P4735 largest XOR and persistence trie tree

 

 

 

 

Maximum sustainable chairman of 01trie similar idea but the tree is not used as a poor version just need to maintain a point to the left

#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);--i)
#define ll long long
#define see(x) (cerr<<(#x)<<'='<<(x)<<endl)
#define inf 0x3f3f3f3f
#define CLR(A,v)  memset(A,v,sizeof A)
//////////////////////////////////
const int N=2e6+10;
int t[N<<5][2],Lmax[N<<5],ncnt,n,m,x,y,z,T[N],s[N];
char ss[2];
void upnode(int k,int i,int pre,int pos)
{
    if(k<0){Lmax[pos]=i;return ;}
    bool c=(s[i]>>k)&1;
    if(pre)t[pos][c^1]=t[pre][c^1];
    t[pos][c]=++ncnt;
    upnode(k-1,i,t[pre][c],t[pos][c]);
    Lmax[pos]=max(Lmax[t[pos][0]],Lmax[t[pos][1]]);
}
int qmax(int k,int val,int l,int pos)
{
    if(k<0)return val^s[Lmax[pos]];
    bool c=(val>>k)&1;
    if(Lmax[t[pos][c^1]]>=l)return qmax(k-1,val,l,t[pos][c^1]);
    else return qmax(k-1,val,l,t[pos][c]);
}
int main()
{
    scanf("%d%d",&n,&m);Lmax[0]=-1;
    T[0]=++ncnt;upnode(23,0,0,T[0]);
    rep(i,1,n)
    {
        int x;scanf("%d",&x);
        s[i]=s[i-1]^x;
        T[i]=++ncnt;
        upnode(23,i,T[i-1],T[i]);
    }int l,r;
    while(m--)
    {
        scanf("%s",ss);
        if(ss[0]=='A')
        {    
            int x;scanf("%d",&x);n++;s[n]=s[n-1]^x;
            T[n]=++ncnt;upnode(23,n,T[n-1],T[n]);
        }
        else 
        {
            scanf("%d%d%d",&l,&r,&x);
            int val=x^s[n];
            printf("%d\n",qmax(23,val,l-1,T[r-1]));
        }
    }
    return 0;
}

View Code