// `计算几何模板` const double eps = 1e-8; const double inf = 1e20; const double pi = acos(-1.0); const int maxp = 1010; //`Compares a double to zero` int sgn(double x){ if(fabs(x) < eps)return 0; if(x < 0)return -1; else return 1; } //square of a double inline double sqr(double x){return x*x;} /* * Point * Point() - Empty constructor * Point(double _x,double _y) - constructor * input() - double input * output() - %.2f output * operator == - compares x and y * operator < - compares first by x, then by y * operator - - return new Point after subtracting curresponging x and y * operator ^ - cross product of 2d points * operator * - dot product * len() - gives length from origin * len2() - gives square of length from origin * distance(Point p) - gives distance from p * operator + Point b - returns new Point after adding curresponging x and y * operator * double k - returns new Point after multiplieing x and y by k * operator / double k - returns new Point after divideing x and y by k * rad(Point a,Point b)- returns the angle of Point a and Point b from this Point * trunc(double r) - return Point that if truncated the distance from center to r * rotleft() - returns 90 degree ccw rotated point * rotright() - returns 90 degree cw rotated point * rotate(Point p,double angle) - returns Point after rotateing the Point centering at p by angle radian ccw */ struct Point{ double x,y; Point(){} Point(double _x,double _y){ x = _x; y = _y; } void input(){ scanf("%lf%lf",&x,&y); } void output(){ printf("%.2f %.2f ",x,y); } bool operator == (Point b)const{ return sgn(x-b.x) == 0 && sgn(y-b.y) == 0; } bool operator < (Point b)const{ return sgn(x-b.x)== 0?sgn(y-b.y)<0:x<b.x; } Point operator -(const Point &b)const{ return Point(x-b.x,y-b.y); } //叉积 double operator ^(const Point &b)const{ return x*b.y - y*b.x; } //点积 double operator *(const Point &b)const{ return x*b.x + y*b.y; } //返回长度 double len(){ return hypot(x,y);//库函数 } //返回长度的平方 double len2(){ return x*x + y*y; } //返回两点的距离 double distance(Point p){ return hypot(x-p.x,y-p.y); } Point operator +(const Point &b)const{ return Point(x+b.x,y+b.y); } Point operator *(const double &k)const{ return Point(x*k,y*k); } Point operator /(const double &k)const{ return Point(x/k,y/k); } //`计算pa 和 pb 的夹角` //`就是求这个点看a,b 所成的夹角` //`测试 LightOJ1203` double rad(Point a,Point b){ Point p = *this; return fabs(atan2( fabs((a-p)^(b-p)),(a-p)*(b-p) )); } //`化为长度为r的向量` Point trunc(double r){ double l = len(); if(!sgn(l))return *this; r /= l; return Point(x*r,y*r); } //`逆时针旋转90度` Point rotleft(){ return Point(-y,x); } //`顺时针旋转90度` Point rotright(){ return Point(y,-x); } //`绕着p点逆时针旋转angle` Point rotate(Point p,double angle){ Point v = (*this) - p; double c = cos(angle), s = sin(angle); return Point(p.x + v.x*c - v.y*s,p.y + v.x*s + v.y*c); } }; /* * Stores two