C. Vladik and Memorable Trip

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

如今菜的连线性dp也想不出来。QAQ

对于当前点，我们不断往前看是否能构成一个新的区间，如果能够成dp[i] = max(dp[i],dp[j-1]+res);这里j代表的就是左端点，res表示新构成的区间的不同的数的异或值。

另外一个情况就是默认不选择当前这个点，那么dp[i] = dp[i-1]。

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5e3+7;
const int inf =1e9;
int n;
int num[MAXN],dp[MAXN];
int first_pos[MAXN],last_pos[MAXN];
bool vis[MAXN];
int main()
{
scanf("%d",&n);
for(int i = 1 ; i <= n ; ++i)
{
scanf("%d",&num[i]);
if(!first_pos[num[i]])first_pos[num[i]] = i;
last_pos[num[i]] = i;
}
for(int i = 1 ; i <= n ; ++i)
{
dp[i] = dp[i-1];
memset(vis,0,sizeof(vis));
int st = first_pos[num[i]];
int res = 0;
for(int j = i ; j >= 1 ; --j)
{
if(!vis[num[j]])
{
if(last_pos[num[j]]>i)break;
st = min(st,first_pos[num[j]]);
res^=num[j];
vis[num[j]] = 1;
}
if(j<=st)dp[i] = max(dp[i] , dp[j-1]+res);//维护区间最值
}
}
printf("%d\n",dp
);
}