题目描述：

vjudge

POJ

题解：

二分答案+半平面交。

半径范围在0到5000之间二分，每次取$mid$然后平移所有直线，判断半平面交面积是否为零。

我的eps值取的是$10^{-12}$，36ms，而且和样例一样。

（大力推荐）

代码：

#include<cmath>
#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 150;
const double eps = 1e-12;
int dcmp(double x)
{
    if(fabs(x)<=eps)return 0;
    return x>0?1:-1;
}
struct Point
{
    double x,y;
    Point(){}
    Point(double x,double y):x(x),y(y){}
    Point operator + (const Point&a)const{return Point(x+a.x,y+a.y);}
    Point operator - (const Point&a)const{return Point(x-a.x,y-a.y);}
    Point operator * (const double&a)const{return Point(x*a,y*a);}
    Point operator / (const double&a)const{return Point(x/a,y/a);}
    double operator * (const Point&a)const{return x*a.x+y*a.y;}
    double operator ^ (const Point&a)const{return x*a.y-y*a.x;}
};
typedef Point Vector;
typedef vector<Point> Pol;
double ang(const Vector&a){return atan2(a.x,a.y);}
double lth(const Vector&a){return sqrt(a*a);}
Vector Vil(const Vector&a){return Vector(-a.y,a.x)/lth(a);}
struct Line
{
    Point p;
    Vector v;
    Line(){}
    Line(Point p,Vector v):p(p),v(v){}
    Line operator + (const Vector&a)const{return Line(p+a,v);}
    bool operator < (const Line&a)const{return ang(v)<ang(a.v);}
};
int n;
Point p[N],tp[N];
Vector vp[N];
Line s0[N],s[N],ts[N];
bool Onleft(Line l,Point p)
{
    return dcmp(l.v^(p-l.p))>0;
}
Point L_L(Line a,Line b)
{
    double t = ((b.p-a.p)^(b.v))/(a.v^b.v);
    return a.p+a.v*t;
}
double S_(Pol&P)
{
    double ans = 0.0;
    for(int i=1,lim=(int)P.size();i<lim;i++)
        ans+=((P[i-1]-P[0])^(P[i]-P[0]));
    return fabs(ans)/2;
}
double bpmj()
{
    int hd,tl;
    ts[hd=tl=0]=s[1];
    for(int i=2;i<=n;i++)
    {
        while(hd<tl&&!Onleft(s[i],tp[tl-1]))tl--;
        while(hd<tl&&!Onleft(s[i],tp[hd]))hd++;
        ts[++tl] = s[i];
        if(!dcmp(s[i].v^ts[tl-1].v))
        {
            tl--;
            if(Onleft(ts[tl],s[i].p))ts[tl]=s[i];
        }
        tp[tl-1]=L_L(ts[tl-1],ts[tl]);
    }
    while(hd<tl&&!Onleft(ts[hd],tp[tl-1]))tl--;
    if(tl-hd<=1)return 0;
    tp[tl]=L_L(ts[hd],ts[tl]);
    Pol P;
    for(int i=hd;i<=tl;i++)P.push_back(tp[i]);
    return S_(P);
}
bool check(double mid)
{
    for(int i=1;i<=n;i++)s[i]=s0[i]+vp[i]*mid;
    return dcmp(bpmj())>0;
}
int main()
{
    while(scanf("%d",&n)&&n!=0)
    {
        for(int i=1;i<=n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);
        for(int i=1;i<n;i++)s0[i]=Line(p[i],p[i+1]-p[i]);
        s0[n]=Line(p[n],p[1]-p[n]);
        sort(s0+1,s0+1+n);
        for(int i=1;i<=n;i++)    
            vp[i]=Vil(s0[i].v);
        double l = 0.0,r = 5000.0;
        while(dcmp(r-l))
        {
            double mid = (l+r)/2;
            if(check(mid))l=mid;
            else r=mid;
        }
        printf("%lf\n",r);
    }
    return 0;
}

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