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)<);
 }

 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)<;
 }

 bool bo(int x){
     int cnt=;
     for (int i=;i<=sum;i++) if (l[i].id<=x) a[++cnt]=l[i];
     int head=,tail=;
     q[]=a[],q[]=a[];
     for (int i=;i<=cnt;i++){
         while (head<tail && check(q[tail-],q[tail],a[i])) tail--;
         while (head<tail && check(q[head+],q[head],a[i])) head++;
         q[++tail]=a[i];
     }
     while (head<tail && check(q[tail-],q[tail],q[head])) tail--;
     while (head<tail && check(q[head+],q[head],q[tail])) head++;
     return tail>head+;
 }

 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=;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){-,(A-C)/B},l[tot].to=(point){,(-A-C)/B},l[tot].id=i;
         C=-y2;
         l[++tot].from=(point){,(-A-C)/B},l[tot].to=(point){-,(A-C)/B},l[tot].id=i;
     }

     for (int i=;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+,l+tot+);
     sum=unique(l+,l+tot+)-l;
     sum--;
     int l=,r=n;
     while (l<=r){
         int mid=(l+r)>>;
         if (bo(mid)) l=mid+;
         else r=mid-;
     }
     printf("%d",r);
     return ;
 }