HDU4454(幾何+三分)
阿新 • • 發佈:2018-12-24
題意:給一個點,一個圓和一個矩形,矩形與圓沒有重疊部分,求從該點出發經過圓上一點再到矩形邊上一點的距離和的最小值。
分析:在區間[0,2*PI]內三分角度即可。
#include <iostream> #include <string.h> #include <stdio.h> #include <iomanip> #include <math.h> using namespace std; const double eps = 1e-9; const double PI = acos(-1.0); struct Point { double x,y; }; struct Line { Point a,b; }; double dist(Point A,Point B) { return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y)); } double cross(Point A,Point B,Point C) { return (B.x-A.x)*(C.y-A.y)-(B.y-A.y)*(C.x-A.x); } double distToLine(Point p,Line s) { Point t = p; t.x += s.a.y - s.b.y; t.y += s.b.x - s.a.x; if(cross(s.a,t,p)*cross(s.b,t,p) > eps) return dist(p,s.a) < dist(p,s.b) ? dist(p,s.a) : dist(p,s.b); return fabs(cross(p,s.a,s.b))/dist(s.a,s.b); } Point O,cir,A,B; Line s[4]; Point p[4]; double r; void Import() { cin>>cir.x>>cir.y>>r; cin>>A.x>>A.y>>B.x>>B.y; if(A.y < B.y) swap(A,B); p[0].x = A.x; p[0].y = B.y; p[1].x = B.x; p[1].y = B.y; p[2].x = B.x; p[2].y = A.y; p[3].x = A.x; p[3].y = A.y; s[0].a = p[0]; s[0].b = p[1]; s[1].a = p[1]; s[1].b = p[2]; s[2].a = p[2]; s[2].b = p[3]; s[3].a = p[3]; s[3].b = p[0]; } double equ(double alpha) { Point tmp; tmp.x = cir.x + r*cos(alpha); tmp.y = cir.y + r*sin(alpha); double d1 = dist(O,tmp); double ans = 99999999; for(int i=0;i<4;i++) ans = min(ans,distToLine(tmp,s[i])); return d1+ans; } double ternarySearch(double l,double r) { while(r-l>eps) { double ll=(2*l+r)/3; double rr=(l+2*r)/3; double ans1=equ(ll); double ans2=equ(rr); if(ans1 > ans2) l=ll; else r=rr; } return l; } void Work() { Import(); cout<<fixed<<setprecision(2)<<equ(ternarySearch(0,2*PI))<<endl; } int main() { while(true) { cin>>O.x>>O.y; if(fabs(O.x)<eps && fabs(O.y)<eps) break; Work(); } return 0; }