题目链接：hdu 4454 Stealing a Cake

题目大意：给定一个起始点s，一个圆形。一个矩形。如今从起点開始，移动到圆形再移动到矩形。求最短距离。

解题思路：在圆周上三分就可以。即对角度[0,2*pi]三分。计算点和矩形的距离能够选点和矩形四条边的距离最短值。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>using namespace std;
const double eps = 1e-9;
const double pi = 4 * atan(1.0);struct point {double x, y;point(double x = 0, double y = 0) {this->x = x;this->y = y;}
}s, o, p[4];double R;inline double distant(point u, point v) {double x = u.x - v.x;double y = u.y - v.y;return sqrt(x*x + y*y);
}inline double handle(point u, point l, point r) {if (fabs(l.x - r.x) < eps) {double a = min(l.y, r.y), b = max(l.y, r.y);if (a <= u.y && u.y <= b)return fabs(u.x - l.x);elsereturn min(distant(u, l), distant(u, r));} else {double a = min(l.x, r.x), b = max(l.x, r.x);if (a <= u.x && u.x <= b)return fabs(u.y - l.y);elsereturn min(distant(u, l), distant(u, r));}
}inline double f(double k) {point u(o.x + R * cos(k), o.y + R * sin(k));double ret = handle(u, p[0], p[3]);for (int i = 0; i < 3; i++)ret = min(ret, handle(u, p[i], p[i+1]));return distant(s, u) + ret;
}double solve (double l, double r) {for (int i = 0; i < 100; i++) {double x1 = l + (r-l) / 3;double x2 = r - (r-l) / 3;if (f(x1) < f(x2))r = x2;elsel = x1;}return f(l);
}void init () {double a, b, c, d;scanf("%lf%lf%lf", &o.x, &o.y, &R);scanf("%lf%lf%lf%lf", &a, &b, &c, &d);p[0].x = min(a, c); p[0].y = min(b, d);p[1].x = min(a, c); p[1].y = max(b, d);p[2].x = max(a, c); p[2].y = max(b, d);p[3].x = max(a, c); p[3].y = min(b, d);/*for (int i = 0; i < 4; i++)printf("%lf %lf\n", p[i].x, p[i].y);*/
}int main () {while (scanf("%lf%lf", &s.x, &s.y) == 2 && (fabs(s.x) > eps || fabs(s.y) > eps)) {init();printf("%.2lf\n", solve(0, 2 * pi));}return 0;
}