首先詢問i相當於詢問a[j]>=a[i],b[j]>=b[i]的j
如果b[j]==b[i],那麼a[j]>a[i],這種情況先用set處理掉
如果b[j]>b[i],那麼a[j]>=a[i],離散化後CDQ分治,用樹狀陣列記錄字首最大值即可
時間複雜度$O(n\log^2n)$
#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; }