cf 1142 C

忽然觉得这个题很有必要写题解。

移项

y-x^2 = bx+c

那么其实就是找有多少条直线上方没有点

所以就是一个上凸壳有多少条直线/点.

绝妙啊！！！！

太妙了啊！！！！

神乎其技卧槽！！！

(我是傻逼)

 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <set>
 using namespace std;
 typedef long long db;
 const double eps=1e-;
 const double pi=acos(-);
 int sign(db k){
     if (k>eps) return ; else if (k<-eps) return -; return ;
 }
 int cmp(db k1,db k2){return sign(k1-k2);}
 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};}
     bool operator == (const point &k1) const{return cmp(x,k1.x)==&&cmp(y,k1.y)==;}
     bool operator <(const point &k1)const {
         int c=cmp(x,k1.x);
         if(c)return c==-;
         return cmp(y,k1.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;}
 int n;
 vector<point> convexHull(vector<point> ps){
     int n = ps.size();if(n<=)return ps;
     sort(ps.begin(),ps.end());
     vector<point> qs(n*);int k=;
     for(int i=;i<n;qs[k++]=ps[i++])
         while (k>&&cross(qs[k-]-qs[k-],ps[i]-qs[k-])<=)--k;
     for(int i=n-,t=k;i>=;qs[k++]=ps[i--])
         while (k>t&&cross(qs[k-]-qs[k-],ps[i]-qs[k-])<=)--k;
     qs.resize(k-);
     return qs;
 }
 void getUDP(vector<point>A,vector<point>&U,vector<point>&D){
     db l=1e7,r=-1e7;
     for (int i=;i<A.size();i++) l=min(l,A[i].x),r=max(r,A[i].x);
     int wherel,wherer;
     for (int i=;i<A.size();i++) if (cmp(A[i].x,l)==) wherel=i;
     for (int i=A.size();i;i--) if (cmp(A[i-].x,r)==) wherer=i-;
     U.clear(); D.clear(); int now=wherel;
     while (){D.push_back(A[now]); if (now==wherer) break; now++; if (now>=A.size()) now=;}
     now=wherel;
     while (){U.push_back(A[now]); if (now==wherer) break; now--; if (now<) now=A.size()-;}
 }
 vector<point>p,u,d;
 db x,y;
 int main(){
     scanf("%d",&n);
     db mx=;
     for(int i=;i<=n;i++){
         scanf("%lld%lld",&x,&y);
         p.push_back(point{x,y-x*x});
         mx+=x;
     }
     p=convexHull(p);
     getUDP(p,u,d);
     set<int>s;
     for(auto p:u)s.insert(p.x);
     printf("%d\n",s.size()-);
 }