[bzoj4765] General calculation ji (block + Tree array +dfs order)

Test instructions

Given a tree of n nodes, the node number is 1 to n, Root, and sum[p] represents the weights and values of all nodes in this subtrees tree with point P as root.
The calculation supports the following two operations: 1 given a two integer u,v, the weight of the modified point U is v. 2 given two integers l,r, calculate sum[l]+sum[l+1]+....+sum[r-1]+sum[r]
N<=10^5,m<=10^5

Exercises

Each block of statistics sum[i] and, this directly find the DFS sequence maintenance tree array Nlogn statistics on the line.

Then ask for a whole block directly with our statistics of sum[i], and then for the corner rest, we use a tree-like array to the line.

We need to maintain a tree array directly on the x-v[i] when modifying.

Then we have to maintain the block's and, we need to put and add the number of ancestors of the block I (including i) * (X-v[i]).

So we preprocess the f[i][j] representing the number of ancestors of J (including J) in Block I. This DFS uses an array to record the ancestor situation (add itself when entering, minus itself when it is rolled out), and for each point sweep over each block statistic.

And then this question to use unsigned long long then also can't open, will Mle

1#include <iostream>2#include <cstring>3#include <cstdio>4#include <cmath>5#include <algorithm>6 using namespacestd;7 Const intn=100100;8 intCnt,head[n];9 intSize[n],id[n],tot;Ten intnum[ -],block[n],f[ -][n]; OneUnsignedLong Longsum[ -],a[n],tr[n],ans,v[n]; A intn,m,r[ -],l[ -],block,root; - structedge{ -     intto,nxt; the}e[n*2]; - voidAddintUintv) { -cnt++; -e[cnt].nxt=Head[u]; +e[cnt].to=v; -head[u]=CNT; + } A voidDFS1 (intUintFA) { atsize[u]=1; -id[u]=++tot; -      for(intI=head[u];i;i=e[i].nxt) { -         intv=e[i].to; -         if(V==FA)Continue; - DFS1 (v,u); insize[u]+=Size[v]; -     } to } + voidDFS2 (intUintFA) { -num[block[u]]++; the      for(intI=1; i<=block[n];i++){ *f[i][u]+=Num[i]; $     }Panax Notoginseng      for(intI=head[u];i;i=e[i].nxt) { -         intv=e[i].to; the         if(V==FA)Continue; + DFS2 (v,u); A     } thenum[block[u]]--; + } - intLowbit (intx) { $     returnx&-x; $ } - voidUpdateintx,unsignedLong LongW) { -      for(inti=x;i<=n;i+=lowbit (i)) { thetr[i]+=W; -     }Wuyi } theUnsignedLong LongQueryintx) { -UnsignedLong Longtmp=0; Wu      for(inti=x;i;i-=lowbit (i)) { -tmp+=Tr[i]; About     } $     returntmp; - } - intMain () { -scanf"%d%d",&n,&m); A      for(intI=1; i<=n;i++){ +scanf"%llu",&v[i]); the     } -      for(intI=1; i<=n;i++){ $         intu,v; thescanf"%d%d",&u,&v); the         if(u==0){ theroot=v; the             Continue; -         } in Add (u,v); the Add (v,u); the     } AboutDFS1 (Root,0); the      for(intI=1; i<=n;i++){ the Update (id[i],v[i]); the     } +      for(intI=1; i<=n;i++){ -A[i]=query (id[i]+size[i]-1)-query (id[i]-1); the     //cout<<i<< "<<id[i]<<" "<<size[i]<<" "<<a[i]<<endl;Bayi     } theblock=sqrt (n); the      for(intI=1; i<=n;i++){ -block[i]= (I-1)/block+1; -sum[block[i]]+=A[i]; the         if(! L[block[i]]) l[block[i]]=i; ther[block[i]]=i; the     } theDFS2 (Root,0); -      while(m--){ the         intk,x,y; thescanf"%d%d%d",&k,&x,&y); the         if(k==1){94              for(intI=1; i<=block[n];i++){ thesum[i]+=f[i][x]* (yv[x]); the             } theUpdate (id[x],y-v[x]);98v[x]=y; About         } -         Else{101ans=0;102             if(block[x]+1>=Block[y]) {103                  for(inti=x;i<=y;i++){104Ans+=query (id[i]+size[i]-1)-query (id[i]-1); the                 }106             }107             Else{108                  for(inti=block[x]+1; i<=block[y]-1; i++){109ans+=Sum[i]; the                 }111                  for(inti=x;i<=r[block[x]];i++){ theAns+=query (id[i]+size[i]-1)-query (id[i]-1);113                 } the                  for(inti=l[block[y]];i<=y;i++){ theAns+=query (id[i]+size[i]-1)-query (id[i]-1); the                 }117             }118printf"%llu\n", ans);119         } -     }121     return 0;122}

[bzoj4765] General calculation ji (block + Tree array +dfs order)