"Mo Team Algorithm" bzoj3289 Mato's file management

MO Team algorithm, offline answer inquiry, by a certain size (sqrt (N*log (n)) the answer block, according to the ① left end of the block ② the right end of the double keyword sorting.

And then the violence shifts.

Use a tree-like array when transferring.

O (N*SQRT (n) *log (n)).

Note: ① adds a number after the number of columns, and the inverse logarithm increases the number of numbers that are larger in the sequence.

② deletes a number after the number of columns, and the inverse logarithm decreases the number of numbers in the series that are larger than it.

③ adds a number to the front of a column number, and the inverse logarithm increases the number of smaller numbers in the sequence.

④ deletes a number in front of a column number, and the inverse logarithm decreases the number of smaller numbers in the sequence.

1#include <cstdio>2#include <algorithm>3#include <cmath>4 using namespacestd;5 structpoint{intx, y;} a[50001];6 structask{intL,r,lb,p;} q[50001];7 BOOL operator< (ConstASK &a,ConstASK &b) {returnA.lb!=b.lb? a.lb<b.lb:a.r<B.R;}8 BOOL operator< (ConstPoint &a,ConstPoint &b) {returna.x<b.x;}9 intn,m,sz,sum,num[50001],b[50001],cnt;Ten intd[50001],ans[50001],res; OneInlineintLowbit (Const int&AMP;X) {returnx& (-x);} AInlinevoidAddintPConst int&AMP;X) { while(p<=n) {d[p]+=x;p+=lowbit (P);}} -InlineintGetsum (intP) {intres=0; while(p) {Res+=d[p];p-=lowbit (p);}returnRes;} - intRes,num;Charc,ch[ A]; theInlineintG () - { -res=0; C='*'; -      while(c<'0'|| C>'9') c=GetChar (); +      while(c>='0'&&c<='9') {res=res*Ten+ (C-'0'); C=GetChar ();} -     returnRes; + } AInlinevoidPLong Longx) at { -num=0;if(!x) {Putchar ('0');p UTS ("");return;} -      while(x>0) ch[++num]=x%Ten, x/=Ten; -      while(Num) Putchar (ch[num--]+ -); -Puts""); - } in voidMakeblock () - { toSz=sqrt ((Double) n4.6);intL,r; +      for(sum=1; sum*sz<n;sum++) -       { theL= (sum-1) *sz+1; r=sum*sz; *          for(inti=l;i<=r;i++) num[i]=sum; $       }Panax Notoginsengl=sz* (sum-1)+1; R=N; -      for(inti=l;i<=r;i++) num[i]=sum; the } + voidLisan () A { the     inten=1; Sort (A +1, a+n+1); b[a[1].y]=1; +      for(intI=2; i<=n;i++) -       { $         if(a[i].x!=a[i-1].x) en++; $b[a[i].y]=en; -       } - } the intMain () - {Wuyin=G (); the      for(intI=1; i<=n;i++) -       { Wua[i].x=G (); -a[i].y=i; About       } $Makeblock (); Lisan (); m=G (); -      for(intI=1; i<=m;i++) -       { -Q[i].l=g (); Q[i].r=G (); Aq[i].lb=NUM[Q[I].L]; +q[i].p=i; the       } -Sort (q+1, q+m+1); $      for(inti=q[1].l;i<=q[1].r;i++) the       { thecnt++; theAdd (B[i],1); theres+= (cnt-getsum (B[i])); -} ans[q[1].p]=Res; in      for(intI=2; i<=m;i++) the       { the         if(q[i].l<q[i-1].l) About           { the              for(intj=q[i-1].l-1; j>=q[i].l;j--) the               { theRes+=getsum (b[j]-1); +Add (B[j],1); -               } thecnt+= (q[i-1].l-q[i].l);Bayi           } the         Else the           { -              for(intj=q[i-1].l;j<q[i].l;j++) -               { theRes-=getsum (b[j]-1); theAdd (b[j],-1); the               } thecnt-= (q[i].l-q[i-1].l); -           } the         if(q[i].r<q[i-1].R) the           { the              for(intj=q[i-1].r;j>q[i].r;j--)94               { theres-= (cnt-getsum (B[j])); thecnt--; theAdd (b[j],-1);98               } About           } -         Else101           {102              for(intj=q[i-1].r+1; j<=q[i].r;j++)103               {104cnt++; theAdd (B[j],1);106res+= (cnt-getsum (B[j]));107               }108           }109ans[q[i].p]=Res; the       }111      for(intI=1; i<=m;i++) P (Ans[i]); the     return 0;113}

"Mo Team Algorithm" bzoj3289 Mato's file management