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;

}