[Codechef] REBXOR

[题目链接]

        https://www.lydsy.com/JudgeOnline/problem.php?id=4260

[算法]

       字典树

       时间复杂度 : O(NlogN)

[代码]

        

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 4e5 + 10;

int n,ans;
int a[MAXN],sufmx[MAXN],premx[MAXN],pre[MAXN],suf[MAXN];

template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; 
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
struct Trie
{
        int tot;
        int child[MAXN * 31][2];
        inline void clear()
        {
                tot = 0;
                memset(child,0,sizeof(child));
        }
        inline void insert(int x)
        {
                int now = 0;
                for (int i = 0; i < 31; i++)
                {
                        int tmp;
                        if (x & (1 << (30 - i))) tmp = 1;
                        else tmp = 0;
                        if (!child[now][tmp]) child[now][tmp] = ++tot;
                        now = child[now][tmp];
                }
        }
        inline int query(int x)
        {
                int now = 0 , ret = 0;
                for (int i = 0; i < 31; i++)
                {
                        int tmp;
                        if (x & (1 << (30 - i))) tmp = 1;
                        else tmp = 0;
                        if (child[now][tmp ^ 1]) 
                        {
                                ret += 1 << (30 - i);
                                now = child[now][tmp ^ 1];
                        } else now = child[now][tmp];
                }
                return ret;
        }
} T;

int main() 
{
        
        read(n);
        for (int i = 1; i <= n; i++) read(a[i]);
        T.clear();
        T.insert(0);
        for (int i = 1; i <= n; i++)
        {
                pre[i] = pre[i - 1] ^ a[i];
                premx[i] = T.query(pre[i]);
                T.insert(pre[i]);
        }
        for (int i = 1; i <= n; i++) premx[i] = max(premx[i - 1],premx[i]);
        T.clear();
        T.insert(0);
        for (int i = n; i >= 1; i--)
        {
                suf[i] = suf[i + 1] ^ a[i];
                sufmx[i] = T.query(suf[i]);
                T.insert(suf[i]);
        }
        for (int i = n; i >= 1; i--) sufmx[i] = max(sufmx[i],sufmx[i + 1]);
        for (int i = 1; i <= n; i++) ans = max(ans,premx[i] + sufmx[i + 1]);
        printf("%d\n",ans);
        
        return 0;
    
}