二分+半平面交——poj1279

/*

二分距离，凸包所有边往左平移这个距离，半平面交后看是否还有核存在

*/

#include<iostream>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<vector>

#include<algorithm>

#include<queue>

using namespace std;

#define N 205

typedef double db;

const db eps=1e-;

const db pi=acos(-1.0);

int sign(db k){

    if (k>eps) return ; else if (k<-eps) return -; return ;

}

int cmp(db k1,db k2){return sign(k1-k2);}

int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=;}// k3 在 [k1,k2] 内

struct point{

    db x,y;

    point(){}

    point(db x,db y):x(x),y(y){}

    point operator + (const point &k1) const{return point(k1.x+x,k1.y+y);}

    point operator - (const point &k1) const{return point(x-k1.x,y-k1.y);}

    point operator * (db k1) const{return point(x*k1,y*k1);}

    point operator / (db k1) const{return point(x/k1,y/k1);}

    int getP() const{return sign(y)==||(sign(y)==&&sign(x)>=);}

    point turn90(){return point(-y,x);}

    db abs(){return sqrt(x*x+y*y);}

    point unit(){db w=abs(); return point(x/w,y/w);}

};

int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);}

db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}

db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}

int comp(point k1,point k2){

    if(k1.getP()==k2.getP())return sign(cross(k1,k2))>;

    return k1.getP()<k2.getP();

}

struct line{

    point p[];

    line(point k1,point k2){p[]=k1; p[]=k2;}

    point& operator [] (int k){return p[k];}

    int include(point k){return sign(cross(p[]-p[],k-p[]))>=;}//k在l左端

    point dir(){return p[]-p[];}

    line push(db eps){//向左边平移eps

        point delta = (p[]-p[]).turn90().unit()*eps;

        return line(p[]+delta,p[]+delta);

    }

};

//输入的点是顺时针：ans<0,逆时针：ans>0

bool judge(vector<point> v){

    double ans=;

    for(int i=;i<v.size()-;i++)

        ans+=cross(v[i]-v[],v[i+]-v[]);

    return ans>;

}

point getLL(point k1,point k2,point k3,point k4){

    db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2);

}

point getLL(line k1,line k2){return getLL(k1[],k1[],k2[],k2[]);}

int parallel(line k1,line k2){return sign(cross(k1.dir(),k2.dir()))==;}

int sameDir(line k1,line k2){return parallel(k1,k2)&&sign(dot(k1.dir(),k2.dir()))==;}

int checkpos(line k1,line k2,line k3){return k3.include(getLL(k1,k2));}//k1,k2交点在 k3 左端

int operator<(line k1,line k2){//按极角排序，角度相同的从左到右排

    if(sameDir(k1,k2))return k2.include(k1[]);

    return comp(k1.dir(),k2.dir());

}

vector<line> getHL(vector<line> L){ // 求半平面交 , 半平面是逆时针方向 , 输出按照逆时针

    sort(L.begin(),L.end()); deque<line> q;

    for (int i=;i<(int)L.size();i++){

        if (i&&sameDir(L[i],L[i-])) continue;

        while (q.size()>&&!checkpos(q[q.size()-],q[q.size()-],L[i])) q.pop_back();

        while (q.size()>&&!checkpos(q[],q[],L[i])) q.pop_front();

        q.push_back(L[i]);

    }

    while (q.size()>&&!checkpos(q[q.size()-],q[q.size()-],q[])) q.pop_back();

    while (q.size()>&&!checkpos(q[],q[],q[q.size()-])) q.pop_front();

    vector<line>ans; for (int i=;i<q.size();i++) ans.push_back(q[i]);

    return ans;

}

vector<point>v;

vector<line>L,LL;

int n;

int judge(db eps){

    LL.clear();

    for(int i=;i<L.size();i++)

        LL.push_back(L[i].push(eps));

    vector<line> res=getHL(LL);

    if(res.size()>)return ;

    return ;

}

int main(){

    while(cin>>n && n){

        v.clear();L.clear();

        for(int i=;i<=n;i++){

            point t;cin>>t.x>>t.y;

            v.push_back(t);

        }

        for(int i=;i<v.size();i++)

            L.push_back(line(v[i],v[(i+)%v.size()]));

        db l=,r=1e9,ans,mid;

        while(r-l>eps){

            mid=(l+r)/;

            if(judge(mid))

                l=mid,ans=mid;

            else r=mid;

        }

        printf("%.6lf\n",ans);

    }

}