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;
}