首页 > 代码库 > POJ 3384 Feng Shui 半平面交
POJ 3384 Feng Shui 半平面交
题目大意:一个人非常信“Feng Shui”,他要在房间里放两个圆形的地毯。这两个地毯之间可以重叠,但是不能折叠,也不能伸到房间的外面。求这两个地毯能够覆盖的最大范围,并输出这两个地毯的圆心。
思路:我们当然希望这两个圆形的地毯离得尽量的远,这样的话两个圆之间的重叠区域就会越小,总的覆盖区域就越大。那我们就先把每一条边向内推进地毯的半径的距离,然后求一次半平面交,这个求出的半平面的交集就是圆心可以取得地方,然后就暴力求出这其中的最远点对就行了。
CODE:
#include <cmath> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAX 110 #define EPS 1e-10 #define DCMP(a) (fabs(a) < EPS) using namespace std; struct Point{ double x,y; Point(double _ = .0,double __ = .0):x(_),y(__) {} Point operator +(const Point &a)const { return Point(x + a.x,y + a.y); } Point operator -(const Point &a)const { return Point(x - a.x,y - a.y); } Point operator *(double a)const { return Point(x * a,y * a); } void Read() { scanf("%lf%lf",&x,&y); } }point[MAX],p[MAX],polygen[MAX]; struct Line{ Point p,v; double alpha; Line(Point _,Point __):p(_),v(__) { alpha = atan2(v.y,v.x); } Line() {} bool operator <(const Line &a)const { return alpha < a.alpha; } }line[MAX],q[MAX]; int points,lines; double adjustment; inline double Cross(const Point &a,const Point &b) { return a.x * b.y - a.y * b.x; } inline double Calc(const Point &a,const Point &b) { return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)); } inline bool OnLeft(const Line &l,const Point &p) { return Cross(l.v,p - l.p) >= 0; } inline void MakeLine(const Point &a,const Point &b) { Point p = a,v = b - a; Point _v(-v.y / Calc(a,b),v.x / Calc(a,b)); p = _v * adjustment + p; line[++lines] = Line(p,v); } inline Point GetIntersection(const Line &a,const Line &b) { Point u = a.p - b.p; double temp = Cross(b.v,u) / Cross(a.v,b.v); return a.p + a.v * temp; } int HalfPlaneIntersection() { int front = 1,tail = 1; q[tail] = line[1]; for(int i = 2;i <= lines; ++i) { while(front < tail && !OnLeft(line[i],p[tail - 1])) --tail; while(front < tail && !OnLeft(line[i],p[front])) ++front; if(DCMP(Cross(line[i].v,q[tail].v))) q[tail] = OnLeft(line[i],q[tail].p) ? q[tail]:line[i]; else q[++tail] = line[i]; if(front < tail) p[tail - 1] = GetIntersection(q[tail],q[tail - 1]); } while(front < tail && !OnLeft(q[front],p[tail - 1])) --tail; p[tail] = GetIntersection(q[tail],q[front]); int re = 0; for(int i = front;i <= tail; ++i) polygen[++re] = p[i]; return re; } pair<Point,Point> GetFarest(int cnt) { double max_length = -1.0; pair<Point,Point> re; for(int i = 1;i <= cnt; ++i) for(int j = i;j <= cnt; ++j) if(Calc(polygen[i],polygen[j]) > max_length) { max_length = Calc(polygen[i],polygen[j]); re.first = polygen[i]; re.second = polygen[j]; } return re; } int main() { cin >> points >> adjustment; for(int i = 1;i <= points; ++i) point[i].Read(); for(int i = points;i > 1; --i) MakeLine(point[i],point[i - 1]); MakeLine(point[1],point[points]); sort(line + 1,line + lines + 1); int cnt = HalfPlaneIntersection(); pair<Point,Point> re = GetFarest(cnt); printf("%.6lf %.6lf %.6lf %.6lf\n",re.first.x,re.first.y,re.second.x,re.second.y); return 0; }
POJ 3384 Feng Shui 半平面交
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。