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$O(n\log^2n)$

 

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#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; }