BZOJ4260 Codechef REBXOR（trie）

　　用trie求出前缀最大区间异或和、后缀最大区间异或和即可。注意空间是nlog的。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 400010
int n,a[N],pre[N],suf[N],trie[N<<][],cnt,ans;
void ins(int x)
{
    int k=;
    for (int j=;~j;j--)
    {
        if (!trie[k][(x&(<<j))>]) trie[k][(x&(<<j))>]=++cnt;
        k=trie[k][(x&(<<j))>];
    }
}
int query(int x)
{
    int k=,s=;
    for (int j=;~j;j--)
    if (trie[k][(x&(<<j))==]) s|=<<j,k=trie[k][(x&(<<j))==];
    else k=trie[k][(x&(<<j))>];
    return s;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4260.in","r",stdin);
    freopen("bzoj4260.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=;i<=n;i++) a[i]=a[i-]^read();
    cnt=;memset(trie,,sizeof(trie));ins();
    for (int i=;i<=n;i++)
    {
        pre[i]=max(pre[i-],query(a[i]));
        ins(a[i]);
    }
    cnt=;memset(trie,,sizeof(trie));ins();
    for (int i=n;i>=;i--)
    {
        suf[i]=max(suf[i+],query(a[i]));
        ins(a[i]);
    }
    for (int i=;i<n;i++) ans=max(ans,pre[i]+suf[i+]);
    cout<<ans;
    return ;
}