标签：return point int k1 凸包 k2 poj3384 line 模板

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-10;
const db pi=acos(-1.0);
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);}
int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}// 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)==1||(sign(y)==0&&sign(x)>=0);}
    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))>0;
    return k1.getP()<k2.getP();
}

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;}//k在l左端 
    point dir(){return p[1]-p[0];}
    line push(db eps){//向左边平移eps 
        point delta = (p[1]-p[0]).turn90().unit()*eps;
        return line(p[0]+delta,p[1]+delta); 
    }
};

//输入的点是顺时针：ans<0,逆时针：ans>0 
bool judge(vector<point> v){
    double ans=0;
    for(int i=1;i<v.size()-1;i++)
        ans+=cross(v[i]-v[0],v[i+1]-v[0]);
    return ans>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);
}
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 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[0]);
    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=0;i<(int)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;
}

vector<point>v;
vector<line>L;
int n;
db r;


int main(){
    cin>>n>>r;
        v.clear();L.clear();
        for(int i=1;i<=n;i++){
            point t;cin>>t.x>>t.y;
            v.push_back(t);
        }
        reverse(v.begin(),v.end());
        for(int i=0;i<v.size();i++)
            L.push_back(line(v[i],v[(i+1)%v.size()]));
        for(int i=0;i<v.size();i++)
            L[i]=L[i].push(r);
        vector<line> res = getHL(L);
        v.clear();
        for(int i=0;i<res.size();i++)
            v.push_back(getLL(res[i],res[(i+1)%res.size()]));
        
        int id1=0,id2=0;
        for(int i=0;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);        
    
}

标签：return,point,int,k1,凸包,k2,poj3384,line,模板
来源： https://www.cnblogs.com/zsben991126/p/12377178.html

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