Most Distant Point from the Sea

Most Distant Point from the Sea

http://poj.org/problem?id=3525

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 6214		Accepted: 2745		Special Judge

Description

The main land of Japan called Honshu is an island surrounded by the sea. In such an island, it is natural to ask a question: “Where is the most distant point from the sea?” The answer to this question for Honshu was found in 1996. The most distant point is located in former Usuda Town, Nagano Prefecture, whose distance from the sea is 114.86 km.

In this problem, you are asked to write a program which, given a map of an island, finds the most distant point from the sea in the island, and reports its distance from the sea. In order to simplify the problem, we only consider maps representable by convex polygons.

Input

The input consists of multiple datasets. Each dataset represents a map of an island, which is a convex polygon. The format of a dataset is as follows.

n
x1		y1
	⋮
xn		yn

Every input item in a dataset is a non-negative integer. Two input items in a line are separated by a space.

n in the first line is the number of vertices of the polygon, satisfying 3 ≤ n ≤ 100. Subsequent n lines are the x- and y-coordinates of the n vertices. Line segments (x_i, y_i)–(x_i+1, y_i+1) (1 ≤ i ≤ n − 1) and the line segment (x_n, y_n)–(x₁, y₁) form the border of the polygon in counterclockwise order. That is, these line segments see the inside of the polygon in the left of their directions. All coordinate values are between 0 and 10000, inclusive.

You can assume that the polygon is simple, that is, its border never crosses or touches itself. As stated above, the given polygon is always a convex one.

The last dataset is followed by a line containing a single zero.

Output

For each dataset in the input, one line containing the distance of the most distant point from the sea should be output. An output line should not contain extra characters such as spaces. The answer should not have an error greater than 0.00001 (10⁻⁵). You may output any number of digits after the decimal point, provided that the above accuracy condition is satisfied.

Sample Input

4
0 0
10000 0
10000 10000
0 10000
3
0 0
10000 0
7000 1000
6
0 40
100 20
250 40
250 70
100 90
0 70
3
0 0
10000 10000
5000 5001
0

Sample Output

5000.000000
494.233641
34.542948
0.353553

求在多边形中画一个圆的最大半径。

反过来想，看多边形最多可以往内减少多少距离，该距离就是圆的半径，二分判断即可

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;
const double eps=1e-8;
const double INF=1e20;
const double PI=acos(-1.0);

int sgn(double x){
    if(fabs(x)<eps) return 0;
    if(x<0) return -1;
    else return 1;
}

inline double sqr(double x){return x*x;}
inline double hypot(double a,double b){return sqrt(a*a+b*b);}
struct Point{
    double x,y;
    Point(){}
    Point(double _x,double _y){
        x=_x;
        y=_y;
    }
    void input(){
        scanf("%lf %lf",&x,&y);
    }
    void output(){
        printf("%.2f %.2f
",x,y);
    }
    bool operator == (const Point &b)const{
        return sgn(x-b.x) == 0 && sgn(y-b.y)== 0;
    }
    bool operator < (const Point &b)const{
        return sgn(x-b.x)==0?sgn(y-b.y)<0:x<b.x;
    }
    Point operator - (const Point &b)const{
        return Point(x-b.x,y-b.y);
    }
    //叉积
    double operator ^ (const Point &b)const{
        return x*b.y-y*b.x;
    }
    //点积
    double operator * (const Point &b)const{
        return x*b.x+y*b.y;
    }
    //返回长度
    double len(){
        return hypot(x,y);
    }
    //返回长度的平方
    double len2(){
        return x*x+y*y;
    }
    //返回两点的距离
    double distance(Point p){
        return hypot(x-p.x,y-p.y);
    }
    Point operator + (const Point &b)const{
        return Point(x+b.x,y+b.y);
    }
    Point operator * (const double &k)const{
        return Point(x*k,y*k);
    }
    Point operator / (const double &k)const{
        return Point(x/k,y/k);
    }

