很久没写题解了= =，来水一发吧= =
首先这道题很明显就是求y=ax^2+bx的是否有值取，每一个式子都代表着两个半平面，然后直接半平面交就行了
借鉴了hzwer的代码，还是特别简洁的说
CODE：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
typedef pair<long double,long double> ii;
typedef pair<ii,ii> line;
#define fi first
#define se second
#define maxn 101000
#define inf 1e15
#define exp 0
inline long double cross(ii x,ii y,ii z) {
return (x.fi-y.fi)*(x.se-z.se)-(x.se-y.se)*(x.fi-z.fi);
}
bool cmp(line x,line y) {return cross(x.fi,x.se,y.fi)>exp;}
line q[maxn*2];
long double tag[maxn*2];
bool cmp1(int x,int y) {if (tag[x]==tag[y]) return cmp(q[x],q[y]);return tag[x]<tag[y];}
int l;
inline void init(){
q[1]=line(ii(-inf,-inf),ii(inf,-inf));
q[2]=line(ii(inf,-inf),ii(inf,inf));
q[3]=line(ii(inf,inf),ii(-inf,inf));
q[4]=line(ii(-inf,inf),ii(-inf,-inf));
l=4;
}
ii inter(line a,line b) {
long double k1,k2,t;
k1=cross(a.fi,b.se,a.se);
k2=cross(a.fi,a.se,b.fi);
t=k2/(k1+k2);
return ii(b.fi.fi+t*(b.se.fi-b.fi.fi),b.fi.se+t*(b.se.se-b.fi.se));
}
bool jud(line a,line b,line t) {
ii p=inter(a,b);
return cross(t.fi,p,t.se)>exp;
}
line a[maxn*2],deq[maxn*2];
int id[maxn*2];
inline bool check(int n) {
int num=0;
for (int i=1;i<=l;i++) if (id[i]<=n*2) a[++num]=q[id[i]];
int l,r;
deq[r=l=1]=a[1];
for (int i=2;i<=n*2;i++) {
while (r>l&&jud(deq[r-1],deq[r],a[i])) r--;
while (l<r&&jud(deq[l+1],deq[l],a[i])) l++;
deq[++r]=a[i];
}
while (l<r&&jud(deq[r-1],deq[r],deq[l])) r--;
while (l<r&&jud(deq[l+1],deq[l],deq[r])) l++;
return r-l>=2;
}
long double cal(long double a,long double b,long double x) {return b/a-a*x;}
int main(){
int n;
scanf("%d",&n);
init();
for (int i=1;i<=n;i++) {
int x1,y1,y2;
scanf("%d%d%d",&x1,&y1,&y2);
q[++l]=line(ii(-1,cal(x1,y1,-1)),ii(1,cal(x1,y1,1)));
q[++l]=line(ii(1,cal(x1,y2,1)),ii(-1,cal(x1,y2,-1)));
}
for (int i=1;i<=n*2+4;i++) {tag[i]=atan2(q[i].se.se-q[i].fi.se,q[i].se.fi-q[i].fi.fi);id[i]=i;}
sort(id+1,id+1+n*2+4,cmp1);
int l=2,r=n+2;
while (l+1<r) {
int mid=(l+r)>>1;
if (check(mid)) l=mid;
else r=mid;
}
printf("%d ",check(r)?r-2:l-2);
return 0;
}