BZOJ3290 : Theresa与数据结构

CANCEL操作可以看作删点，X坐标可以离散化

将询问按Z坐标差分，转化成两个求Z<=某个数的和的询问

将操作CDQ分治

每次将前一半的修改、后一半的查询按照Z坐标排序

然后扫描线，每到一个询问就把所有Z<=它的修改都加入

树状数组套Treap维护

时间复杂度$O(nlog^3n)$

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#define N 600010
using std::sort;
int n,m,Q,i,j,x,y,z,r,t1,t2,fin[N],L[N],li,k,q[N],t,pos[N],T,C,D,ans;
char ch;
struct P{
  int x1,y1,x2,y2,z,t,id;
  P(){}
  P(int _x1,int _y1,int _x2,int _y2,int _z,int _t,int _id){x1=_x1,y1=_y1,x2=_x2,y2=_y2,z=_z,t=_t,id=_id;}
}a[N],b[N],c[N];
struct node{
  int p,val,v,sum;node*l,*r;
  node(){val=v=sum=p=0;l=r=NULL;}
  inline void up(){sum=v+l->sum+r->sum;}
}*blank=new(node),*bit[N],pool[N<<1],*cur=pool;
inline void Rotatel(node*&x){node*y=x->r;x->r=y->l;x->up();y->l=x;y->up();x=y;}
inline void Rotater(node*&x){node*y=x->l;x->l=y->r;x->up();y->r=x;y->up();x=y;}
void Ins(node*&x){
  if(x==blank){
    x=cur++;x->val=C;x->l=x->r=blank;x->v=x->sum=D;x->p=std::rand();
    return;
  }
  x->sum+=D;
  if(C==x->val){x->v+=D;return;}
  if(C<x->val){
    Ins(x->l);
    if(x->l->p>x->p)Rotater(x);
  }else{
    Ins(x->r);
    if(x->r->p>x->p)Rotatel(x);
  }
}
void Ask(node*&x,int a,int b){
  if(x==blank)return;
  if(C<=a&&b<=D){ans+=x->sum;return;}
  if(C<=x->val&&x->val<=D)ans+=x->v;
  if(C<x->val)Ask(x->l,a,x->val-1);
  if(D>x->val)Ask(x->r,x->val+1,b);
}
inline void read(int&a){char c;while(!(((c=getchar())>='0')&&(c<='9')));a=c-'0';while(((c=getchar())>='0')&&(c<='9'))(a*=10)+=c-'0';}
inline int lower(int x){
  int l=1,r=k,t,mid;
  while(l<=r)if(L[mid=(l+r)>>1]<=x)l=(t=mid)+1;else r=mid-1;
  return t;
}
inline bool cmp(P a,P b){return a.z<b.z;}
inline void add(int x){for(;x<=k;Ins(bit[x]),x+=x&-x)if(pos[x]<T)pos[x]=T,bit[x]=blank;}
inline int sum(int x){for(ans=0;x;x-=x&-x)if(pos[x]==T)Ask(bit[x],0,10000000);return ans;}
void solve(int l,int r){
  if(l==r)return;
  int mid=(l+r)>>1;
  solve(l,mid),solve(mid+1,r);
  for(t1=0,i=l;i<=mid;i++)if(!a[i].id)b[++t1]=a[i];
  for(t2=0,i=r;i>mid;i--)if(a[i].id)c[++t2]=a[i];
  if(!t1||!t2)return;
  sort(b+1,b+t1+1,cmp),sort(c+1,c+t2+1,cmp);
  for(T++,cur=pool,i=j=1;i<=t2;i++){
    while(j<=t1&&b[j].z<=c[i].z)C=b[j].y1,D=b[j].t,add(b[j].x1),j++;
    C=c[i].y1,D=c[i].y2,fin[c[i].id]+=c[i].t*(sum(c[i].x2)-sum(c[i].x1-1));
  }
}
int main(){
  blank->l=blank->r=blank;
  read(n);
  while(n--)read(x),read(y),read(z),a[++m]=P(x,y,x,y,z,1,0);
  read(n);
  for(i=1;i<=n;i++){
    while(!(((ch=getchar())=='A')||(ch=='Q')||(ch=='C')));
    if(ch=='A')read(x),read(y),read(z),a[q[++t]=++m]=P(x,y,x,y,z,1,0);
    if(ch=='C'){
      while((ch=getchar())!='L');
      a[++m]=a[q[t--]],a[m].t=-1;
    }
    if(ch=='Q')read(x),read(y),read(z),read(r),a[++m]=P(x,y,x+r,y+r,z+r,1,++Q),a[++m]=P(x,y,x+r,y+r,z-1,-1,Q);
  }
  for(i=1;i<=m;i++)L[++li]=a[i].x1,L[++li]=a[i].x2;
  sort(L+1,L+li+1);
  for(i=1;i<=li;i++)if(i==1||L[i-1]!=L[i])L[++k]=L[i];
  for(i=1;i<=m;i++)a[i].x1=lower(a[i].x1),a[i].x2=lower(a[i].x2);
  solve(1,m);
  for(i=1;i<=Q;i++)printf("%d
",fin[i]);
  return 0;
}