极角排序+stl——hdu6731

 碰到极角排序不能无脑就用双指针写，要思考是否有更加好的写法

本题由于有三点共线，所以双指针扫描会有问题，由于只要求直角个数，所以我们可以直接用lower_bound二分去找这个就可以

或者也可以用map重载point的=,<,>运算

#include<bits/stdc++.h>
using namespace std;
#define N 10005
#define db long long

const double 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;
    int id;
    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 getP() const{return sign(y)==1||(sign(y)==0&&sign(x)>=0);}
    point turn90(){return (point){-y,x};}
};
db cross(const point &k1,const 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;}
bool operator <(const point &k1,const point &k2){
    if(k1.getP()==k2.getP())return sign(cross(k1,k2))>0;
    return k1.getP()<k2.getP();
}

int n,q,ans[N],tot;
point p[N],pp[N],o;

int main(){
    while(scanf("%d%d",&n,&q)!=EOF){
    for(int i=1;i<=n+q;i++){
        scanf("%lld%lld",&p[i].x,&p[i].y);
        p[i].id=0;
    }
    for(int i=1;i<=n+q;i++)ans[i]=0;
    
    for(int i=1+n;i<=q+n;i++){
        o=p[i];
        for(int j=1;j<=n;j++)pp[j]=p[j]-o;
        sort(pp+1,pp+1+n);
        for(int j=1;j<=n;j++){
            point now=pp[j].turn90();
            int posl=lower_bound(pp+1,pp+1+n,now)-pp;
            int posr=upper_bound(pp+1,pp+1+n,now)-pp;
            ans[i]+=posr-posl;
        }
    }
    for(int i=1;i<=n;i++){
        o=p[i];
        tot=0;
        for(int j=1;j<=n;j++)if(j!=i)pp[++tot]=p[j]-o;
        sort(pp+1,pp+1+tot);
        
        for(int j=n+1;j<=n+q;j++){
            point now=p[j]-o;
            now=now.turn90();
            int posl=lower_bound(pp+1,pp+1+tot,now)-pp;
            int posr=upper_bound(pp+1,pp+1+tot,now)-pp;
            ans[j]+=posr-posl;
            
            now=now.turn90();
            now=now.turn90();
            posl=lower_bound(pp+1,pp+1+tot,now)-pp;
            posr=upper_bound(pp+1,pp+1+tot,now)-pp;
            ans[j]+=posr-posl;
        }
    }
        
        for(int i=1+n;i<=q+n;i++)
            cout<<ans[i]<<"\n";
    }
}
/*
4 4
0 0
1 1
1 0
0 1
2 0
0 2
2 1
1 2
*/