极角排序——ICPC Latin American Regional Contests 2019 D

这题还是按角度进行极角排序简单点

/*
按亮度从大到小排序，从大的给小的连一条向量，极角排序后所有向量在两个象限内，那么可行，反之不可行
*/
#include<bits/stdc++.h>
using namespace std;
#define N 1005 

typedef double db;
const db eps=1e-8;
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);}
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,rad;
    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};}
    int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;}
    point turn(db k1){return (point){x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1)};}
    point turn90(){return (point){-y,x};}
    bool operator < (const point k1) const{
        int a=cmp(x,k1.x);
        if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1;
    }
    db abs(){return sqrt(x*x+y*y);}
    db abs2(){return x*x+y*y;}
    db dis(point k1){return ((*this)-k1).abs();}
    db getw(){return atan2(y,x);}
    point getdel(){if (sign(x)==-1||(sign(x)==0&&sign(y)==-1)) return (*this)*(-1); else return (*this);}
    int getP() const{return sign(y)==1||(sign(y)==0&&sign(x)>=0);}
};
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){
    return k1.rad<k2.rad;
}

point v[N*N],k1,k2;
int n,tot;
struct Node{
    int x,y,w;
}p[N];
int cmp1(Node a,Node b){return a.w>b.w;}
db rads[N];

int main(){
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>p[i].x>>p[i].y>>p[i].w;
    sort(p+1,p+1+n,cmp1);
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++)
            if(p[i].w>p[j].w)v[++tot]=(point){1.0*p[j].x-p[i].x,1.0*p[j].y-p[i].y};
    for(int i=1;i<=tot;i++)
        v[i].rad=atan2(v[i].y,v[i].x);

    sort(v+1,v+1+tot,comp);//极角排序

    if(tot<=2){
        puts("Y");return 0;
    }

    //判断是否所有向量都在两个象限内
    int flag=0;
    for(int i=2;i<=tot;i++){
        db rad1=v[i-1].rad;
        db rad2=v[i].rad;
        if(sign(rad2-rad1-pi)>=0)
            flag=1;
    }

    db rad1=v[tot].rad;
    db rad2=v[1].rad;
    if(rad2>rads[2])rad2-=2*pi;
    if(sign(rad1-rad2-pi)<=0)
        flag=1;
    if(flag)puts("Y");
    else puts("N");

}
/*

8

-9 -1 3

-3 -10 1

8 -9 1

1 7 2

-8 -5 2

1 -4 2

2 -8 3

6 -4 3

*/

原文地址：https://www.cnblogs.com/zsben991126/p/12670249.html