[POJ3237] 树的维护 题解

一眼树链剖分或 \(LCT\)，由于在学后者所以就写了。

取反操作相当于把 \(min,max\) 取反后交换，所以要维护 \(min,max,val\)。

时间复杂度 \(O(m\log n)\)。

#include<bits/stdc++.h>
#define fa(x) lct[x].fa
#define fl(x) lct[x].fl
#define mx(x) lct[x].mx
#define mn(x) lct[x].mn
#define id(x) lct[x].id
#define val(x) lct[x].val
#define sn(x,i) lct[x].sn[i]
#define inf 1000000000
using namespace std;
const int N=20005;
struct node{
    int sn[2],fa,fl,mx,mn,val,id;
}lct[N];int n,m,tp,st[N];
int check(int x){
    return sn(fa(x),0)!=x&&sn(fa(x),1)!=x;
}int chksn(int x){
    return sn(fa(x),1)==x;
}void push_up(int x){
    mx(x)=max({mx(sn(x,0)),mx(sn(x,1)),x>n?val(x):-inf});
    mn(x)=min({mn(sn(x,0)),mn(sn(x,1)),x>n?val(x):inf});
}void push_down(int x){
    if(!x) return;
    if(fl(x)){
        fl(sn(x,0))^=1,fl(sn(x,1))^=1;
        swap(sn(x,0),sn(x,1)),fl(x)=0;
    }if(!id(x)) return;id(x)=0,
    swap(mx(sn(x,0)),mn(sn(x,0)));
    swap(mx(sn(x,1)),mn(sn(x,1)));
    id(sn(x,0))^=1,id(sn(x,1))^=1;
    val(sn(x,0))=-val(sn(x,0));
    val(sn(x,1))=-val(sn(x,1));
    mx(sn(x,0))=-mx(sn(x,0));
    mn(sn(x,0))=-mn(sn(x,0));
    mx(sn(x,1))=-mx(sn(x,1));
    mn(sn(x,1))=-mn(sn(x,1));
}void rotate(int x){
    int y=fa(x),z=fa(y),k=chksn(x);
    if(!check(y))
        sn(z,chksn(y))=x;
    fa(x)=z,fa(y)=x,fa(sn(x,1-k))=y;
    sn(y,k)=sn(x,1-k),sn(x,1-k)=y;
    push_up(y);
}void splay(int x){
    st[tp=1]=x;
    for(int i=x;!check(i);i=fa(i)) st[++tp]=fa(i);
    while(tp) push_down(st[tp--]);
    while(!check(x)){
        int y=fa(x),z=fa(y);
        if(!check(y))
            rotate(chksn(x)!=chksn(y)?x:y);
        rotate(x);
    }push_up(x);
}void access(int x){
    for(int i=0;x;i=x,x=fa(x))
        splay(x),sn(x,1)=i,push_up(x);
}void mk(int x){
    access(x),splay(x),fl(x)^=1;
}void split(int x,int y){
    mk(x),access(y),splay(y);
}void link(int x,int y){
    mk(x),access(y),fa(x)=y;
}int main(){
    ios::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    cin>>n,mx(n)=-inf,mn(n)=inf;
    mx(0)=-inf,mn(0)=inf;
    for(int i=1;i<n;i++){
        mx(i)=-inf,mn(i)=inf;
        int x,y,z;cin>>x>>y>>z;
        mx(n+i)=mn(n+i)=val(n+i)=z;
        link(x,n+i),link(n+i,y);
    }while(1){
        string s;int x,y;cin>>s;
        if(s=="DONE") break;
        cin>>x>>y;
        if(s=="NEGATE"){
            split(x,y),id(y)^=1;
            swap(mx(y),mn(y)),val(y)=-val(y);
            mx(y)=-mx(y),mn(y)=-mn(y);
        }if(s=="CHANGE")
            mk(n+x),mn(n+x)=mx(n+x)=val(n+x)=y;
        if(s=="QUERY") split(x,y),cout<<mx(y)<<"\n";
    }return 0;
}