官方题解，内附我不会的\(O(nlogn)\)做法
http://www.csc.kth.se/~austrin/icpc/finals2018solutions.pdf
这题题意很简单．

给你一个可以不凸的多边形，要求找一个点距离所有多边形的顶点最小距离最大．

首先这个东西很像是半平面交，因为半平面交可以做

给你一个可以不凸的多边形，要求找一个点距离所有多边形的边最小距离最大．

然而直接上半平面交是搞不出来的．
观察一下性质，显然最优取值点在某一点对的垂直平分线上．

然后大力
https://en.wikipedia.org/wiki/Voronoi_diagram
一发．
发现自己不会写找出维诺图的\(O(nlogn)\)算法．
这时候就有两种思考：
１．乱搞，卡精度
２．观察到数据范围很小，直接暴力半平面交找出维诺图的顶点．（然而我并没有想到这个）

考虑乱搞，怎么乱搞呢，首先要利用结论，垂直平分线上的结论．
那么我随机生成一个点，让他随机沿垂直平分线走一个随机的距离．实践证明，这样做很不理想．
考虑抱精度，那么设一个步长，每次乘上一个alpha，搞一搞精度．
还是不能过．

因为点会走到维诺图外，所以只沿维诺图的某条边所在直线走，不一定可以走到维诺图的最优定点上．
考虑随机一个角度，多随机几次，在那个角度上走步长．

然后发现被近乎一条直线的多边形卡爆了，那么考虑再加一种策略，向一个随机顶点走．
然后还是被卡爆了．

考虑借鉴遗传算法的思想，在下一次的方向中借鉴上一次的较优解．
然后就勉强可以水过．

#include <bits/stdc++.h>using namespace std;
template<class T>
bool Chkmin(T &x,T y){return x>y?(x=y,1):0;
}
template<class T>
bool Chkmax(T &x,T y){return x<y?(x=y,1):0;
}
template<class T>
T sqr(T x){return x*x;
}
typedef double ld;
typedef long long i64;
const ld INF=1e18;
const ld EPS=1e-8;
struct point{ld x,y;point operator +(const point &_) const{return {x+_.x,y+_.y};}point operator -(const point &_) const{return {x-_.x,y-_.y};}point operator /(const ld &_){return {x/_,y/_};}point operator *(const ld &_){return {x*_,y*_};}ld cross(const point &_) const{return x*_.y-y*_.x;}ld dot(const point &_) const{return x*_.x+y*_.y;}point rotate() const{return {-y,x};}ld dis(){return sqrt(sqr(x)+sqr(y));}
};
ostream& operator <<(ostream & os,const point &_){return os<<"{"<<_.x<<","<<_.y<<"}";
}
struct segment{point s,t;bool strict_cross(const segment &_) const{return (t-s).cross(_.s-s)*(t-s).cross(_.t-s)<0&&(_.t-_.s).cross(s-_.s)*(_.t-_.s).cross(t-_.s)<0;}bool operator &(const segment &_) const{//cerr<<"&"<<s<<" "<<t<<",,"<<_.s<<" "<<_.t<<endl;return strict_cross(_);}
};
struct polygon{vector<point> data;size_t next(size_t val) const{return val==size()-1?0:val+1;}size_t size() const{return data.size();}segment edge(int x) const{return segment({data[x],data[next(x)]});}point& operator [](const size_t &x){return data[x];}
};
bool in(const point &test,const polygon &con){static const point add({42348.23712341,34435.61431236});int counter=0;for (int i=0; i<con.size(); ++i)if (segment({test,test+add})&con.edge(i)) ++counter;return counter&1;
}
