【模板】凸包向内推进求不严格的半平面交——poj3384

想不明白这题写严格的半平面交为什么会错

/*
凸包所有边向内推进r
*/

#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);}
    db abs(){return sqrt(x*x+y*y);}
    db abs2(){return x*x+y*y;}
    db dis(point k1){return ((*this)-k1).abs();}
    int getP() const{return sign(y)==||(sign(y)==&&sign(x)>=);}
    point turn90(){return point(-y,x);}
    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;
int n;
db r;

int main(){
    cin>>n>>r;
        v.clear();L.clear();
        for(int i=;i<=n;i++){
            point t;cin>>t.x>>t.y;
            v.push_back(t);
        }
        reverse(v.begin(),v.end());
        for(int i=;i<v.size();i++)
            L.push_back(line(v[i],v[(i+)%v.size()]));
        for(int i=;i<v.size();i++)
            L[i]=L[i].push(r);
        vector<line> res = getHL(L);
        v.clear();
        for(int i=;i<res.size();i++)
            v.push_back(getLL(res[i],res[(i+)%res.size()]));

        int id1=,id2=;
        for(int i=;i<v.size();i++)
            for(int j=i;j<v.size();j++){
                if(v[i].dis(v[j])>v[id1].dis(v[id2]))
                    id1=i,id2=j;
            }
        if(v[id1].x>v[id2].x)swap(id1,id2);
        printf("%.4f %.4f %.4f %.4f\n",v[id1].x,v[id1].y,v[id2].x,v[id2].y);        

}