[校内训练19_09_02]A

题意

给出N 个形如$f_i(x) = a_i x^2 + b_i x $的二次函数。

有Q 次询问，每次给出一个x，询问$max{\{f_i(x)\}}$。$N,Q \leq 5*10^5$。

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思考

首先将x大于0还是小于0分类，对于某一类全都除以x，那么就得到了一些直线。最优的答案一定在某条最上方或最下方的直线上，半平面交或凸包即可。

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代码

 // luogu-judger-enable-o2
 #pragma GCC optimize 2
 #include<bits/stdc++.h>
 using namespace std;
 typedef long long int ll;
 typedef long double ld;
 const int maxn=5E5+;
 const int base=;
 const ld eps=1E-;
 const ld inf=1E17;
 int n,m,T;
 ll a[maxn],b[maxn],ans[maxn];
 int idR[maxn],idL[maxn];
 ld rightPlane[maxn],leftPlane[maxn];
 int q[maxn];
 struct pt
 {
     double x,y;
     pt(double a=,double b=):x(a),y(b){}
     pt operator+(const pt&A){return pt(x+A.x,y+A.y);}
     pt operator-(const pt&A){return pt(x-A.x,y-A.y);}
     pt operator*(ld d){return pt(x*d,y*d);}
     double operator*(const pt&A){return x*A.y-y*A.x;}
     void out(){cout<<"("<<x<<","<<y<<")"<<endl;}
 };
 struct line
 {
     pt A,B;
     int id,b;
     int slope;
     line(){}
     line(pt a,pt b):A(a),B(b){}
 }s[maxn];
 inline ll max(ll x,ll y)
 {
     return x>y?x:y;
 }
 inline ll min(ll x,ll y)
 {
     return x<y?x:y;
 }
 inline ll get(ll x,ll a,ll b)
 {
     return (a*x+b)*x;
 }
 inline int read()
 {
     bool flag=;
     char ch=getchar();
     if(ch=='-')
         flag^=;
     while(!isdigit(ch))
     {
         ch=getchar();
         if(ch=='-')
             flag^=;
     }
     int sum=ch-'';
     ch=getchar();
     while(isdigit(ch))
     {
         sum=sum*+ch-'';
         ch=getchar();
     }
     return flag?-sum:sum;
 }
 void write(ll x)
 {
     if(x>=)
         write(x/);
     putchar(''+x%);
 }
 inline void writen(ll x)
 {
     if(x<)
     {
         putchar('-');
         write(-x);
     }
     else
         write(x);
     putchar('\n');
 }
 inline int cross(pt A,pt B)
 {
     ld d=A*B;
     if(abs(d)<=eps)
         return ;
     return d>?:-;
 }
 inline pt intersection(line a,line b)
 {
     pt A=b.B-b.A,B=a.B-a.A,C=b.A-a.A;
     if(cross(A,B)==)
         return pt(inf,inf);
     ld d=-(B*C)/(B*A);
     return b.A+A*d;
 }
 inline int onClockwise(line a,line b,line c)
 {
     return cross(a.B-a.A,intersection(b,c)-a.A)==-;
 }
 bool cmpR(line A,line B)
 {
     if(A.slope==B.slope)
         return A.b<B.b;
     return A.slope<B.slope;
 }
 bool cmpL(line A,line B)
 {
     if(A.slope==B.slope)
         return A.b>B.b;
     return A.slope<B.slope;
 }
 void initR()
 {
     sort(s+,s+n+,cmpR);
     int L=,R=;
     for(int i=;i<=n;++i)
     {
         while(i!=n&&s[i].slope==s[i+].slope)
             ++i;
         while(R->=L&&onClockwise(s[i],s[q[R]],s[q[R-]]))
             --R;
         q[++R]=i;
     }
     for(int i=;i<=R;++i)
         idR[i]=s[q[i]].id;
     for(int i=;i<R;++i)
         rightPlane[i]=intersection(s[q[i]],s[q[i+]]).x;
     int pos=;
     for(int i=-base;i<=base;++i)
     {
         while(pos<R&&rightPlane[pos]<ld(i))
             ++pos;
         ans[i+base]=get(i,s[q[pos]].slope,s[q[pos]].b);
     }
 }
 void initL()
 {
     sort(s+,s+n+,cmpL);
     int L=,R=;
     for(int i=;i<=n;++i)
     {
         while(i!=n&&s[i].slope==s[i+].slope)
             ++i;
         while(R->=L&&onClockwise(s[i],s[q[R]],s[q[R-]]))
             --R;
         q[++R]=i;
     }
     for(int i=;i<=R;++i)
         idL[i]=s[q[i]].id;
     for(int i=;i<R;++i)
         leftPlane[i]=intersection(s[q[i]],s[q[i+]]).x;
     int pos=;
     for(int i=base;i>=-base;--i)
     {
         while(pos<R&&ld(i)<leftPlane[pos])
             ++pos;
         ans[i+base]=max(ans[i+base],get(i,s[q[pos]].slope,s[q[pos]].b));
     }
 }
 int main()
 {
 //    freopen("A.in","r",stdin);
 //    freopen("A.out","w",stdout);
     ios::sync_with_stdio(false);
     n=read(),T=read();
     for(int i=;i<=n;++i)
     {
         a[i]=read(),b[i]=read();
         s[i].A=pt(,b[i]);
         s[i].B=pt(,a[i]+b[i]);
         s[i].slope=a[i];
         s[i].b=b[i];
         s[i].id=i;
     }
     initR();
     for(int i=;i<=n;++i)
         swap(s[i].A,s[i].B);
     initL();
     while(T--)
     {
         int x=read();
         writen(ans[x+base]);
     }
     return ;
 }