[BZOJ4523] Routing table

For each query, we can use an array to maintain the insertion time at the first match length of I to calculate how many times it changed in [L,r]

Since the length is now monotonous, the selection will change when and only when the time is monotonous,

So we can change the number of times in [1,x] by the monotone stack calculation.

Ans=solve (R)-solve (L-1)

For multiple queries, we can insert queries into the trie tree, implementing multiple queries to share a maintenance array

1#include <bits/stdc++.h>2 using namespacestd;3typedefLong Longll;4 #defineMAXN 10000055 #defineMaxnd 150000006 structnode{ll X;intL,r;} A[MAXN],Q[MAXN];7 intn,aa,qq;8 inttrie[maxnd][2],nd,mark[maxnd],sta[ +];9 intRead () {Ten     inttmp=0;CharCh=0; One      while(ch<'0'|| Ch>'9') ch=GetChar (); A      while(ch>='0'&&ch<='9') tmp=tmp*Ten+ch-'0', ch=GetChar (); -     returntmp; - } the voidInsert (ll x) { -     intp=0; -      for(intI= +; i;i--){ -         intT=x>> (I-1) &1; +         if(!trie[p][t]) trie[p][t]=++nd; -p=Trie[p][t]; +     } A } at voidinit () { -n=read (); -     Charop[3]; -      for(intI=1; i<=n;i++){ -scanf"%s", op); -         if(op[0]=='A'){ inaa++; -              for(intj=0;j<4; j + +) toA[aa].x= (a[aa].x<<8)|read (); +A[aa].l=read (); -         } the         Else{ *qq++; $              for(intj=0;j<4; j + +)Panax NotoginsengQ[qq].x= (q[qq].x<<8)|read (); -Q[qq].l=read (), q[qq].r=read (); the Insert (q[qq].x); +         } A     } the } + voidGetmark (intPOS) { -     intp=0; $      for(intI= +;i> +-a[pos].l;i--){ $         intT=a[pos].x>> (I-1) &1; -         if(!trie[p][t])return; -p=Trie[p][t]; the     } -     if(!mark[p]) mark[p]=Pos;Wuyi } the intGetans (ll X,intrng) { -     intp=0, top=0; Wu      for(intI= +; i;i--){ -         intT=x>> (I-1) &1; Aboutp=Trie[p][t]; $         if(mark[p]&&mark[p]<=rng) { -              while(Top&&mark[p]<sta[top]) top--; -sta[++top]=Mark[p]; -         } A     } +     returntop; the } - voidsolve () { $      for(intI=1; i<=aa;i++) the Getmark (i); the      for(intI=1; i<=qq;i++){ the         intAns1=getans (q[i].x,q[i].l-1); the         intAns2=Getans (Q[I].X,Q[I].R); -printf"%d\n", ans2-ans1); in     } the } the intMain () { About init (); the solve (); the     return 0; the}

View Code

[BZOJ4523] Routing table