Luo Gu P4735 solution to a problem

If you want in-depth study can be persistent 0-1Trie trees portal .

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Description:

A given number of columns \ (\ {A_N \} \) , supports two modes of operation:

• Add the number of columns in a number of tail \ (X \) , the number of column length becomes \ (n-+. 1 \) ;
• Given the closed interval \ ([l, r] \ ) and a number \ (X \) , requirements:

\[ \max_{i=l}^{r}\left \{\left(\bigoplus_{j=i}^{n}a_j \right)\bigoplus x\right \} \]

Method:

Defined \ (Xorsum_i \) of \ (\ bigoplus_. 1 = {I}} ^ {n-a_i \) , i.e. the prefix and XOR. Obviously, we can get
\ [\ left (\ bigoplus_ { i = pos} ^ {n} a_i \ right) \ bigoplus x = Xorsum_ {pos-1} \ bigoplus Xorsum_n \ bigoplus x \]

NOTE : \ (X \ Bigoplus X = 0 \) , \ (X \ Bigoplus 0 = X \)

We found \ (Xorsum_n \ bigoplus x \) is a fixed value , we only need to maintain \ (Xorsum_ {pos-1} \) can.

May consider using persistence 0-1Trie tree maintenance. Chairman of the tree the same way, we build \ (n + 1 \) version of 0-1Trie tree, when the query using greedy ideas can be.

Persistence segment tree can also support " prefix and " thinking, we need only last the first \ (r \) version of 0-1Trie trees look \ (l \) to position.

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Code:

#include<bits/stdc++.h>
#define Maxn 600010
#define Maxdep 23
#define getchar()(p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char buf[1<<21],*p1=buf,*p2=buf;
inline void read(int &x)
{
    int f=1;x=0;char s=getchar();
    while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
    while(s>='0'&&s<='9'){x=x*10+s-'0';s=getchar();}
    x*=f;
}
int n,m;
int sum[Maxn];
struct trie
{
    trie *chd[2];
    int symbl;
    trie()
    {
        for(int i=0;i<2;i++) chd[i]=NULL;
        symbl=0;
    }
}*root[Maxn],tree[Maxn<<5],*tail;
void Init(){tail=tree;} 
void build(trie *&p,int dep)
{
    p=new (tail++)trie();
    if(dep<0) return ;
    build(p->chd[0],dep-1);
}
void update(trie *&p,trie *flag,int dep,int i)
{
    p=new (tail++)trie();
    if(flag) *p=*flag;
    if(dep<0) return (void)(p->symbl=i);
    int tmp=(sum[i]>>dep)&1;//判断是1还是0 
    if(!tmp) update(p->chd[0],flag?flag->chd[0]:NULL,dep-1,i);
    else update(p->chd[1],flag?flag->chd[1]:NULL,dep-1,i);
    if(p->chd[0]) p->symbl=std::max(p->symbl,p->chd[0]->symbl);
    if(p->chd[1]) p->symbl=std::max(p->symbl,p->chd[1]->symbl);
}
int query(trie *p,int x,int dep,int limit)
{
    if(dep<0) return sum[p->symbl]^x;
    int tmp=(x>>dep)&1;
    if(p->chd[tmp^1]&&p->chd[tmp^1]->symbl>=limit) return query(p->chd[tmp^1],x,dep-1,limit);
    return query(p->chd[tmp],x,dep-1,limit);
}
signed main()
{
    Init();
    read(n),read(m);
    build(root[0],Maxdep);
    for(int i=1,x;i<=n;i++)
    {
        read(x);
        sum[i]=sum[i-1]^x;
        update(root[i],root[i-1],Maxdep,i);
    } 
    for(int i=1;i<=m;i++)
    {
        char ch=getchar();
        while(ch!='A'&&ch!='Q') ch=getchar();
        if(ch=='A')
        {
            int x;
            read(x);
            n++;
            sum[n]=sum[n-1]^x;
            update(root[n],root[n-1],Maxdep,n); 
            continue;
        }   
        if(ch=='Q')
        {
            int l,r,x;
            read(l),read(r),read(x);
            int ans=query(root[r-1],sum[n]^x,Maxdep,l-1);
            printf("%d\n",ans);
            continue;
        }
    }
    return 0;
}