point neareast(const point &test,polygon &con){ld nowdis=(test-con[0]).dis();point rvalue=con[0];for (size_t i=1; i<con.size(); ++i)if (Chkmin(nowdis,(test-con[i]).dis())) rvalue=con[i];return rvalue;
}
const int C=10000;
const int T=1000;
point rec[T];
bool b[T];
ld Rand(){return (((i64)rand()<<31)+rand());
}
int mxx,mix,mxy,miy;
int RandIntX(){return rand()%(mxx-mix+1)+mix;
}
int RandIntY(){return rand()%(mxy-miy+1)+miy;
}
polygon con;
point trans(ld x,point y){return y*(x/y.dis());
}
point pp(){int x,y;do{x=rand()%con.size();y=rand()%con.size();}while (x==y);return (con[x]-con[y]).rotate();
}
point conclude(int TT){point x({RandIntX(),RandIntY()});//bug herebool fi=1;for (ld step=C/2; step>EPS; step*=0.8){//cerr<<x<<" "<<step<<endl;rec[1]=x+trans(step,x-rec[0]);rec[0]=x;if (fi)for (int i=2,t; i<TT; ++i){if ((t=rand())&1){rec[i]=x+trans(step,pp());//optimize}else if ((t>>1)&1) rec[i]=x+trans(step,(con[rand()%con.size()]-x));else{ld alpha=Rand();rec[i]=x+point({step*cos(alpha),step*sin(alpha)});}}elsefor (int i=2,t; i<TT; ++i)if ((t=rand())&1){ld alpha=Rand();rec[i]=x+point({step*cos(alpha),step*sin(alpha)});}else if ((t>>1)&1) rec[i]=x+trans(step,(con[rand()%con.size()]-x));else{rec[i]=x+trans(step,pp());//optimize}int counter=0;for (int i=0; i<TT; ++i)if (in(rec[i],con)){//bug//cerr<<"inint"<<i<<" "<<rec[i]<<endl;if (fi){//cerr<<x<<endl;fi=0;step=C/2;}b[i]=1;++counter;}else b[i]=0;//getchar();//exit(0);if (counter){ld nowdis=-1;for (int i=0; i<TT; ++i)if (b[i]&&Chkmax(nowdis,(rec[i]-neareast(rec[i],con)).dis())){//cerr<<"UPD"<<nowdis<<endl;x=rec[i];}}else{ld nowdis=INF;for (int i=0; i<TT; ++i)if (Chkmin(nowdis,(rec[i]-neareast(rec[i],con)).dis())){x=rec[i];}}}return x;
}
int main(){int n;cin>>n;con.data.resize(n);mxx=-10000;mix=10000;mxy=-10000;miy=10000;for (int i=0; i<n; ++i){int x,y;cin>>x>>y;con[i]=point({x,y});mix=min(mix,x);mxx=max(mxx,x);miy=min(miy,y);mxy=max(mxy,y);}int TIME=clock();ld ans=0;while (clock()-TIME<CLOCKS_PER_SEC){int p=rand()%15+5;//cerr<<"P"<<p<<endl;point tmp=conclude(p);ans=max(ans,(tmp-neareast(tmp,con)).dis());}while (clock()-TIME<CLOCKS_PER_SEC*1.5){int p=rand()%200;if (p>100) p=rand()%200;if (p>100) p=rand()%200;if (p>100) p=rand()%200;if (p>100) p=rand()%200;p=max(p,30);point tmp=conclude(p);ans=max(ans,(tmp-neareast(tmp,con)).dis());//cerr<<"amns"<<ans<<endl;}cout<<fixed<<setprecision(10)<<ans;
}

