On the tree

LCA: Tree section, multiplication, Tarjan

• Tree section - O (logn) (n < = 10 ^ 6)

• https://www.cnblogs.com/cangT-Tlan/p/8846408.html

• The tree is divided into a chain of several, to maintain a data structure of each chain

• #include<bits/stdc++.h>
  #define ll long long
  #define rll register ll
  #define M 0x3f3f3f
  #define For(i,l,r) for(int i=l;i<=r;i++)
  using namespace std; 
  ll n,m,s,head[M],a[M],x,y,z,tot,k,t;
  ll fa[M],d[M],top[M],size[M],id[M],ril[M];
  struct node1{
      ll to,nxt;
  }e[M<<1];
  struct node2{
      ll l,r,sum,flag;
  }tree[M<<1];
  
  inline ll read(){
      ll f=1,sum=0;
      char ch=getchar();
      while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
      while(isdigit(ch)){sum=(sum<<1)+(sum<<3)+(ch^48);ch=getchar();}
      return f*sum;
  }
  
  inline void add(ll x,ll y){
      e[++tot].to=y;
      e[tot].nxt=head[x];
      head[x]=tot;
  }
  
  inline void dfs1(ll u){//Dfs first pass through the tree, the depth d pretreatment, size this point node number, its parent node FA 
      d [U] = d [FA [U]] + . 1 ; 
      size [U] = . 1 ;
       for (I = RLL head [U]; I; I = E [I] .nxt) {
           IF ! (E [I] .to = FA [U]) { 
              FA [E [I] .to] = U; 
              DFS1 (E [I ] .to); 
              size [U] + = size [E [I] .to]; 
          } 
      } 
  } 
  
  inline void DFS2 (U LL) { // the number of child nodes according to the severity of side division number, ensure that only a point in a chain. Usually only the heavy side and light side less than. 
      T = LL 0 ; // Top i.e. points where the edges for this vertex heavy  
      IF Top [U] = (Top [U]!) U;
       for(RLL head I = [U]; I; I = E [I] .nxt) {
           IF (! E [I] .to FA = [U] && size [E [I] .to]> T) T = E [I] .to; 
      } 
      IF (T) { 
          Top [T] = Top [U]; 
          DFS2 (T); 
      } 
      for (RLL head I = [U]; I; I = E [I] .nxt) {
           iF (!! E [I] .to FA = [U] && E [I] .to = T) DFS2 (E [I] .to); 
      } 
  } 
  
  inline LCA LL (LL X, Y LL) { // when two points are located on a heavy chain that is the same operation is ended. Otherwise skip to point deeper depth where a point on the heavy chain, shallow depth position at the end point of the operation is also desired LCA
       the while (Top [X]! = Top [Y]) {
           IF (D [Top [X ]] < D [Top [Y]]) the swap (X, Y); 
          X = FA [Top [X]]; 
      } 
      IF(d[x]>d[y]) swap(x,y);
      return x;
  }
  
  int main(){
      n=read(),m=read(),s=read();
      For(i,1,n-1) {x=read(),y=read(),add(x,y),add(y,x);}
      dfs1(s),dfs2(s);
      For(i,1,m){x=read(),y=read(),printf("%lld\n",lca(x,y));}
      return 0;
  }

• topic:

1. Hamster find sugar