pid=4454" target="_blank" style="">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);

        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;

}