points * Line() - Empty constructor * Line(Point _s,Point _e) - Line through _s and _e * operator == - checks if two points are same * Line(Point p,double angle) - one end p , another end at angle degree * Line(double a,double b,double c) - Line of equation ax + by + c = 0 * input() - inputs s and e * adjust() - orders in such a way that s < e * length() - distance of se * angle() - return 0 <= angle < pi * relation(Point p) - 3 if point is on line * 1 if point on the left of line * 2 if point on the right of line * pointonseg(double p) - return true if point on segment * parallel(Line v) - return true if they are parallel * segcrossseg(Line v) - returns 0 if does not intersect * returns 1 if non-standard intersection * returns 2 if intersects * linecrossseg(Line v) - line and seg * linecrossline(Line v) - 0 if parallel * 1 if coincides * 2 if intersects * crosspoint(Line v) - returns intersection point * dispointtoline(Point p) - distance from point p to the line * dispointtoseg(Point p) - distance from p to the segment * dissegtoseg(Line v) - distance of two segment * lineprog(Point p) - returns projected point p on se line * symmetrypoint(Point p) - returns reflection point of p over se * */ struct Line{ Point s,e; Line(){} Line(Point _s,Point _e){ s = _s; e = _e; } bool operator ==(Line v){ return (s == v.s)&&(e == v.e); } //`根据一个点和倾斜角angle确定直线,0<=angle<pi` Line(Point p,double angle){ s = p; if(sgn(angle-pi/2) == 0){ e = (s + Point(0,1)); } else{ e = (s + Point(1,tan(angle))); } } //ax+by+c=0 Line(double a,double b,double c){ if(sgn(a) == 0){ s = Point(0,-c/b); e = Point(1,-c/b); } else if(sgn(b) == 0){ s = Point(-c/a,0); e = Point(-c/a,1); } else{ s = Point(0,-c/b); e = Point(1,(-c-a)/b); } } void input(){ s.input(); e.input(); } void adjust(){ if(e < s)swap(s,e); } //求线段长度 double length(){ return s.distance(e); } //`返回直线倾斜角 0<=angle<pi` double angle(){ double k = atan2(e.y-s.y,e.x-s.x); if(sgn(k) < 0)k += pi; if(sgn(k-pi) == 0)k -= pi; return k; } //`点和直线关系` //`1 在左侧` //`2 在右侧` //`3 在直线上` int relation(Point p){ int c = sgn((p-s)^(e-s)); if(c < 0)return 1; else if(c > 0)return 2; else return 3; } // 点在线段上的判断 bool pointonseg(Point p){ return sgn((p-s)^(e-s)) == 0 && sgn((p-s)*(p-e)) <= 0; } //`两向量平行(对应直线平行或重合)` bool parallel(Line v){ return sgn((e-s)^(v.e-v.s)) == 0; } //`两线段相交判断` //`2 规范相交` //`1 非规范相交` //`0 不相交` int segcrossseg(Line v){ int d1 = sgn((e-s)^(v.s-s)); int d2 = sgn((e-s)^(v.e-s)); int d3 = sgn((v.e-v.s)^(s-v.s)); int d4 = sgn((v.e-v.s)^(e-v.s)); if( (d1^d2)==-2 && (d3^d4)==-2 )return 2; return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) || (d2==0 && sgn((v.e-s)*(v.e-e))<=0) || (d3==0 && sgn((s-v.s)*(s-v.e))<=0) || (d4==0 && sgn((e-v.s)*(e-v.e))<=0); } //`直线和线段相交判断` //`-*this line -v seg` //`2 规范相交` //`1 非规范相交` //`0 不相交` int linecrossseg(Line