    //计算pa和pb的夹角
    //就是求这个点看a,b所成的夹角
    ///LightOJ1202
    double rad(Point a,Point b){
        Point p=*this;
        return fabs(atan2(fabs((a-p)^(b-p)),(a-p)*(b-p)));
    }
    //化为长度为r的向量
    Point trunc(double r){
        double l=len();
        if(!sgn(l)) return *this;
        r/=l;
        return Point(x*r,y*r);
    }
    //逆时针转90度
    Point rotleft(){
        return Point(-y,x);
    }
    //顺时针转90度
    Point rotright(){
        return Point(y,-x);
    }
    //绕着p点逆时针旋转angle
    Point rotate(Point p,double angle){
        Point v=(*this) -p;
        double c=cos(angle),s=sin(angle);
        return Point(p.x+v.x*c-v.y*s,p.y+v.x*s+v.y*c);
    }
};

struct Line{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e){
        s=_s;
        e=_e;
    }
    bool operator==(Line v){
        return (s==v.s)&&(e==v.e);
    }
    //根据一个点和倾斜角angle确定直线，0<=angle<pi
    Line(Point p,double angle){
        s=p;
        if(sgn(angle-PI/2)==0){
            e=(s+Point(0,1));
        }
        else{
            e=(s+Point(1,tan(angle)));
        }
    }
    //ax+by+c=0;
    Line(double a,double b,double c){
        if(sgn(a)==0){
            s=Point(0,-c/b);
            e=Point(1,-c/b);
        }
        else if(sgn(b)==0){
            s=Point(-c/a,0);
            e=Point(-c/a,1);
        }
        else{
            s=Point(0,-c/b);
            e=Point(1,(-c-a)/b);
        }
    }
    void input(){
        s.input();
        e.input();
    }
    void adjust(){
        if(e<s) swap(s,e);
    }
    //求线段长度
    double length(){
        return s.distance(e);
    }
    //返回直线倾斜角 0<=angle<pi
    double angle(){
        double k=atan2(e.y-s.y,e.x-s.x);
        if(sgn(k)<0) k+=PI;
        if(sgn(k-PI)==0) k-=PI;
        return k;
    }
    //点和直线的关系
    //1 在左侧
    //2 在右侧
    //3 在直线上
    int relation(Point p){
        int c=sgn((p-s)^(e-s));
        if(c<0) return 1;
        else if(c>0) return 2;
        else return 3;
    }
    //点在线段上的判断
    bool pointonseg(Point p){
        return sgn((p-s)^(e-s))==0&&sgn((p-s)*(p-e))<=0;
    }
    //两向量平行(对应直线平行或重合)
    bool parallel(Line v){
        return sgn((e-s)^(v.e-v.s))==0;
    }
    //两线段相交判断
    //2 规范相交
    //1 非规范相交
    //0 不相交
    int segcrossseg(Line v){
        int d1=sgn((e-s)^(v.s-s));
        int d2=sgn((e-s)^(v.e-s));
        int d3=sgn((v.e-v.s)^(s-v.s));
        int d4=sgn((v.e-v.s)^(e-v.s));
        if((d1^d2)==-2&&(d3^d4)==-2) return 2;
        return (d1==0&&sgn((v.s-s)*(v.s-e))<=0||
                d2==0&&sgn((v.e-s)*(v.e-e))<=0||
                d3==0&&sgn((s-v.s)*(s-v.e))<=0||
                d4==0&&sgn((e-v.s)*(e-v.e))<=0);
    }
    //直线和线段相交判断
    //-*this line -v seg
    //2 规范相交
    //1 非规范相交
    //0 不相交
    int linecrossseg(Line v){
        int d1=sgn((e-s)^(v.s-s));
        int d2=sgn((e-s)^(v.e-s));
        if((d1^d2)==-2) return 2;
        return (d1==0||d2==0);
    }
    //两直线关系
    //0 平行
    //1 重合
    //2 相交
    int linecrossline(Line v){
        if((*this).parallel(v))
            return v.relation(s)==3;
        return 2;
    }
    //求两直线的交点
    //要保证两直线不平行或重合
    Point crosspoint(Line v){
        double a1=(v.e-v.s)^(s-v.s);
        double a2=(v.e-v.s)^(e-v.s);
        return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
    }
    //点到直线的距离
    double dispointtoline(Point p){
        return fabs((p-s)^(e-s))/length();
    }
    //点到线段的距离
    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);
    }
    //返回线段到线段的距离
    //前提是两线段不相交，相交距离就是0了
    double dissegtoseg(Line v){
        return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));
    }
    //返回点P在直线上的投影
    Point lineprog(Point p){
        return s+(((e-s)*((e-s)*(p-s)))/((e-s).len2()));
    }
    //返回点P关于直线的对称点
    Point symmetrypoint(Point p){
        Point q=lineprog(p);
        return Point(2*q.x-p.x,2*q.y-p.y);
    }
};

