poj 3525

多邊形內最大半徑圓。

哇沒有枉費了我自閉了這么些天，大概五天前我看到這種題可能毫無思路抓耳撓腮舉手投降什么的，現在已經能1A了哇。

還是先玩一會計算幾何，刷個幾百道

嗯這個半平面交+二分就闊以解決。雖然隊友說他施展三分套三分*****

想象一下，如果一個多邊形能放進去半徑為r的圓，那么在每條邊向里平移r之后，他的內核一定不為空。

所以我們可以二分r，然后求半平面交，平移操作其實很好處理。

1A了很開森，去快樂的玩耍惹。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cmath>
#include <deque>
using namespace std;
typedef double db;
const db eps=1e-6;
const db pi=acos(-1);
int sign(db k){
    if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
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)==1||(sign(y)==0&&sign(x)==-1);}
};
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))>0);
}
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[2];
    line(point k1,point k2){p[0]=k1;p[1]=k2;}
    point &operator[](int k){ return p[k];}
    int include(point k){ return sign(cross(p[1]-p[0],k-p[0])>0);}
    point dir(){ return p[1]-p[0];}
    line push(db eps){//向左手邊平移eps
        //const db eps=1e-6;
        point delta=(p[1]-p[0]).turn90().unit()*eps;
        return {p[0]-delta,p[1]-delta};
    }
};
point getLL(line k1,line k2){
    return getLL(k1[0],k1[1],k2[0],k2[1]);
}
int parallel(line k1,line k2){ return sign(cross(k1.dir(),k2.dir()))==0;}
int sameDir(line k1,line k2){
    return parallel(k1,k2)&&sign(dot(k1.dir(),k2.dir()))==1;
}
int operator <(line k1,line k2){
    if(sameDir(k1,k2))return k2.include(k1[0]);
    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=0;i<L.size();i++){
        if(i&&sameDir(L[i],L[i-1]))continue;
        while (q.size()>1&&!checkpos(q[q.size()-2],q[q.size()-1],L[i]))q.pop_back();
        while (q.size()>1&&!checkpos(q[1],q[0],L[i]))q.pop_front();
        q.push_back(L[i]);
    }
    while (q.size()>2&&!checkpos(q[q.size()-2],q[q.size()-1],q[0]))q.pop_back();
    while (q.size()>2&&!checkpos(q[1],q[0],q[q.size()-1]))q.pop_front();
    vector<line> ans;for(int i=0;i<q.size();i++)ans.push_back(q[i]);
    return ans;
}
point p[1551];
int n;
bool cw(){//時針
    db s=0;
    for(int i=1;i<n-1;i++){
        s+=cross(p[i]-p[0],p[i+1]-p[0]);
    }
    return s>0;
}
vector<line> L,tmp;
bool check(db x){
    tmp.clear();
    for(int i=0;i<L.size();i++){
        tmp.push_back(L[i].push(-x));
    }
    tmp = getHL(tmp);
    if(tmp.size()>=3)
        return true;
    return false;
}
int main(){
    //freopen("3525.in","r",stdin);
    while (scanf("%d",&n)&&n){
        for(int i=0;i<n;i++){
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        if(!cw())reverse(p,p+n);
        for(int i=0;i<n;i++){
            L.push_back(line(p[i],p[(i+1)%n]));
        }
        db l = 0,r=100000.0;
        while (l+0.0000001<r){
            db mid = (l+r)/2;
            if(check(mid))
                l=mid;
            else
                r=mid;
        }
        printf("%.7f\n",l);
        L.clear();
    }
}

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