v){ int d1 = sgn((e-s)^(v.s-s)); int d2 = sgn((e-s)^(v.e-s)); if((d1^d2)==-2) return 2; return (d1==0||d2==0); } //`两直线关系` //`0 平行` //`1 重合` //`2 相交` int linecrossline(Line v){ if((*this).parallel(v)) return v.relation(s)==3; return 2; } //`求两直线的交点` //`要保证两直线不平行或重合` Point crosspoint(Line v){ double a1 = (v.e-v.s)^(s-v.s); double a2 = (v.e-v.s)^(e-v.s); return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1)); } //点到直线的距离 double dispointtoline(Point p){ return fabs((p-s)^(e-s))/length(); } //点到线段的距离 double dispointtoseg(Point p){ if(sgn((p-s)*(e-s))<0 || sgn((p-e)*(s-e))<0) return min(p.distance(s),p.distance(e)); return dispointtoline(p); } //`返回线段到线段的距离` //`前提是两线段不相交，相交距离就是0了` double dissegtoseg(Line v){ return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e))); } //`返回点p在直线上的投影` Point lineprog(Point p){ return s + ( ((e-s)*((e-s)*(p-s)))/((e-s).len2()) ); } //`返回点p关于直线的对称点` Point symmetrypoint(Point p){ Point q = lineprog(p); return Point(2*q.x-p.x,2*q.y-p.y); } }; //圆 struct circle{ Point p;//圆心 double r;//半径 circle(){} circle(Point _p,double _r){ p = _p; r = _r; } circle(double x,double y,double _r){ p = Point(x,y); r = _r; } //`三角形的外接圆` //`需要Point的+ / rotate() 以及Line的crosspoint()` //`利用两条边的中垂线得到圆心` //`测试：UVA12304` circle(Point a,Point b,Point c){ Line u = Line((a+b)/2,((a+b)/2)+((b-a).rotleft())); Line v = Line((b+c)/2,((b+c)/2)+((c-b).rotleft())); p = u.crosspoint(v); r = p.distance(a); } //`三角形的内切圆` //`参数bool t没有作用，只是为了和上面外接圆函数区别` //`测试：UVA12304` circle(Point a,Point b,Point c,bool t){ Line u,v; double m = atan2(b.y-a.y,b.x-a.x), n = atan2(c.y-a.y,c.x-a.x); u.s = a; u.e = u.s + Point(cos((n+m)/2),sin((n+m)/2)); v.s = b; m = atan2(a.y-b.y,a.x-b.x) , n = atan2(c.y-b.y,c.x-b.x); v.e = v.s + Point(cos((n+m)/2),sin((n+m)/2)); p = u.crosspoint(v); r = Line(a,b).dispointtoseg(p); } //输入 void input(){ p.input(); scanf("%lf",&r); } //输出 void output(){ printf("%.2lf %.2lf %.2lf ",p.x,p.y,r); } bool operator == (circle v){ return (p==v.p) && sgn(r-v.r)==0; } bool operator < (circle v)const{ return ((p<v.p)||((p==v.p)&&sgn(r-v.r)<0)); } //面积 double area(){ return pi*r*r; } //周长 double circumference(){ return 2*pi*r; } //`点和圆的关系` //`0 圆外` //`1 圆上` //`2 圆内` int relation(Point b){ double dst = b.distance(p); if(sgn(dst-r) < 0)return 2; else if(sgn(dst-r)==0)return 1; return 0; } //`线段和圆的关系` //`比较的是圆心到线段的距离和半径的关系` int relationseg(Line v){ double dst = v.dispointtoseg(p); if(sgn(dst-r) < 0)return 2; else if(sgn(dst-r) == 0)return 1; return 0; } //`直线和圆的关系` //`比较的是圆心到直线的距离和半径的关系` int relationline(Line v){ double dst = v.dispointtoline(p); if(sgn(dst-r) < 0)return 2; else if(sgn(dst-r) == 0)return 1; return 0; } //`两圆的关系` //`5 相离` //`4 外切` //`3 相交` //`2 内切` //`1 内含` //`需要Point的distance` //`测试：UVA12304` int relationcircle(circle v){ double d = p.distance(v.p); if(sgn(d-r-v.r) > 