struct circle{
    Point p;
    double r;
    circle(){}
    circle(Point _p,double _r){
        p=_p;
        r=_r;
    }

    circle(double x,double y,double _r){
        p=Point(x,y);
        r=_r;
    }

    circle(Point a,Point b,Point c){///三角形的外接圆
        Line u=Line((a+b)/2,((a+b)/2)+((b-a).rotleft()));
        Line v=Line((b+c)/2,((b+c)/2)+((c-b).rotleft()));
        p=u.crosspoint(v);
        r=p.distance(a);
    }

    circle(Point a,Point b,Point c,bool t){///三角形的内切圆
        Line u,v;
        double m=atan2(b.y-a.y,b.x-a.x),n=atan2(c.y-a.y,c.x-a.x);
        u.s=a;
        u.e=u.s+Point(cos((n+m)/2),sin((n+m)/2));
        v.s=b;
        m=atan2(a.y-b.y,a.x-b.x),n=atan2(c.y-b.y,c.x-b.x);
        v.e=v.s+Point(cos((n+m)/2),sin((n+m)/2));
        p=u.crosspoint(v);
        r=Line(a,b).dispointtoseg(p);
    }

    void input(){
        p.input();
        scanf("%lf",&r);
    }

    void output(){
        printf("%.2f %.2f %.2f
",p.x,p.y,r);
    }

    bool operator==(circle v){
        return (p==v.p)&&sgn(r-v.r)==0;
    }

    bool operator<(circle v)const{
        return ((p<v.p)||((p==v.p)&&sgn(r-v.r)<0));
    }

    double area(){
        return PI*r*r;
    }

    double circumference(){ ///周长
        return 2*PI*r;
    }

    int relation(Point b){///点和圆的关系  0圆外  1圆上  2圆内
        double dst=b.distance(p);
        if(sgn(dst-r)<0) return 2;
        else if(sgn(dst-r)==0) return 1;
        return 0;
    }

    int relationseg(Line v){///线段和圆的关系，比较的是圆心到线段的距离和半径的关系
        double dst=v.dispointtoseg(p);
        if(sgn(dst-r)<0) return 2;
        else if(sgn(dst-r)==0) return 1;
        return 0;
    }

    int relationline(Line v){///直线和圆的关系，比较的是圆心到直线的距离和半径的关系
        double dst=v.dispointtoline(p);
        if(sgn(dst-r)<0) return 2;
        else if(sgn(dst-r)==0) return 1;
        return 0;
    }

    int relationcircle(circle v){///两圆的关系  5相离 4外切 3相交 2内切 1内含
        double d=p.distance(v.p);
        if(sgn(d-r-v.r)>0) return 5;
        if(sgn(d-r-v.r)==0) return 4;
        double l=fabs(r-v.r);
        if(sgn(d-r-v.r)<0&&sgn(d-l)>0) return 3;
        if(sgn(d-l)==0) return 2;
        if(sgn(d-l)<0)  return 1;
    }

    int pointcrosscircle(circle v,Point &p1,Point &p2){///求两个圆的交点，0没有交点 1一个交点 2两个交点
        int rel=relationcircle(v);
        if(rel == 1 || rel == 5) return 0;
        double d=p.distance(v.p);
        double l=(d*d+r*r-v.r*v.r)/2*d;
        double h=sqrt(r*r-l*l);
        Point tmp=p+(v.p-p).trunc(l);
        p1=tmp+((v.p-p).rotleft().trunc(h));
        p2=tmp+((v.p-p).rotright().trunc(h));
        if(rel == 2 || rel == 4) return 1;
        return 2;
    }

