CF811C：Vladik and Memorable Trip（dp）

C. Vladik and Memorable Trip

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Vladik often travels by trains. He remembered some of his trips especially well and I would like to tell you about one of these trips:

Vladik is at initial train station, and now n people (including Vladik) want to get on the train. They are already lined up in some
order, and for each of them the city code ai is
known (the code of the city in which they are going to).

Train chief selects some number of disjoint segments of the original sequence of people (covering entire sequence by segments is not necessary). People who are in the same segment will be in the
same train carriage. The segments are selected in such way that if at least one person travels to the city x, then all people who
are going to city x should be in the same railway carriage. This means that they can’t belong to different segments. Note, that all
people who travel to the city x, either go to it and in the same railway carriage, or do not go anywhere at all.

Comfort of a train trip with people on segment from position l to position r is
equal to XOR of all distinct codes of cities for people on the segment from position l to
position r. XOR operation also
known as exclusive OR.

Total comfort of a train trip is equal to sum of comfort for each segment.

Help Vladik to know maximal possible total comfort.

Input

First line contains single integer n (1 ≤ n ≤ 5000) —
number of people.

Second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 5000),
where ai denotes
code of the city to which i-th person is going.

Output

The output should contain a single integer — maximal possible total comfort.

Examples

input

6
4 4 2 5 2 3

output

14

input

9
5 1 3 1 5 2 4 2 5

output

9

Note

In the first test case best partition into segments is: [4, 4] [2, 5, 2] [3],
answer is calculated as follows: 4 + (2 xor5) + 3 = 4 + 7 + 3 = 14
In the second test case best partition into segments is: 5 1 [3] 1 5 [2, 4, 2] 5,
answer calculated as follows: 3 + (2 xor 4) = 3 + 6 = 9.

题意：给n个数，要求任意选出若干个不相交区间，规定同一种数字只能出现在同一个区间内，每个区间的价值为里面不同的数的异或值，问怎么分配区间使得总价值和最大。

思路：普通的线性dp，预处理每种数的左右界即可。

# include <bits/stdc++.h>
using namespace std;

const int maxn = 5e3+3;
int a[maxn], l[maxn], r[maxn], dp[maxn], vis[maxn];
int main()
{
int n;
scanf("%d",&n);
for(int i=1; i<=n; ++i)
{
scanf("%d",&a[i]);
if(!l[a[i]])
l[a[i]] = i;
r[a[i]] = i;
}
for(int i=1; i<=n; ++i)
{
dp[i] = dp[i-1];
int t = 0, h = l[a[i]];
memset(vis, 0, sizeof(vis));
for(int j=i; j>0; --j)
{
if(r[a[j]] > i) break;
h = min(h, l[a[j]]);
if(!vis[a[j]]) t^=a[j], vis[a[j]] = 1;
if(h >= j) dp[i] = max(dp[i], dp[j-1]+t);
}
}
printf("%d\n",dp
);
return 0;
}