0)return 5; if(sgn(d-r-v.r) == 0)return 4; double l = fabs(r-v.r); if(sgn(d-r-v.r)<0 && sgn(d-l)>0)return 3; if(sgn(d-l)==0)return 2; if(sgn(d-l)<0)return 1; } //`求两个圆的交点，返回0表示没有交点，返回1是一个交点，2是两个交点` //`需要relationcircle` //`测试：UVA12304` int pointcrosscircle(circle v,Point &p1,Point &p2){ int rel = relationcircle(v); if(rel == 1 || rel == 5)return 0; double d = p.distance(v.p); double l = (d*d+r*r-v.r*v.r)/(2*d); double h = sqrt(r*r-l*l); Point tmp = p + (v.p-p).trunc(l); p1 = tmp + ((v.p-p).rotleft().trunc(h)); p2 = tmp + ((v.p-p).rotright().trunc(h)); if(rel == 2 || rel == 4) return 1; return 2; } //`求直线和圆的交点，返回交点个数` int pointcrossline(Line v,Point &p1,Point &p2){ if(!(*this).relationline(v))return 0; Point a = v.lineprog(p); double d = v.dispointtoline(p); d = sqrt(r*r-d*d); if(sgn(d) == 0){ p1 = a; p2 = a; return 1; } p1 = a + (v.e-v.s).trunc(d); p2 = a - (v.e-v.s).trunc(d); return 2; } //`得到过a,b两点，半径为r1的两个圆` int gercircle(Point a,Point b,double r1,circle &c1,circle &c2){ circle x(a,r1),y(b,r1); int t = x.pointcrosscircle(y,c1.p,c2.p); if(!t)return 0; c1.r = c2.r = r; return t; } //`得到与直线u相切，过点q,半径为r1的圆` //`测试：UVA12304` int getcircle(Line u,Point q,double r1,circle &c1,circle &c2){ double dis = u.dispointtoline(q); if(sgn(dis-r1*2)>0)return 0; if(sgn(dis) == 0){ c1.p = q + ((u.e-u.s).rotleft().trunc(r1)); c2.p = q + ((u.e-u.s).rotright().trunc(r1)); c1.r = c2.r = r1; return 2; } Line u1 = Line((u.s + (u.e-u.s).rotleft().trunc(r1)),(u.e + (u.e-u.s).rotleft().trunc(r1))); Line u2 = Line((u.s + (u.e-u.s).rotright().trunc(r1)),(u.e + (u.e-u.s).rotright().trunc(r1))); circle cc = circle(q,r1); Point p1,p2; if(!cc.pointcrossline(u1,p1,p2))cc.pointcrossline(u2,p1,p2); c1 = circle(p1,r1); if(p1 == p2){ c2 = c1; return 1; } c2 = circle(p2,r1); return 2; } //`同时与直线u,v相切，半径为r1的圆` //`测试：UVA12304` int getcircle(Line u,Line v,double r1,circle &c1,circle &c2,circle &c3,circle &c4){ if(u.parallel(v))return 0;//两直线平行 Line u1 = Line(u.s + (u.e-u.s).rotleft().trunc(r1),u.e + (u.e-u.s).rotleft().trunc(r1)); Line u2 = Line(u.s + (u.e-u.s).rotright().trunc(r1),u.e + (u.e-u.s).rotright().trunc(r1)); Line v1 = Line(v.s + (v.e-v.s).rotleft().trunc(r1),v.e + (v.e-v.s).rotleft().trunc(r1)); Line v2 = Line(v.s + (v.e-v.s).rotright().trunc(r1),v.e + (v.e-v.s).rotright().trunc(r1)); c1.r = c2.r = c3.r = c4.r = r1; c1.p = u1.crosspoint(v1); c2.p = u1.crosspoint(v2); c3.p = u2.crosspoint(v1); c4.p = u2.crosspoint(v2); return 4; } //`同时与不相交圆cx,cy相切，半径为r1的圆` //`测试：UVA12304` int getcircle(circle cx,circle cy,double r1,circle &c1,circle &c2){ circle x(cx.p,r1+cx.r),y(cy.p,r1+cy.r); int t = x.pointcrosscircle(y,c1.p,c2.p); if(!t)return 0; c1.r = c2.r = r1; return t; } //`过一点作圆的切线(先判断点和圆的关系)` //`测试：UVA12304` int tangentline(Point q,Line &u,Line &v){ int x = relation(q); if(x == 2)return 