bzoj2732: [HNOI2012]射箭 半平面交

这题乍一看与半平面交并没有什么卵联系，然而每个靶子都可以转化为两个半平面。

scanf("%lf%lf%lf",&x,&ymin,&ymax);

于是乎就有ymin<=ax^2+bx<=ymax。（因为抛物线一定经过点（0,0），所以c=0）

考虑前一个有ax^2+bx>=ymin  <=>  ax^2+bx-ymin>=0。

#define A x^2

#define B x

#define C ymin

#define x' a

#define y' b

于是乎Ax'+By'+c>=0

这个式子貌似在哪见过的样子

于是乎上半平面交。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define pi (acos(-1.0))
#define maxn 200100
#define double long double
const long long inf=1e15;
int n,tot,sum;
 
double sqr(double x){return x*x;}
 
struct point{
    double x,y;
}p[maxn];
 
struct line{
    point from,to;
    int id;
    double slope;
}l[maxn],q[maxn],a[maxn];
 
point operator -(point a,point b){return(point){a.x-b.x,a.y-b.y};}
point operator +(point a,point b){return(point){a.x+b.x,a.y+b.y};}
double operator *(point a,point b){return a.x*b.y-a.y*b.x;}
bool operator ==(line a,line b){return a.slope==b.slope;}
bool operator <(line a,line b){
    return a.slope<b.slope||(a.slope==b.slope && (a.to-a.from)*(b.to-a.from)<0); 
}
 
point getpoint(line a,line b){
    double t1=(b.to-a.from)*(a.to-a.from),t2=(a.to-a.from)*(b.from-a.from);
    double t=t1/(t1+t2);
    return (point){(b.from.x-b.to.x)*t+b.to.x,(b.from.y-b.to.y)*t+b.to.y};
}
 
bool check(line a,line b,line c){
    point d=getpoint(a,b);
    return (c.to-c.from)*(d-c.from)<0;
}
 
bool bo(int x){
    int cnt=0;
    for (int i=1;i<=sum;i++) if (l[i].id<=x) a[++cnt]=l[i];
    int head=1,tail=2;
    q[1]=a[1],q[2]=a[2];
    for (int i=3;i<=cnt;i++){
        while (head<tail && check(q[tail-1],q[tail],a[i])) tail--;
        while (head<tail && check(q[head+1],q[head],a[i])) head++;
        q[++tail]=a[i];
    }
    while (head<tail && check(q[tail-1],q[tail],q[head])) tail--;
    while (head<tail && check(q[head+1],q[head],q[tail])) head++;
    return tail>head+1;
}
 
int main(){
    scanf("%d",&n);
    l[++tot].to=(point){-inf,inf},l[tot].from=(point){inf,inf};
    l[++tot].to=(point){inf,inf},l[tot].from=(point){inf,-inf};
    l[++tot].to=(point){inf,-inf},l[tot].from=(point){-inf,-inf};
    l[++tot].to=(point){-inf,-inf},l[tot].from=(point){-inf,inf};
    for (int i=1;i<=n;i++){
        double x,y1,y2;
        scanf("%llf%llf%llf",&x,&y1,&y2);
        double A=sqr(x),B=x,C=-y1;
        l[++tot].from=(point){-1,(A-C)/B},l[tot].to=(point){1,(-A-C)/B},l[tot].id=i;
        C=-y2;
        l[++tot].from=(point){1,(-A-C)/B},l[tot].to=(point){-1,(A-C)/B},l[tot].id=i;
    }
     
     
    for (int i=1;i<=tot;i++) l[i].slope=atan2(l[i].to.y-l[i].from.y,l[i].to.x-l[i].from.x);
    sort(l+1,l+tot+1);
    sum=unique(l+1,l+tot+1)-l;
    sum--;
    int l=1,r=n;
    while (l<=r){
        int mid=(l+r)>>1;
        if (bo(mid)) l=mid+1;
        else r=mid-1;
    }
    printf("%d",r);
    return 0;
}