poj 3525

多边形内最大半径圆。

哇没有枉费了我自闭了这么些天，大概五天前我看到这种题可能毫无思路抓耳挠腮举手投降什么的，现在已经能1A了哇。

还是先玩一会计算几何，刷个几百道

嗯这个半平面交+二分就阔以解决。虽然队友说他施展三分套三分*****

想象一下，如果一个多边形能放进去半径为r的圆，那么在每条边向里平移r之后，他的内核一定不为空。

所以我们可以二分r，然后求半平面交，平移操作其实很好处理。

1A了很开森，去快乐的玩耍惹。

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <deque>
 using namespace std;
 typedef double db;
 const db eps=1e-;
 const db pi=acos(-);
 int sign(db k){
     if (k>eps) return ; else if (k<-eps) return -; return ;
 }
 int cmp(db k1,db k2){return sign(k1-k2);}
 struct point{
     db x,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};}
     db abs(){return sqrt(x*x+y*y);}
     point unit(){db w=abs(); return point{x/w,y/w};}
     point turn90(){ return point{-y,x};}
     db getP()const { return sign(y)==||(sign(y)==&&sign(x)==-);}
 };
 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;}
 db rad(point k1,point k2){ return atan2(cross(k1,k2),dot(k1,k2));}
 int compareangle(point k1,point k2){
     return k1.getP()<k2.getP()||(k1.getP()==k2.getP()&&sign(cross(k1,k2))>);
 }
 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);
 }
 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[])>);}
     point dir(){ return p[]-p[];}
     line push(db eps){//向左手边平移eps
         //const db eps=1e-6;
         point delta=(p[]-p[]).turn90().unit()*eps;
         return {p[]-delta,p[]-delta};
     }
 };
 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 operator <(line k1,line k2){
     if(sameDir(k1,k2))return k2.include(k1[]);
     return compareangle(k1.dir(),k2.dir());
 }
 int checkpos(line k1,line k2,line k3){ return k3.include(getLL(k1,k2));}
 vector<line> getHL(vector<line> &L){
     sort(L.begin(),L.end());deque<line> q;
     for(int i=;i<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;
 }
 point p[];
 int n;
 bool cw(){//时针
     db s=;
     for(int i=;i<n-;i++){
         s+=cross(p[i]-p[],p[i+]-p[]);
     }
     return s>;
 }
 vector<line> L,tmp;
 bool check(db x){
     tmp.clear();
     for(int i=;i<L.size();i++){
         tmp.push_back(L[i].push(-x));
     }
     tmp = getHL(tmp);
     if(tmp.size()>=)
         return true;
     return false;
 }
 int main(){
     //freopen("3525.in","r",stdin);
     while (scanf("%d",&n)&&n){
         for(int i=;i<n;i++){
             scanf("%lf%lf",&p[i].x,&p[i].y);
         }
         if(!cw())reverse(p,p+n);
         for(int i=;i<n;i++){
             L.push_back(line(p[i],p[(i+)%n]));
         }
         db l = ,r=100000.0;
         while (l+0.0000001<r){
             db mid = (l+r)/;
             if(check(mid))
                 l=mid;
             else
                 r=mid;
         }
         printf("%.7f\n",l);
         L.clear();
     }
 }