0; if(x == 1){ u = Line(q,q + (q-p).rotleft()); v = u; return 1; } double d = p.distance(q); double l = r*r/d; double h = sqrt(r*r-l*l); u = Line(q,p + ((q-p).trunc(l) + (q-p).rotleft().trunc(h))); v = Line(q,p + ((q-p).trunc(l) + (q-p).rotright().trunc(h))); return 2; } //`求两圆相交的面积` double areacircle(circle v){ int rel = relationcircle(v); if(rel >= 4)return 0.0; if(rel <= 2)return min(area(),v.area()); double d = p.distance(v.p); double hf = (r+v.r+d)/2.0; double ss = 2*sqrt(hf*(hf-r)*(hf-v.r)*(hf-d)); double a1 = acos((r*r+d*d-v.r*v.r)/(2.0*r*d)); a1 = a1*r*r; double a2 = acos((v.r*v.r+d*d-r*r)/(2.0*v.r*d)); a2 = a2*v.r*v.r; return a1+a2-ss; } //`求圆和三角形pab的相交面积` //`测试：POJ3675 HDU3982 HDU2892` double areatriangle(Point a,Point b){ if(sgn((p-a)^(p-b)) == 0)return 0.0; Point q[5]; int len = 0; q[len++] = a; Line l(a,b); Point p1,p2; if(pointcrossline(l,q[1],q[2])==2){ if(sgn((a-q[1])*(b-q[1]))<0)q[len++] = q[1]; if(sgn((a-q[2])*(b-q[2]))<0)q[len++] = q[2]; } q[len++] = b; if(len == 4 && sgn((q[0]-q[1])*(q[2]-q[1]))>0)swap(q[1],q[2]); double res = 0; for(int i = 0;i < len-1;i++){ if(relation(q[i])==0||relation(q[i+1])==0){ double arg = p.rad(q[i],q[i+1]); res += r*r*arg/2.0; } else{ res += fabs((q[i]-p)^(q[i+1]-p))/2.0; } } return res; } }; /* * n,p Line l for each side * input(int _n) - inputs _n size polygon * add(Point q) - adds a point at end of the list * getline() - populates line array * cmp - comparision in convex_hull order * norm() - sorting in convex_hull order * getconvex(polygon &convex) - returns convex hull in convex * Graham(polygon &convex) - returns convex hull in convex * isconvex() - checks if convex * relationpoint(Point q) - returns 3 if q is a vertex * 2 if on a side * 1 if inside * 0 if outside * convexcut(Line u,polygon &po) - left side of u in po * gercircumference() - returns side length * getarea() - returns area * getdir() - returns 0 for cw, 1 for ccw * getbarycentre() - returns barycenter * */ struct polygon{ int n; Point p[maxp]; Line l[maxp]; void input(int _n){ n = _n; for(int i = 0;i < n;i++) p[i].input(); } void add(Point q){ p[n++] = q; } void getline(){ for(int i = 0;i < n;i++){ l[i] = Line(p[i],p[(i+1)%n]); } } struct cmp{ Point p; cmp(const Point &p0){p = p0;} bool operator()(const Point &aa,const Point &bb){ Point a = aa, b = bb; int d = sgn((a-p)^(b-p)); if(d == 0){ return sgn(a.distance(p)-b.distance(p)) < 0; } return d > 0; } }; //`进行极角排序` //`首先需要找到最左下角的点` //`需要重载号好Point的 < 操作符(min函数要用) ` void norm(){ Point mi = p[0]; for(int i = 1;i < n;i++)mi = min(mi,p[i]); sort(p,p+n,cmp(mi)); } //`得到凸包` //`得到的凸包里面的点编号是0$sim$n-1的` //`两种凸包的方法` //`注意如果有影响，要特判下所有点共点，或者共线的特殊情况` //`测试 LightOJ1203 LightOJ1239` void getconvex(polygon &convex){ sort(p,p+n); convex.n = n; for(int i = 0;i < min(n,2);i++){ convex.p[i] = p[i]; } if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判 if(n <= 2)return; int &top = convex.n; top = 1; for(int i = 2;i < n;i++){ while(top && sgn((convex.p[top]-p[i])^(convex.p[top-1]-p[i])) <= 0) top--; convex.p[++top] = p[i]; } int temp = top; convex.p[++top] = p[n-2]; for(int i = n-3;i >= 0;i--){ while(top != temp && sgn((convex.p[top]-p[i])^(convex.p[top-1]-p[i])) <= 0) top--; convex.p[++top] = p[i]; } if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判 convex.norm();//`原来得到的是顺时针的点，排序后逆时针` } //`得到凸包的另外一种方法` //`测试 LightOJ1203 LightOJ1239` void Graham(polygon &convex){ norm(); int &top = convex.n; top = 0; if(n == 1){ top = 1; convex.p[0] = p[0]; return; } if(n == 2){ top = 2; convex.p[0] = p[0]; convex.p[1] = p[1]; if(convex.p[0] == convex.p[1])top--; return; } convex.p[0] = p[0]; convex.p[1] = p[1]; top = 2; for(int i = 2;i < n;i++){ while( top > 1 && sgn((convex.p[top-1]-convex.p[top-2])^(p[i]-convex.p[top-2])) <= 0 ) top--; convex.p[top++] = p[i]; } if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判 } //`判断是不是凸的` bool isconvex(){ bool s[2]; memset(s,false,sizeof(s)); for(int i = 0;i < n;i++){ int j = (i+1)%n; int k = (j+1)%n; s[sgn((p[j]-p[i])^(p[k]-p[i]))+1] = true; if(s[0] && s[2])return false; } return true; } //`判断点和任意多边形的关系` //` 3 点上` //` 2 边上` //` 1 内部` //` 0 外部` int relationpoint(Point q){ for(int i = 0;i < n;i++){ if(p[i] == q)return 3; } getline(); for(int i = 0;i < n;i++){ if(l[i].pointonseg(q))return 2; } int cnt = 0; for(int i = 0;i < n;i++){ int j = (i+1)%n; int k = sgn((q-p[j])^(p[i]-p[j])); int u = sgn(p[i].y-q.y); int v = sgn(p[j].y-q.y); if(k > 0 && u < 0 && v >= 0)cnt++; if(k < 0 && v < 0 && u >= 0)cnt--; } return cnt != 0; } //`直线u切割凸多边形左侧` //`注意直线方向` //`测试：HDU3982` void convexcut(Line u,polygon &po){ int &top = po.n;//注意引用 top = 0; for(int i = 0;i < n;i++){ int d1 = sgn((u.e-u.s)^(p[i]-u.s)); int d2 = sgn((u.e-u.s)^(p[(i+1)%n]-u.s)); if(d1 >= 0)po.p[top++] = p[i]; if(d1*d2 < 0)po.p[top++] = u.crosspoint(Line(p[i],p[(i+1)%n])); } } //`得到周长` //`测试 LightOJ1239` double getcircumference(){ double sum = 0; for(int i = 0;i < n;i++){ sum += p[i].distance(p[(i+1)%n]); } return sum; } //`得到面积` double getarea(){ double sum = 0; for(int i = 0;i < n;i++){ sum += (p[i]^p[(i+1)%n]); } return fabs(sum)/2; } //`得到方向` //` 1 表示逆时针，0表示顺时针` bool getdir(){ double sum = 0; for(int i = 0;i < n;i++) sum += (p[i]^p[(i+1)%n]); if(sgn(sum) > 0)return 1; return 0; } //`得到重心` Point getbarycentre(){ Point ret(0,0); double area = 0; for(int i = 1;i < n-1;i++){ double tmp = (p[i]-p[0])^(p[i+1]-p[0]); if(sgn(tmp) == 0)continue; area += tmp; ret.x += (p[0].x+p[i].x+p[i+1].x)/3*tmp; ret.y += (p[0].y+p[i].y+p[i+1].y)/3*tmp; } if(sgn(area)) ret = ret/area; return ret; } //`多边形和圆交的面积` //`测试：POJ3675 HDU3982 HDU2892` double areacircle(circle c){ double ans = 0; for(int i = 0;i < n;i++){ int j = (i+1)%n; if(sgn( (p[j]-c.p)^(p[i]-c.p) ) >= 0) ans += c.areatriangle(p[i],p[j]); else ans -= c.areatriangle(p[i],p[j]); } return fabs(ans); } //`多边形和圆关系` //` 2 圆完全在多边形内` //` 1 圆在多边形里面，碰到了多边形边界` //` 0 其它` int relationcircle(circle c){ getline(); int x = 2; if(relationpoint(c.p) != 1)return 0;//圆心不在内部 for(int i = 0;i < n;i++){ if(c.relationseg(l[i])==2)return 0; if(c.relationseg(l[i])==1)x = 1; } return x; } }; //`AB X AC` double cross(Point A,Point B,Point C){ return (B-A)^(C-A); } //`AB*AC` double dot(Point A,Point B,Point C){ return (B-A)*(C-A); } //`最小矩形面积覆盖` //` A 必须是凸包(而且是逆时针顺序)` //` 测试 UVA 10173` double minRectangleCover(polygon A){ //`要特判A.n < 3的情况` if(A.n < 3)return 0.0; A.p[A.n] = A.p[0]; double ans = -1; int r = 1, p = 1, q; for(int i = 0;i < A.n;i++){ //`卡出离边A.p[i] - A.p[i+1]最远的点` while( sgn( cross(A.p[i],A.p[i+1],A.p[r+1]) - cross(A.p[i],A.p[i+1],A.p[r]) ) >= 0 ) r = (r+1)%A.n; //`卡出A.p[i] - A.p[i+1]方向上正向n最远的点` while(sgn( dot(A.p[i],A.p[i+1],A.p[p+1]) - dot(A.p[i],A.p[i+1],A.p[p]) ) >= 0 ) p = (p+1)%A.n; if(i == 0)q = p; //`卡出A.p[i] - A.p[i+1]方向上负向最远的点` while(sgn(dot(A.p[i],A.p[i+1],A.p[q+1]) - dot(A.p[i],A.p[i+1],A.p[q])) <= 0) q = (q+1)%A.n; double d = (A.p[i] - A.p[i+1]).len2(); double tmp = cross(A.p[i],A.p[i+1],A.p[r]) * (dot(A.p[i],A.p[i+1],A.p[p]) - dot(A.p[i],A.p[i+1],A.p[q]))/d; if(ans < 0 || ans > tmp)ans = tmp; } return ans; } //`直线切凸多边形` //`多边形是逆时针的，在q1q2的左侧` //`测试:HDU3982` vector<Point> convexCut(const vector<Point> &ps,Point q1,Point q2){ vector<Point>qs; int n = ps.size(); for(int i = 0;i < n;i++){ Point p1 = ps[i], p2 = ps[(i+1)%n]; int d1 = sgn((q2-q1)^(p1-q1)), d2 = sgn((q2-q1)^(p2-q1)); if(d1 >= 0) qs.push_back(p1); if(d1 * d2 < 0) qs.push_back(Line(p1,p2).crosspoint(Line(q1,q2))); } return qs; } //`半平面交` //`测试 POJ3335 POJ1474 POJ1279` //*************************** struct halfplane:public Line{ double angle; halfplane(){} //`表示向量s->e逆时针(左侧)的半平面` halfplane(Point _s,Point _e){ s = _s; e = _e; } halfplane(Line v){ s = v.s; e = v.e; } void calcangle(){ angle = atan2(e.y-s.y,e.x-s.x); } bool operator <(const halfplane &b)const{ return angle < b.angle; } }; struct halfplanes{ int n; halfplane hp[maxp]; Point p[maxp]; int que[maxp]; int st,ed; void push(halfplane tmp){ hp[n++] = tmp; } //去重 void unique(){ int m = 1; for(int i = 1;i < n;i++){ if(sgn(hp[i].angle-hp[i-1].angle) != 0) hp[m++] = hp[i]; else if(sgn( (hp[m-1].e-hp[m-1].s)^(hp[i].s-hp[m-1].s) ) > 0) hp[m-1] = hp[i]; } n = m; } bool halfplaneinsert(){ for(int i = 0;i < n;i++)hp[i].calcangle(); sort(hp,hp+n); unique(); que[st=0] = 0; que[ed=1] = 1; p[1] = hp[0].crosspoint(hp[1]); for(int i = 2;i < n;i++){ while(st<ed && sgn((hp[i].e-hp[i].s)^(p[ed]-hp[i].s))<0)ed--; while(st<ed && sgn((hp[i].e-hp[i].s)^(p[st+1]-hp[i].s))<0)st++; que[++ed] = i; if(hp[i].parallel(hp[que[ed-1]]))return