    int pointcrossline(Line v,Point &p1,Point &p2){///求直线和圆的交点，返回交点的个数
        if(!(*this).relationline(v)) return 0;
        Point a=v.lineprog(p);
        double d=v.dispointtoline(p);
        d=sqrt(r*r-d*d);
        if(sgn(d)==0) {
            p1=a;
            p2=a;
            return 1;
        }
        p1=a+(v.e-v.s).trunc(d);
        p2=a-(v.e-v.s).trunc(d);
        return 2;
    }

    int getcircle(Point a,Point b,double r1,circle &c1,circle &c2){///得到过a，b两点，半径为r1的两个圆
        circle x(a,r1),y(b,r1);
        int t=x.pointcrosscircle(y,c1.p,c2.p);
        if(!t) return 0;
        c1.r=c2.r=r;
        return t;
    }

    int getcircle(Line u,Point q,double r1,circle &c1,circle &c2){///得到与直线u相切，过点q，半径为r1的圆
        double dis = u.dispointtoline(q);
        if(sgn(dis-r1*2)>0) return 0;
        if(sgn(dis)==0) {
            c1.p=q+((u.e-u.s).rotleft().trunc(r1));
            c2.p=q+((u.e-u.s).rotright().trunc(r1));
            c1.r=c2.r=r1;
            return 2;
        }
        Line u1=Line((u.s+(u.e-u.s).rotleft().trunc(r1)),(u.e+(u.e-u.s).rotleft().trunc(r1)));
        Line u2=Line((u.s+(u.e-u.s).rotright().trunc(r1)),(u.e+(u.e-u.s).rotright().trunc(r1)));
        circle cc=circle(q,r1);
        Point p1,p2;
        if(!cc.pointcrossline(u1,p1,p2)) cc.pointcrossline(u2,p1,p2);
        c1=circle(p1,r1);
        if(p1==p2){
            c2=c1;
            return 1;
        }
        c2=circle(p2,r1);
        return 2;
    }

    int getcircle(Line u,Line v,double r1,circle &c1,circle &c2,circle &c3,circle &c4){///同时与直线u，v相切，半径为r1的圆
        if(u.parallel(v)) return 0;///两直线平行
        Line u1=Line(u.s+(u.e-u.s).rotleft().trunc(r1),u.e+(u.e-u.s).rotleft().trunc(r1));
        Line u2=Line(u.s+(u.e-u.s).rotright().trunc(r1),u.e+(u.e-u.s).rotright().trunc(r1));
        Line v1=Line(v.s+(v.e-v.s).rotleft().trunc(r1),v.e+(v.e-v.s).rotleft().trunc(r1));
        Line v2=Line(v.s+(v.e-v.s).rotright().trunc(r1),v.e+(v.e-v.s).rotright().trunc(r1));
        c1.r=c2.r=c3.r=c4.r=r1;
        c1.p=u1.crosspoint(v1);
        c2.p=u1.crosspoint(v2);
        c3.p=u2.crosspoint(v1);
        c4.p=u2.crosspoint(v2);
        return 4;
    }

    int getcircle(circle cx,circle cy,double r1,circle &c1,circle &c2){///同时与不相交圆 cx,cy相切，半径为r1的圆
        circle x(cx.p,r1+cx.r),y(cy.p,r1+cy.r);
        int t=x.pointcrosscircle(y,c1.p,c2.p);
        if(!t) return 0;
        c1.r=c2.r=r1;
        return t;
    }

    int tangentline(Point q,Line &u,Line &v){///过一点作圆的切线（先判断点和圆的关系）
        int x=relation(q);
        if(x==2) return 0;
        if(x==1){
            u=Line(q,q+(q-p).rotleft());
            v=u;
            return 1;
        }
        double d=p.distance(q);
        double l=r*r/d;
        double h=sqrt(r*r-l*l);
        u=Line(q,p+((q-p).trunc(l)+(q-p).rotleft().trunc(h)));
        v=Line(q,p+((q-p).trunc(l)+(q-p).rotright().trunc(h)));
        return 2;
    }

