Codeforces 282E Sausage Maximization (Trie)

题意

给一些数字，求前缀异或和后缀异或的最大值，前缀和后缀不能有交集。

思路

枚举每一个后缀，如果用朴素算法就要枚举每一个前缀求最大值。
我们可以后缀往后枚举的时候，从之前的前缀里找到异或最大的，然后前缀加一个数的长度之后继续插入。

代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define LL long long
#define Lowbit(x) ((x)&(-x))
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1|1
#define MP(a, b) make_pair(a, b)
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int maxn = 1e5 + 10;
const double eps = 1e-8;
const double PI = acos(-1.0);
typedef pair<int, int> pii;
struct Trie
{
    int next[2];
    Trie() {next[0] = next[1] = 0;}
}trie[64*maxn];

LL a[maxn];
LL tot;

void insert(LL s)
{
    int step = 0;
    for (int i = 63; i >= 0; i--)
    {
        int num = (((1LL << i) & s) != 0);
        if (!trie[step].next[num])
            trie[step].next[num] = ++tot;
        step = trie[step].next[num];
    }
}

LL query(LL s)
{
    LL res = 0;
    int step = 0;
    for (int i = 63; i >= 0; i--)
    {
        int num = (((1LL << i) & s) == 0);
        if (!trie[step].next[num])
            num ^= 1;
        res = res * 2LL + (LL)num;
        step = trie[step].next[num];
    }
    return res;
}


int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    while (scanf("%d", &n) != EOF)
    {
        LL post = 0, pre = 0;
        for (int i = 0; i < n; i++)
            scanf("%lld", &a[i]), post ^= a[i];
        insert(0LL);
        LL ans = post;
        for (int i = 0; i < n; i++)
        {
            pre ^= a[i], post ^= a[i];
            insert(pre);
            LL t = query(post);
            ans = max(ans,  t ^ post);
        }
        printf("%lld\n", ans);
    }
    return 0;
}