve(); } }

上一篇:Android内核解读-Android系统的开机启动过程
下一篇:【学习笔记】matlab算法实现贝叶斯判别classify函数

相关文章

相关评论