    double areacircle(circle v){///求两圆相交的面积
        int rel=relationcircle(v);
        if(rel >= 4) return 0.0;
        if(rel <= 2) return min(area(),v.area());
        double d=p.distance(v.p);
        double hf=(r+v.r+d)/2.0;
        double ss=2*sqrt(hf*(hf-r)*(hf-v.r)*(hf-d));
        double a1=acos((r*r+d*d-v.r*v.r)/(2.0*r*d));
        a1=a1*r*r;
        double a2=acos((v.r*v.r+d*d-r*r)/(2.0*v.r*d));
        a2=a2*v.r*v.r;
        return a1+a2-ss;
    }

    double areatriangle(Point a,Point b){///求圆和三角形pab的相交面积
        if(sgn((p-a)^(p-b))==0) return 0.0;
        Point q[5];
        int len=0;
        q[len++]=a;
        Line l(a,b);
        Point p1,p2;
        if(pointcrossline(l,q[1],q[2])==2){
            if(sgn((a-q[1])*(b-q[1]))<0) q[len++]=q[1];
            if(sgn((a-q[2])*(b-q[2]))<0) q[len++]=q[2];
        }
        q[len++]=b;
        if(len==4 && sgn((q[0]-q[1])*(q[2]-q[1]))>0) swap(q[1],q[2]);
        double res=0;
        for(int i=0;i<len-1;i++){
            if(relation(q[i])==0||relation(q[i+1])==0){
                double arg=p.rad(q[i],q[i+1]);
                res+=r*r*arg/2.0;
            }
            else{
                res+=fabs((q[i]-p)^(q[i+1]-p))/2.0;
            }
        }
        return res;
    }
};

struct polygon{
    int n;
    Point p[100010];
    Line l[100010];
    void input(int _n){
        n=_n;
        for(int i=0;i<n;i++){
            p[i].input();
        }
    }

    void add(Point q){
        p[n++]=q;
    }

    void getline(){
        for(int i=0;i<n;i++){
            l[i]=Line(p[i],p[(i+1)%n]);
        }
    }

    struct cmp{
        Point p;
        cmp(const Point &p0){p=p0;}
        bool operator()(const Point &aa,const Point &bb){
            Point a=aa,b=bb;
            int d=sgn((a-p)^(b-p));
            if(d==0){
                return sgn(a.distance(p)-b.distance(p))<0;
            }
            return d>0;
        }
    };

    void norm(){///进行极角排序
        Point mi=p[0];
        for(int i=1;i<n;i++){
            mi=min(mi,p[i]);
        }
        sort(p,p+n,cmp(mi));
    }

    void getconvex(polygon &convex){///得到第一种凸包的方法，编号为0～n-1，可能要特判所有点共点或共线的特殊情况
        sort(p,p+n);
        convex.n=n;
        for(int i=0;i<min(n,2);i++){
            convex.p[i]=p[i];
        }
        if(convex.n==2&&(convex.p[0]==convex.p[1])) convex.n--;
        if(n<=2) return;
        int &top=convex.n;
        top=1;
        for(int i=2;i<n;i++){
            while(top&&sgn((convex.p[top]-p[i])^(convex.p[top-1]-p[i]))<0) top--;
            convex.p[++top]=p[i];
        }
        int temp=top;
        convex.p[++top]=p[n-2];
        for(int i=n-3;i>=0;i--){
            while(top!=temp&&sgn((convex.p[top]-p[i])^(convex.p[top-1]-p[i]))<=0) top--;
            convex.p[++top]=p[i];
        }
        if(convex.n==2&&(convex.p[0]==convex.p[1])) convex.n--;
        convex.norm();
    }

    void Graham(polygon &convex){///得到凸包的第二种方法
        norm();
        int &top=convex.n;
        top=0;
        if(n==1){
            top=1;
            convex.p[0]=p[0];
            return;
        }
        if(n==2){
            top=2;
            convex.p[0]=p[0];
            convex.p[1]=p[1];
            if(convex.p[0]==convex.p[1]) top--;
            return;
        }
        convex.p[0]=p[0];
        convex.p[1]=p[1];
        top=2;
        for(int i=2;i<n;i++){
            while(top>1&&sgn((convex.p[top-1]-convex.p[top-2])^(p[i]-convex.p[top-2]))<=0) top--;
            convex.p[top++]=p[i];
        }
        if(convex.n==2 && (convex.p[0]==convex.p[1])) convex.n--;
    }

