给定一个起始点，一个矩形，一个圆，三者互不相交。求从起始点->圆->矩形的最短距离。

自己画一画就知道距离和会是凹函数，不过不是一个凹函数。按与水平向量夹角为圆心角求圆上某点坐标，[0, PI] , [PI, 2*pi]两个区间的点会有两个凹函数。所以要做两次三分才行。

#include<algorithm>

#include<iostream>

#include<fstream>

#include<sstream>

#include<cstring>

#include<cstdlib>

#include<string>

#include<vector>

#include<cstdio>

#include<queue>

#include<stack>

#include<cmath>

#include<map>

#include<set>

#define FF(i, a, b) for(int i=a; i<b; i++)

#define FD(i, a, b) for(int i=a; i>=b; i--)

#define REP(i, n) for(int i=0; i<n; i++)

#define CLR(a, b) memset(a, b, sizeof(a))

#define LL long long

#define PB push_back

#define eps 1e-10

#define debug puts("**debug**");

using namespace std;

const double PI = acos(-1);

struct Point

{

    double x, y;

    Point(double x=0, double y=0):x(x), y(y){}

};

typedef Point Vector;

Vector operator + (Vector a, Vector b) { return Vector(a.x+b.x, a.y+b.y); }

Vector operator - (Vector a, Vector b) { return Vector(a.x-b.x, a.y-b.y); }

Vector operator * (Vector a, double p) { return Vector(a.x*p, a.y*p); }

Vector operator / (Vector a, double p) { return Vector(a.x/p, a.y/p); }

bool operator < (const Point& a, const Point& b)

{

    return a.x < b.x || (a.x == b.x && a.y < b.y);

}

int dcmp(double x)

{

    if(fabs(x) < eps) return 0; return x < 0 ? -1 : 1;

}

bool operator == (const Point& a, const Point& b)

{

    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;

}

double Dot(Vector a, Vector b) { return a.x*b.x + a.y*b.y; }

double Length(Vector a) { return sqrt(Dot(a, a)); }

double Cross(Vector a, Vector b) { return a.x*b.y - a.y*b.x; }

double DistanceToSegment(Point p, Point a, Point b)

{

    if(a == b) return Length(p-a);

    Vector v1 = b-a, v2 = p-a, v3 = p-b;

    if(dcmp(Dot(v1, v2)) < 0) return Length(v2);

    else if(dcmp(Dot(v1, v3)) > 0) return Length(v3);

    else return fabs(Cross(v1, v2)) / Length(v1);

}

struct Circle

{

    Point c;

    double r;

    Circle(){}

    Circle(Point c, double r):c(c), r(r){}

    Point point(double a) //根据圆心角求点坐标

    {

        return Point(c.x+cos(a)*r, c.y+sin(a)*r);

    }

}o;

Point p, p1, p2, p3, p4, s;

double a, b, c, d;

double Calc(double x)

{

    p = o.point(x);

    double d1 = DistanceToSegment(p, p1, p2),

    d2 = DistanceToSegment(p, p2, p3),

    d3 = DistanceToSegment(p, p3, p4),

    d4 = DistanceToSegment(p, p4, p1);

    //点p到矩形最近距离加上s到p距离

    return min(min(d1, d2), min(d3, d4)) + Length(s-p);

}

double solve()

{

    double L, R, m, mm, mv, mmv;

    L = 0; R = PI;

    while (L + eps < R)

    {

        m = (L + R) / 2;

        mm = (m + R) / 2;

        mv = Calc(m);

        mmv = Calc(mm);

        if (mv <= mmv) R = mm; //三分法求最大值时改为mv>=mmv

        else L = m;

    }

    double ret = Calc(L);

    L = PI; R = 2*PI;

    while (L + eps < R)

    {

        m = (L + R) / 2;

        mm = (m + R) / 2;

        mv = Calc(m);

        mmv = Calc(mm);

        if (mv <= mmv) R = mm;

        else L = m;

    }

    return min(ret, Calc(L));

}

int main()

{

    while(scanf("%lf%lf", &s.x, &s.y))

    {

        if(s.x == 0 && s.y == 0) break;

        scanf("%lf%lf%lf", &o.c.x, &o.c.y, &o.r);

        scanf("%lf%lf%lf%lf", &a, &b, &c, &d);

        //确定矩形四个点坐标，左上点开始 逆时针

        double maxx, maxy, minx, miny;

        maxx = max(a, c); maxy = max(b, d);

        minx = min(a, c); miny = min(b, d);

        p1 = Point(minx, maxy);

        p2 = Point(minx, miny);

        p3 = Point(maxx, miny);

        p4 = Point(maxx, maxy);

        double ans = solve();

        printf("%.2f\n", ans);

    }

    return 0;

}