HDU-4454 Stealing a Cake

題意： 給定一個點， 圓和矩形。 求這個點到圓和再從圓到矩形的最短距離之和。
分析： 很明顯這個距離是一個凹函式， 我們要求這個極值點， 這裡用到三分， 標準解法， 要注意的是， 需要分為兩個部分[0, pi]和[pi, 2*pi]。是因為他的函式影象是這樣的。

所以我們要分成兩個部分分別求極值， 然後取最小值即可。
程式碼：

#include <bits/stdc++.h>

using namespace std;

const double pi = acos(-1.0);
const double eps =
 
 1e-8;
const double inf = 0x3f3f3f3f;
int sgn(double x)
{
    if (fabs(x) < eps)
        return 0;
    else if (x < 0)
        return -1;
    else
        return 1;
}

struct Point
{
    double x, y;
    Point() {}
    Point(double _x, double _y)
    {
        x = _x;
        y = _y;
    }

    Point operator
 
+(const Point b) const
    {
        return Point(x + b.x, y + b.y);
    }

    Point operator-(const Point b) const
    {
        return Point(x - b.x, y - b.y);
    }

    double operator^(const Point b) const
    {
        return x * b.y - y * b.x;
    }

    double operator*(const Point b) const
 

    {
        return x * b.x + y * b.y;
    }

    double distance(Point p)
    {
        return hypot(x - p.x, y - p.y);
    }
};

struct Line
{
    Point s, e;
    Line() {}
    Line(Point _s, Point _e)
    {
        s = _s;
        e = _e;
    }

    double length()
    {
        return s.distance(e);
    }

    double dispointtoline(Point p)
    {
        return fabs((p - s) ^ (e - s)) / length();
    }
    double dispointtoseg(Point p)
    {
        if (sgn((p - s) * (e - s)) < 0 || sgn((p - e) * (s - e)) < 0)
            return min(p.distance(s), p.distance(e));
        return dispointtoline(p);
    }
};
Point st, c1, t1, t2;
double r;

Point getPoint(double angle)
{
    double xx = c1.x + r * cos(angle);
    double yy = c1.y + r * sin(angle);
    return Point(xx, yy);
}
double getdist(Point p)
{
    Line l1(t1, Point(t1.x, t2.y));
    Line l2(t1, Point(t2.x, t1.y));
    Line l3(t2, Point(t1.x, t2.y));
    Line l4(t2, Point(t2.x, t1.y));
    double res = inf;
    res = min(res, l1.dispointtoseg(p));
    res = min(res, l2.dispointtoseg(p));
    res = min(res, l3.dispointtoseg(p));
    res = min(res, l4.dispointtoseg(p));
    return res;
}
double solve()
{
    double l, r, mid, midmid;
    double ans = inf;
    Point p1, p2;
    double dis1, dis2;
    l = 0, r = pi;
    while (r - l >= eps)
    {
        mid = (l + r) / 2;
        midmid = (mid + r) / 2;
        p1 = getPoint(mid);
        p2 = getPoint(midmid);
        dis1 = p1.distance(st) + getdist(p1);
        dis2 = p2.distance(st) + getdist(p2);

        if (dis1 > dis2)
            l = mid;
        else
            r = midmid;
    }
    Point tt = getPoint(l);
    ans = tt.distance(st) + getdist(tt);
    l = pi;
    r = 2 * pi;
    while (r - l >= eps)
    {
        mid = (l + r) / 2;
        midmid = (mid + r) / 2;
        p1 = getPoint(mid);
        p2 = getPoint(midmid);
        dis1 = p1.distance(st) + getdist(p1);
        dis2 = p2.distance(st) + getdist(p2);

        if (dis1 > dis2)
            l = mid;
        else
            r = midmid;
    }
    tt = getPoint(l);
    ans = min(ans, tt.distance(st) + getdist(tt));
    return ans;
}

int main()
{

    while (cin >> st.x >> st.y)
    {
        if (sgn(st.x) == 0 && sgn(st.y) == 0)
            break;
        cin >> c1.x >> c1.y >> r;
        cin >> t1.x >> t1.y >> t2.x >> t2.y;

        printf("%.2f\n", solve());
    }
    return 0;
}