    bool inconvex(){///判断是不是凸的
        bool s[3];
        memset(s,false,sizeof(s));
        for(int i=0;i<n;i++){
            int j=(i+1)%n;
            int k=(j+1)%n;
            s[sgn((p[j]-p[i])^(p[k]-p[i]))+1]=true;
            if(s[0]&&s[2]) return false;
        }
        return true;
    }

    int relationpoint(Point q){///判断点和任意多边形的关系  3点上 2边上 1内部 0外部
        for(int i=0;i<n;i++){
            if(p[i]==q) return 3;
        }
        getline();
        for(int i=0;i<n;i++){
            if(l[i].pointonseg(q)) return 2;
        }
        int cnt=0;
        for(int i=0;i<n;i++){
            int j=(i+1)%n;
            int k=sgn((q-p[j])^(p[i]-p[j]));
            int u=sgn(p[i].y-q.y);
            int v=sgn(p[j].y-q.y);
            if(k>0&&u<0&&v>=0) cnt++;
            if(k<0&&v<0&&u>=0) cnt--;
        }
        return cnt!=0;
    }

    void convexcnt(Line u,polygon &po){///直线u切割凸多边形左侧  注意直线方向
        int &top=po.n;
        top=0;
        for(int i=0;i<n;i++){
            int d1=sgn((u.e-u.s)^(p[i]-u.s));
            int d2=sgn((u.e-u.s)^(p[(i+1)%n]-u.s));
            if(d1>=0) po.p[top++]=p[i];
            if(d1*d2<0)po.p[top++]=u.crosspoint(Line(p[i],p[(i+1)%n]));
        }
    }

    double getcircumference(){///得到周长
        double sum=0;
        for(int i=0;i<n;i++){
            sum+=p[i].distance(p[(i+1)%n]);
        }
        return sum;
    }

    double getarea(){///得到面积
        double sum=0;
        for(int i=0;i<n;i++){
            sum+=(p[i]^p[(i+1)%n]);
        }
        return fabs(sum)/2;
    }

    bool getdir(){///得到方向 1表示逆时针  0表示顺时针
        double sum=0;
        for(int i=0;i<n;i++){
            sum+=(p[i]^p[(i+1)%n]);
        }
        if(sgn(sum)>0) return 1;
        return 0;
    }

    Point getbarycentre(){///得到重心
        Point ret(0,0);
        double area=0;
        for(int i=1;i<n-1;i++){
            double tmp=(p[i]-p[0])^(p[i+1]-p[0]);
            if(sgn(tmp)==0) continue;
            area+=tmp;
            ret.x+=(p[0].x+p[i].x+p[i+1].x)/3*tmp;
            ret.y+=(p[0].y+p[i].y+p[i+1].y)/3*tmp;
        }
        if(sgn(area)) ret =ret/area;
        return ret;
    }

    double areacircle(circle c){///多边形和圆交的面积
        double ans=0;
        for(int i=0;i<n;i++){
            int j=(i+1)%n;
            if(sgn((p[j]-c.p)^(p[i]-c.p))>=0) ans+=c.areatriangle(p[i],p[j]);
            else ans-=c.areatriangle(p[i],p[j]);
        }
        return fabs(ans);
    }

    int relationcircle(circle c){///多边形和圆的关系 2圆完全在多边形内  1圆在多边形里面，碰到了多边形的边界  0其他
        getline();
        int x=2;
        if(relationpoint(c.p)!=1) return 0;
        for(int i=0;i<n;i++){
            if(c.relationseg(l[i])==2) return 0;
            if(c.relationseg(l[i])==1) x=1;
        }
        return x;
    }
};

