学步园

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。现在从起点开始，移动到圆形再移动到矩形，求最短距离。

解题思路：在圆周上三分即可。即对角度[0,2*pi]三分，计算点和矩形的距离可以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);
        else
            return 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);
        else
            return 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;
        else
            l = 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;
}