BZOJ2773 : ispiti

首先询问i相当于询问a[j]>=a[i],b[j]>=b[i]的j

如果b[j]==b[i]，那么a[j]>a[i]，这种情况先用set处理掉

如果b[j]>b[i]，那么a[j]>=a[i]，离散化后CDQ分治，用树状数组记录前缀最大值即可

时间复杂度$O(n\log^2n)$

#include<cstdio>
#include<set>
#include<algorithm>
#define N 200010
using namespace std;
typedef pair<int,int> PI;
int n,q,cnt,i,j,x,y,t1,t2,bit[N],pos[N],T,ans[N],L[N];
char op;
set<PI>Set[N];
set<PI>::iterator it;
struct P{int x,y,id;P(){}P(int _x,int _y,int _id){x=_x,y=_y,id=_id;}}a[N],b[N],c[N],stu[N];
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=cnt,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.x>b.x;}
inline int merge(int x,int y){
  if(!x)return y;
  if(!y)return x;
  if(stu[x].y==stu[y].y)return stu[x].x<stu[y].x?x:y;
  return stu[x].y<stu[y].y?x:y;
}
inline void add(int x,int y){for(;x<=n;x+=x&-x)if(pos[x]<T)pos[x]=T,bit[x]=y;else bit[x]=merge(bit[x],y);}
inline int ask(int x){int t=0;for(;x;x-=x&-x)if(pos[x]==T)t=merge(t,bit[x]);return t;}
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<0)b[t1++]=a[i];
  for(t2=0,i=r;i>mid;i--)if(a[i].id>0)c[t2++]=a[i];
  if(!t2)return;
  sort(b,b+t1,cmp),sort(c,c+t2,cmp);
  for(T++,i=j=0;i<t2;i++){
    while(j<t1&&b[j].x>=c[i].x)add(b[j].y,-b[j].id),j++;
    ans[c[i].id]=merge(ans[c[i].id],ask(c[i].y-1));
  }
}
int main(){
  read(n);
  for(i=1;i<=n;i++){
    while(!(((op=getchar())=='D')||(op=='P')));
    read(x);
    if(op=='D')read(y),L[++cnt]=y,stu[cnt]=a[i]=P(x,y,-cnt);else a[i]=stu[x],a[i].id=++q;
  }
  sort(L+1,L+cnt+1);
  for(i=1;i<=n;i++){
    a[i].y=cnt-lower(a[i].y)+1;
    if(a[i].id<0)Set[a[i].y].insert(PI(a[i].x,-a[i].id));
    else{
      it=Set[a[i].y].lower_bound(PI(a[i].x+1,0));
      if(it!=Set[a[i].y].end())ans[a[i].id]=it->second;
    }
  }
  solve(1,n);
  for(i=1;i<=q;i++)if(!ans[i])puts("NE");else printf("%d\n",ans[i]);
  return 0;
}