struct halfplane:public Line{
    double angle;
    halfplane(){}
    halfplane(Point _s,Point _e){///表示向量s->e逆时针（左侧）的半平面
        s=_s;
        e=_e;
    }
    halfplane(Line v){
        s=v.s;
        e=v.e;
    }
    void calcangle(){
        angle=atan2(e.y-s.y,e.x-s.x);
    }
    bool operator<(const halfplane &b)const{
        return angle<b.angle;
    }
};

struct halfplanes{
    int n;
    halfplane hp[2020];
    Point p[2020];
    int que[2020];
    int st,ed;
    void push(halfplane tmp){
        hp[n++]=tmp;
    }

    void unique(){///去重
        int m=1;
        for(int i=1;i<n;i++){
            if(sgn(hp[i].angle-hp[i-1].angle)!=0) hp[m++]=hp[i];
            else if(sgn((hp[m-1].e-hp[m-1].s)^(hp[i].s-hp[m-1].s))>0) hp[m-1]=hp[i];
        }
        n=m;
    }
    bool halfplaneinsert(){
        for(int i=0;i<n;i++) hp[i].calcangle();
        sort(hp,hp+n);
        unique();
        que[st=0]=0;
        que[ed=1]=1;
        p[1]=hp[0].crosspoint(hp[1]);
        for(int i=2;i<n;i++){
            while(st<ed&&sgn((hp[i].e-hp[i].s)^(p[ed]-hp[i].s))<0) ed--;
            while(st<ed&&sgn((hp[i].e-hp[i].s)^(p[st+1]-hp[i].s))<0) st++;
            que[++ed]=i;
            if(hp[i].parallel(hp[que[ed-1]])) return false;
            p[ed]=hp[i].crosspoint(hp[que[ed-1]]);
        }
        while(st<ed&&sgn((hp[que[st]].e-hp[que[st]].s)^(p[ed]-hp[que[st]].s))<0) ed--;
        while(st<ed&&sgn((hp[que[ed]].e-hp[que[ed]].s)^(p[st+1]-hp[que[ed]].s))<0) st++;
        if(st+1>=ed) return false;
        return true;
    }

    void getconvex(polygon &con){///得到最后半平面交得到的凸多边形，要先调用halfplaneinsert()且返回true
        p[st]=hp[que[st]].crosspoint(hp[que[ed]]);
        con.n=ed-st+1;
        for(int j=st,i=0;j<=ed;i++,j++){
            con.p[i]=p[j];
        }
    }
};

bool Check(Line a,Line b){
    if(sgn((a.s-a.e)^(b.s-a.e))*sgn((a.s-a.e)^(b.e-a.e))>0) return false;
    if(sgn((b.s-b.e)^(a.s-b.e))*sgn((b.s-b.e)^(a.e-b.e))>0) return false;
    if(sgn(max(a.s.x,a.e.x)-min(b.s.x,b.e.x))>=0&&sgn(max(b.s.x,b.e.x)-min(a.s.x,a.e.x))>=0
     &&sgn(max(a.s.y,a.e.y)-min(b.s.y,b.e.y))>=0&&sgn(max(b.s.y,b.e.y)-min(a.s.y,a.e.y))>=0)
        return true;
    else return false;
}

halfplanes hps;
int n;
Point p[105];
polygon poly;

void change(Point a,Point b,Point &c,Point &d,double p)///将线段ab往左移动距离p
{
    double len = a.distance(b);
    double dx = (a.y - b.y)*p/len;
    double dy = (b.x - a.x)*p/len;
    c.x = a.x + dx; c.y = a.y + dy;
    d.x = b.x + dx; d.y = b.y + dy;
}

int main(){
    int n;
    while(~scanf("%d",&n)){
        if(!n) break;
        for(int i=0;i<n;i++){
            p[i].input();
        }
        double L=0,R=5005,mid;
        Point aa,bb;
        while(R-L>eps){
            mid=(L+R)/2;
            hps.n=0;
            for(int i=0;i<n;i++){
                change(p[i],p[(i+1)%n],aa,bb,mid);
                hps.push(halfplane(aa,bb));
            }
            if(hps.halfplaneinsert()){
                L=mid;
            }
            else{
                R=mid;
            }
        }
        printf("%.6f
",L);
    }
    return 0;
}