题目大意:给出一个点,再给出都处于这个点之下的一些圆,求这个点光源照到这些圆上之后所得到的阴影的并集。
思路:求出每一个圆关于那个点的切线,每一个圆可以处理出来两个切线,这两个切线在x轴上交点的中间部分就是要求的阴影。最后将所有的阴影部分取并输出。
关于求切线,我是利用方向向量解方程做的。应该有更简洁的方法吧。。
CODE:
#include <cmath> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAX 510 #define EPS 1e-8 #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); } Point operator /(double a)const { return Point(x / a,y / a); } void Read() { scanf("%lf%lf",&x,&y); } }start; struct Circle{ Point o; double r; void Read() { o.Read(); scanf("%lf",&r); } }circle; struct Complex{ double x,y; Complex(double _ = .0,double __ = .0):x(_),y(__) {} bool operator <(const Complex &a)const { if(x == a.x) return y < a.y; return x < a.x; } }ans[MAX]; int cnt; 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 Point Solve(double a1,double b1,double c1,double a2,double b2,double c2) { double y = ((a1 * c2) - (a2 * c1)) / ((a1 * b2) - (a2 * b1)); double x = (c1 - b1 * y) / a1; return Point(x,y); } inline Complex Work(const Point &p,const Circle &circle) { double d = Calc(circle.o,p),_d = sqrt(d * d - circle.r * circle.r); Point v = Solve(circle.r,-_d,p.x - circle.o.x,_d,circle.r,p.y - circle.o.y); Point tangency = circle.o + v * circle.r; Point u = tangency - p; Point center = p + ((u / (p.y - tangency.y)) * p.y); double first = center.x; v = Solve(circle.r,_d,p.x - circle.o.x,-_d,circle.r,p.y - circle.o.y); tangency = circle.o + v * circle.r; u = tangency - p; center = p + ((u / (p.y - tangency.y)) * p.y); double second = center.x; return Complex(second,first); } int main() { while(scanf("%d",&cnt),cnt) { start.Read(); for(int i = 1; i <= cnt; ++i) { circle.Read(); ans[i] = Work(start,circle); } sort(ans + 1,ans + cnt + 1); double l = ans[1].x,r = ans[1].y; for(int i = 2; i <= cnt; ++i) { if(ans[i].x > r) { printf("%.2f %.2f\n",l,r); l = ans[i].x,r = ans[i].y; } else r = max(r,ans[i].y); } printf("%.2f %.2f\n\n",l,r); } return 0; }
时间: 2024-10-10 04:06:00