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)

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <algorithm>
 4 #include <vector>
 5 #include <set>
 6 using namespace std;
 7 typedef long long db;
 8 const double eps=1e-6;
 9 const double pi=acos(-1);
10 int sign(db k){
11     if (k>eps) return 1; else if (k<-eps) return -1; return 0;
12 }
13 int cmp(db k1,db k2){return sign(k1-k2);}
14 struct point{
15     db x,y;
16     point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
17     point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
18     point operator * (db k1) const{return (point){x*k1,y*k1};}
19     point operator / (db k1) const{return (point){x/k1,y/k1};}
20     bool operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;}
21     bool operator <(const point &k1)const {
22         int c=cmp(x,k1.x);
23         if(c)return c==-1;
24         return cmp(y,k1.y)==-1;
25     }
26 };
27 db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
28 db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
29 int n;
30 vector<point> convexHull(vector<point> ps){
31     int n = ps.size();if(n<=1)return ps;
32     sort(ps.begin(),ps.end());
33     vector<point> qs(n*2);int k=0;
34     for(int i=0;i<n;qs[k++]=ps[i++])
35         while (k>1&&cross(qs[k-1]-qs[k-2],ps[i]-qs[k-2])<=0)--k;
36     for(int i=n-2,t=k;i>=0;qs[k++]=ps[i--])
37         while (k>t&&cross(qs[k-1]-qs[k-2],ps[i]-qs[k-2])<=0)--k;
38     qs.resize(k-1);
39     return qs;
40 }
41 void getUDP(vector<point>A,vector<point>&U,vector<point>&D){
42     db l=1e7,r=-1e7;
43     for (int i=0;i<A.size();i++) l=min(l,A[i].x),r=max(r,A[i].x);
44     int wherel,wherer;
45     for (int i=0;i<A.size();i++) if (cmp(A[i].x,l)==0) wherel=i;
46     for (int i=A.size();i;i--) if (cmp(A[i-1].x,r)==0) wherer=i-1;
47     U.clear(); D.clear(); int now=wherel;
48     while (1){D.push_back(A[now]); if (now==wherer) break; now++; if (now>=A.size()) now=0;}
49     now=wherel;
50     while (1){U.push_back(A[now]); if (now==wherer) break; now--; if (now<0) now=A.size()-1;}
51 }
52 vector<point>p,u,d;
53 db x,y;
54 int main(){
55     scanf("%d",&n);
56     db mx=0;
57     for(int i=1;i<=n;i++){
58         scanf("%lld%lld",&x,&y);
59         p.push_back(point{x,y-x*x});
60         mx+=x;
61     }
62     p=convexHull(p);
63     getUDP(p,u,d);
64     set<int>s;
65     for(auto p:u)s.insert(p.x);
66     printf("%d\n",s.size()-1);
67 }

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