hdu 4454 三分*****

 

参考自：http://www.cnblogs.com/kuangbin/archive/2012/11/08/2761425.html

很容易想到三分应该可以

此题明显是满足凸性的，用三分法很简单解决了。

在圆上分成两个半圆，0-Pi,Pi-2Pi,做两次三分法就解决了。（不过那个矩形只有对角线，是怎么确定下来的我就不明白，，，）

//============================================================================
// Name        : HDU4454.cpp
// Author      : kuangbin
// Version     :
// Copyright   : Your copyright notice
// Description : 三分法
//============================================================================

#include <iostream>
#include <math.h>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
const double eps=1e-6;
const double PI=acos(-1.0);
struct Point
{
    double x,y;
    Point(double xx=0,double yy=0):x(xx),y(yy){}
    Point operator -(const Point p1)
    {
        return Point(x-p1.x,y-p1.y);
    }
    double operator ^(const Point p1)
    {
        return x*p1.x+y*p1.y;
    }
};
inline int sign(double d)
{
    if(d>eps)return 1;
    else if(d<-eps)return -1;
    else return 0;
}
double dis(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double cross(Point a,Point b)
{
    return a.x*b.y-a.y*b.x;
}

double ptoline(Point tp,Point st,Point ed)//求点到线段的距离
{
    double t1=dis(tp,st);
    double t2=dis(tp,ed);
    double ans=min(t1,t2);
    if(sign((ed-st)^(tp-st))>=0 && sign((st-ed)^(tp-ed))>=0)//锐角
    {
        double h=fabs(cross(st-tp,ed-tp))/dis(st,ed);
        ans=min(ans,h);
    }
    return ans;
}
double xx1,yy1,xx2,yy2;

double ptoRec(Point tp)
{
    Point p1(xx1,yy1);
    Point p2(xx1,yy2);
    Point p3(xx2,yy1);
    Point p4(xx2,yy2);
    double ans=ptoline(tp,p1,p2);//矩形四条边
    ans=min(ans,ptoline(tp,p2,p4));
    ans=min(ans,ptoline(tp,p4,p3));
    ans=min(ans,ptoline(tp,p3,p1));
    return ans;
}
double x,y,R;
double xx0,yy0;
Point get_point(double A)
{
    return Point(x+R*cos(A),y+R*sin(A));
}
double solve()
{
    double l,r,mid,midmid;
    Point p0(xx0,yy0);
    Point p;
    l=0;r=PI;
    while(r-l>=1e-8)
    {
        mid=(r+l)/2;
        midmid=(mid+r)/2;
        Point p1=get_point(mid);
        Point p2=get_point(midmid);
        double t1=ptoRec(p1)+dis(p1,p0);
        double t2=ptoRec(p2)+dis(p2,p0);
        if(t1>t2)l=mid;
        else r=midmid;
    }
    p=get_point(l);
    double ans1=dis(p,p0)+ptoRec(p);
    l=PI;r=2*PI;
    while(r-l>=1e-8)
    {
        mid=(r+l)/2;
        midmid=(mid+r)/2;
        Point p1=get_point(mid);
        Point p2=get_point(midmid);
        double t1=ptoRec(p1)+dis(p1,p0);
        double t2=ptoRec(p2)+dis(p2,p0);
        if(t1>t2)l=mid;
        else r=midmid;
    }
    p=get_point(l);
    double ans2=dis(p,p0)+ptoRec(p);
    return min(ans1,ans2);
}

int main() {
    freopen("in.txt","r",stdin);
    while(scanf("%lf%lf",&xx0,&yy0)==2)
    {
        if(xx0==0 && yy0==0)break;
        scanf("%lf%lf%lf",&x,&y,&R);
        scanf("%lf%lf%lf%lf",&xx1,&yy1,&xx2,&yy2);
        printf("%.2lf
",solve());
    }
    return 0;
}