考虑正经算法．
半平面交怎么求维诺图呢？
把某一个顶点作为线段端点的垂直平分线搞出来，然后暴力半平面交，最后的点集就是维诺图上在该点附近的顶点．
结论:维诺图的顶点数和边数都为\(O(n)\)级别．
注意维诺图的边和多边形的交点也有可能是答案．
复杂度\(O(n^2logn)\)

#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optmize(3)
#include <bits/stdc++.h>
using namespace std;
template<class T>
bool Chkmin(T &x,T y){return x>y?(x=y,1):0;
}
template<class T>
bool Chkmax(T &x,T y){return x<y?(x=y,1):0;
}
template<class T>
T sqr(T x){return x*x;
}
typedef double ld;
typedef long long i64;
const ld INF=1e18;
const ld EPS=1e-9;
struct point{ld x,y;point operator +(const point &_) const{return {x+_.x,y+_.y};}point operator -(const point &_) const{return {x-_.x,y-_.y};}point operator /(const ld &_) const{return {x/_,y/_};}point operator *(const ld &_) const{return {x*_,y*_};}ld cross(const point &_) const{return x*_.y-y*_.x;}ld dot(const point &_) const{return x*_.x+y*_.y;}bool operator ==(const point &_) const{return x==_.x&&y==_.y;}bool operator !=(const point &_) const{return x!=_.x||y!=_.y;}point rotate() const{return {-y,x};}ld dis() const{return sqrt(sqr(x)+sqr(y));}ld sqrdis() const{return x*x+y*y;}point makelong(ld len=3e4) const{ld t=len/(dis());return (*this)*t;}
};
ld atan2(const point &x){return atan2(x.y,x.x);
}
ostream& operator <<(ostream & os,const point &_){return os<<"{"<<_.x<<","<<_.y<<"}";
}
struct segment{point s,t;bool strict_cross(const segment &_) const{return (t-s).cross(_.s-s)*(t-s).cross(_.t-s)<0&&(_.t-_.s).cross(s-_.s)*(_.t-_.s).cross(t-_.s)<0;}bool operator &(const segment &_) const{//cerr<<"&"<<s<<" "<<t<<",,"<<_.s<<" "<<_.t<<endl;return strict_cross(_);}bool onright(const point &_) const{return (_-s).cross(t-s)>0;}point generator(const ld &_) const{return s+(t-s)*_;}point intersect(const segment &_) const{ld t1=(t-s).cross(_.t-_.s);ld t2=(_.t-_.s).cross(s-_.s);return generator(t2/t1);}bool on(const point &_) const{//cerr<<(_-s).dot(_-t)<<" "<<fabs((_-s).cross(_-t))<<endl;return (_-s).dot(_-t)<0&&fabs((_-s).cross(_-t))<EPS;}bool line_on(const point &_) const{//cerr<<(_-s).dot(_-t)<<" "<<fabs((_-s).cross(_-t))<<endl;return (_-s).dot(_-t)<0;}
};
ld atan2(const segment &x){return atan2(x.t-x.s);
}
ld ans;
struct polygon:vector<point>{size_t next(size_t val) const{return val==size()-1?0:val+1;}segment edge(int x) const{return segment({(*this)[x],(*this)[next(x)]});}ld neareast(const point &p) const{//cerr<<"ne"<<p<<endl;ld ret=INF;ld c=ans*ans;for (auto i:*this)if ((ret=min(ret,(i-p).sqrdis()))<=c) break;//if (sqrt(ret)>30000) cerr<<"SUCC"<<p<<" "<<sqrt(ret)<<endl;return sqrt(ret);}bool in(const point &p) const{for (size_t i=0; i<this->size(); ++i){auto j=edge(i);if (j.on(p)) return 1;}static const point v={43333.3435499546546L,74354.4534534512434L};int ret=0;for (size_t i=0; i<this->size(); ++i){auto j=edge(i);ret+=segment{p,p+v}&j;}return ret&1;}
};
typedef vector<segment> vs;
polygon con;
struct Segment:segment{ld tt;Segment(){}Segment(const segment &x):segment(x){tt=atan2(x);}
};
polygon halfplaneintersect(vs v){vector<Segment> vp(v.begin(),v.end());stable_sort(vp.begin(),vp.end(),[](const Segment &x,const Segment &y)->bool{if (x.tt+EPS<y.tt) return 1;if (y.tt+EPS<x.tt) return 0;return x.onright(y.s);});//for (auto i:v) cerr<<i.s<<" "<<i.t<<endl;//exit(0);static const int N=2005;static Segment q[N];static point p[N];int l=0,r=-1;for (int i=0; i<vp.size(); ++i){//cerr<<"Line: "<<"\t"<<v[i].s<<" "<<v[i].t<<" "<<atan2(v[i])<<endl;while (l<r&&vp[i].onright(p[r])) --r;while (l<r&&vp[i].onright(p[l+1])) ++l;if (l<=r&&vp[i].tt<q[r].tt+EPS) continue;q[++r]=vp[i];if (l<r){//cerr<<"intersect \t \t \t"<<l<<" "<<r<<" "<<p[r]<<endl;p[r]=q[r-1].intersect(q[r]);}//cerr<<"this"<<l<<" "<<r<<endl;}while (l<r&&q[l].onright(p[r])) --r;while (l<r&&q[r].onright(p[l+1])) ++l;//cerr<<"atlast \t"<<l<<" "<<r<<endl;p[l]=q[l].intersect(q[r]);polygon ret;ret.assign(p+l,p+r+1);for (int i=l; i<=r; ++i){segment t1=q[i];for (int j=0; j<con.size(); ++j){segment t2=con.edge(j);point tmp=t1.intersect(t2);//cerr<<"tmp"<<tmp<<" "<<t1.s<<" "<<t1.t<<" "<<t1.on(tmp)<<endl;//cerr<<"init"<<endl;if (t2.line_on(tmp))ans=max(ans,con.neareast(tmp));}}//for (auto i:ret) cerr<<i<<"|";//cerr<<endl;return ret;
}
segment perpendicular_bisector(point x,point y){point a=(x+y)/2,b=(y-x).rotate().makelong();return {a-b,a+b};
}
void solve(polygon result){//cerr<<"RE"<<result.size()<<endl;for (auto i:result){//cerr<<i<<endl;if (con.in(i)){//cerr<<"I"<<i<<endl;ans=max(ans,con.neareast(i));}}
}
void test(){segment a2({{0,0},{1,1}});segment a1({{1,0},{-1,0}});cerr<<a1.intersect(a2)<<endl;
}
int main(){//test();int n;cin>>n;con.resize(n);for (int i=0; i<n; ++i){int x,y;cin>>x>>y;con[i]={x,y};}for (auto i:con){//cerr<<"BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"<<i<<endl;vs tmp;for (auto j:con)if (i!=j){tmp.push_back(perpendicular_bisector(i,j));//cerr<<tmp.back().s<<" "<<tmp.back().t<<" "<<j<<endl;}solve(halfplaneintersect(tmp));//return 0;}cout<<fixed<<setprecision(10)<<ans;
}