离散化+圆直线交点+转化——icpc cerc 2019 D

题目明明写的是线段和圆。。实际上是直线和圆，白白讨论了很多情况。。

这种转化老套路了

#include<bits/stdc++.h>
using namespace std;
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;
    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;}
    // 逆时针旋转 
    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();}
    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;}
db rad(point k1,point k2){return atan2(cross(k1,k2),dot(k1,k2));} 
struct circle{
    point o; db r;
    int inside(point k){
        if(sign(k.dis(o)-r)<=0)return 1;
        return 0;
    }
};
point proj(point k1,point k2,point q){ // q 到直线 k1,k2 的投影 
    point k=k2-k1; return k1+k*(dot(q-k1,k)/k.abs2());
}
vector<point> getCL(circle k1,point k2,point k3){ // 求直线和圆的交点，沿着 k2->k3 方向给出 , 相切给出两个 
    point k=proj(k2,k3,k1.o); db d=k1.r*k1.r-(k-k1.o).abs2();
    if (sign(d)==-1) return {};
    point del=(k3-k2).unit()*sqrt(max((db)0.0,d)); return {k-del,k+del};
}

point A,B;
circle c;
int n;
db R;

vector<pair<point,int> >v;

//离源点远的排在后面 
bool cmp1(pair<point,int> a, pair<point,int> b){
    if(a.first==b.first)return a.second>b.second;
    if(a.first.x==b.first.x)return a.first.y<b.first.y;
    return a.first.x<b.first.x;
}

int main(){
    //freopen("01.in","r",stdin);
    cin>>n>>R>>B.x>>B.y;
    A.x=A.y=0;c.r=R;
    for(int i=1;i<=n;i++){
        scanf("%lf%lf",&c.o.x,&c.o.y);
        vector<point> res=getCL(c,A,B);
        if(!res.size())continue;
        v.push_back(make_pair(res[0],1));
        v.push_back(make_pair(res[1],-1));
    }
    sort(v.begin(),v.end(),cmp1);
    int ans=0,now=0;
    for(int i=0;i<v.size();i++){
        now+=v[i].second;
        ans=max(ans,now);
    }
    now=0;
    for(int i=v.size()-1;i>=0;i--){
        now+=v[i].second;
        ans=max(ans,now);
    } 
    cout<<ans<<'
';
}