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设(x,y)为Q的查询点,分类讨论如下:
1、y>0: 最大化a*x+b*y,维护一个上凸壳三分即可
2、y<0:最大化a*x+b*y 维护一个下凸壳三分即可
我们考虑对时间建出一棵线段树
对于每个区间,如果满了就做出两个凸壳
总时间复杂度是O(n*log^2n)
之后我们考虑查询,每个区间最多被分解为log(n)个区间
在每个区间的凸壳上三分最优解即可
至于优化,可以设定一个阈值,当区间长度小于阈值时不用做凸壳,查询时直接暴力就可以了
#include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> using namespace std; typedef long long LL; const int maxn=400010; const LL oo=1LL<<60; char s,Q; int n,cnt,a,b,T,TA,TB; LL ans; struct Point{ int x,y; Point(int x=0,int y=0):x(x),y(y){} }p[maxn],tmp[maxn],now,A[4000010],B[4000010]; typedef Point Vector; Vector operator -(const Point &A,const Point &B){return Vector(A.x-B.x,A.y-B.y);} Vector operator +(const Point &A,const Point &B){return Vector(A.x+B.x,A.y+B.y);} inline LL Dot(const Point &A,const Point &B){return 1LL*A.x*B.x+1LL*A.y*B.y;} inline LL Cross(const Point &A,const Point &B){return 1LL*A.x*B.y-1LL*A.y*B.x;} struct Seg_Tree{ int L,R; int AL,AR; int BL,BR; }t[maxn<<2]; bool cmp1(const Point &a,const Point &b){ if(a.x==b.x)return a.y>b.y; return a.x<b.x; } bool cmp2(const Point &a,const Point &b){ if(a.x==b.x)return a.y<b.y; return a.x<b.x; } inline void decode(int &x){x=(s==‘E‘?x:x=x^(ans&0x7fffffff));} inline void Get(char &ch){ ch=getchar(); while(ch<‘!‘)ch=getchar(); } inline void read(int &num){ num=0;int f=1;char ch;Get(ch); if(ch==‘-‘)f=-1,ch=getchar(); while(ch>=‘0‘&&ch<=‘9‘)num=num*10+ch-‘0‘,ch=getchar(); num*=f; } void build(int o,int L,int R){ t[o].L=L;t[o].R=R; if(L==R)return; int mid=(L+R)>>1; build(o<<1,L,mid); build(o<<1|1,mid+1,R); } void add(int o){ int L=t[o].L,R=t[o].R; int mid=(L+R)>>1; if(R==cnt&&R-L>=70){ T=0; for(int i=L;i<=R;++i)tmp[++T]=p[i]; sort(tmp+1,tmp+T+1,cmp1); A[++TA]=tmp[1];t[o].AL=TA; for(int i=2;i<=T;++i){ if(tmp[i].x!=tmp[i-1].x){ while(TA>t[o].AL&&Cross(tmp[i]-A[TA],A[TA]-A[TA-1])<=0)TA--; A[++TA]=tmp[i]; } }t[o].AR=TA; sort(tmp+1,tmp+T+1,cmp2); B[++TB]=tmp[1];t[o].BL=TB; for(int i=2;i<=T;++i){ if(tmp[i].x!=tmp[i-1].x){ while(TB>t[o].BL&&Cross(tmp[i]-B[TB],B[TB]-B[TB-1])>=0)TB--; B[++TB]=tmp[i]; } }t[o].BR=TB; } if(L==R)return; if(cnt<=mid)add(o<<1); else add(o<<1|1); } LL Get_A(int L,int R){ while(R-L>=3){ int m1=(L+L+R)/3,m2=(R+R+L)/3; if(Dot(now,A[m1])>Dot(now,A[m2]))R=m2; else L=m1; } LL ans=-oo; for(int i=L;i<=R;++i)ans=max(ans,Dot(now,A[i])); return ans; } LL Get_B(int L,int R){ while(R-L>=3){ int m1=(L+L+R)/3,m2=(R+R+L)/3; if(Dot(now,B[m1])>Dot(now,B[m2]))R=m2; else L=m1; } LL ans=-oo; for(int i=L;i<=R;++i)ans=max(ans,Dot(now,B[i])); return ans; } LL Get_ask(int o){ LL ans=-oo; int L=t[o].L,R=t[o].R; int mid=(L+R)>>1; if(a<=L&&R<=b){ if(R-L<70)for(int i=L;i<=R;++i)ans=max(ans,Dot(now,p[i])); else if(now.y>0)ans=Get_A(t[o].AL,t[o].AR); else ans=Get_B(t[o].BL,t[o].BR); return ans; } if(b<=mid)return Get_ask(o<<1); else if(a>mid)return Get_ask(o<<1|1); else return max(Get_ask(o<<1),Get_ask(o<<1|1)); } int main(){ read(n);Get(s);build(1,1,n); for(int i=1;i<=n;++i){ Get(Q); if(Q==‘A‘){ ++cnt; read(p[cnt].x);read(p[cnt].y); decode(p[cnt].x);decode(p[cnt].y); add(1); }else{ read(now.x);read(now.y); decode(now.x);decode(now.y); read(a);read(b); decode(a);decode(b); ans=Get_ask(1); printf("%lld\n",ans); } }return 0; }
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原文地址:http://www.cnblogs.com/joyouth/p/5350714.html