false; p[ed]=hp[i].crosspoint(hp[que[ed-1]]); } while(st<ed && sgn((hp[que[st]].e-hp[que[st]].s)^(p[ed]-hp[que[st]].s))<0)ed--; while(st<ed && sgn((hp[que[ed]].e-hp[que[ed]].s)^(p[st+1]-hp[que[ed]].s))<0)st++; if(st+1>=ed)return false; return true; } //`得到最后半平面交得到的凸多边形` //`需要先调用halfplaneinsert() 且返回true` void getconvex(polygon &con){ p[st] = hp[que[st]].crosspoint(hp[que[ed]]); con.n = ed-st+1; for(int j = st,i = 0;j <= ed;i++,j++) con.p[i] = p[j]; } }; //*************************** const int maxn = 1010; struct circles{ circle c[maxn]; double ans[maxn];//`ans[i]表示被覆盖了i次的面积` double pre[maxn]; int n; circles(){} void add(circle cc){ c[n++] = cc; } //`x包含在y中` bool inner(circle x,circle y){ if(x.relationcircle(y) != 1)return 0; return sgn(x.r-y.r)<=0?1:0; } //圆的面积并去掉内含的圆 void init_or(){ bool mark[maxn] = {0}; int i,j,k=0; for(i = 0;i < n;i++){ for(j = 0;j < n;j++) if(i != j && !mark[j]){ if( (c[i]==c[j])||inner(c[i],c[j]) )break; } if(j < n)mark[i] = 1; } for(i = 0;i < n;i++) if(!mark[i]) c[k++] = c[i]; n = k; } //`圆的面积交去掉内含的圆` void init_add(){ int i,j,k; bool mark[maxn] = {0}; for(i = 0;i < n;i++){ for(j = 0;j < n;j++) if(i != j && !mark[j]){ if( (c[i]==c[j])||inner(c[j],c[i]) )break; } if(j < n)mark[i] = 1; } for(i = 0;i < n;i++) if(!mark[i]) c[k++] = c[i]; n = k; } //`半径为r的圆，弧度为th对应的弓形的面积` double areaarc(double th,double r){ return 0.5*r*r*(th-sin(th)); } //`测试SPOJVCIRCLES SPOJCIRUT` //`SPOJVCIRCLES求n个圆并的面积，需要加上init\_or()去掉重复圆（否则WA）` //`SPOJCIRUT 是求被覆盖k次的面积，不能加init\_or()` //`对于求覆盖多少次面积的问题，不能解决相同圆，而且不能init\_or()` //`求多圆面积并，需要init\_or,其中一个目的就是去掉相同圆` void getarea(){ memset(ans,0,sizeof(ans)); vector<pair<double,int> >v; for(int i = 0;i < n;i++){ v.clear(); v.push_back(make_pair(-pi,1)); v.push_back(make_pair(pi,-1)); for(int j = 0;j < n;j++) if(i != j){ Point q = (c[j].p - c[i].p); double ab = q.len(),ac = c[i].r, bc = c[j].r; if(sgn(ab+ac-bc)<=0){ v.push_back(make_pair(-pi,1)); v.push_back(make_pair(pi,-1)); continue; } if(sgn(ab+bc-ac)<=0)continue; if(sgn(ab-ac-bc)>0)continue; double th = atan2(q.y,q.x), fai = acos((ac*ac+ab*ab-bc*bc)/(2.0*ac*ab)); double a0 = th-fai; if(sgn(a0+pi)<0)a0+=2*pi; double a1 = th+fai; if(sgn(a1-pi)>0)a1-=2*pi; if(sgn(a0-a1)>0){ v.push_back(make_pair(a0,1)); v.push_back(make_pair(pi,-1)); v.push_back(make_pair(-pi,1)); v.push_back(make_pair(a1,-1)); } else{ v.push_back(make_pair(a0,1)); v.push_back(make_pair(a1,-1)); } } sort(v.begin(),v.end()); int cur = 0; for(int j = 0;j < v.size();j++){ if(cur && sgn(v[j].first-pre[cur])){ ans[cur] += areaarc(v[j].first-pre[cur],c[i].r); ans[cur] += 0.5*(Point(c[i].p.x+c[i].r*cos(pre[cur]),c[i].p.y+c[i].r*sin(pre[cur]))^Point(c[i].p.x+c[i].r*cos(v[j].first),c[i].p.y+c[i].r*sin(v[j].first))); } cur += v[j].second; pre[cur] = v[j].first; } } for(int i = 1;i < n;i++) ans[i] -= ans[i+1]; } };