HDU4454(几何+三分)
题目：http://acm.hdu.edu.cn/showproblem.php?pid=4454
题意：给一个点，一个圆和一个矩形，矩形与圆没有重叠部分，求从该点出发经过圆上一点再到矩形边上一点的距离和的最小值。
分析：在区间[0,2*PI]内三分角度即可。
#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;}