[刷题]Codeforces 811C

哇咔咔，终于期末考完啦，终于放暑假啦，我终于又有时间学习啦！！(o゜▽゜)o☆

Description

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

input64 4 2 5 2 3output14input95 1 3 1 5 2 4 2 5output9

Note

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

In the second test case best partition into segments is:51[3]15[2, 4, 2]5, answer calculated as follows: 3 + (2xor4) = 3 + 6 = 9.

Key

给一个n个元素（可含有重名元素）的序列，按照原来的顺序分割出若干段，要求所有重名元素要么全部出现在一个段里，要么全部不在段里。问分割出的所有段的价值和。其中每个段的价值为段内所有不同名元素按照名字的二进制异或。

明显的DP问题，但当时还是没写出来。。。果然一个月没写就生疏了。

当时就觉得这个异或有猫腻。什么意思呢？对于两个二进制数x和y，x+y≥x⊕y（就相当于异或比加法少一个进位情况）。也就是说，每个段中不同名的元素越少越好。

先预处理一下，每种元素只需要知道其最左侧与最右侧的元素位置即可。

对于这个序列中的每一个元素：有选中与不选中（或无法选中）两种情况。若试图选中当前元素，则要求出包含此类元素的段的最小区间[l,r]及其价值Value。若r大于当前元素位置，则不选中，DPi←DPi−1；否则DPi←max(DPi−1,DPl−1+Value)。

Code

#include<iostream>#include<cstring>#include<algorithm>#define l(t) s[t][0]#define r(t) s[t][1]using namespace std;const int maxn = 5000 + 10;int n;int a[maxn];int s[maxn][2];int tv[maxn];int dp[maxn];bool vis[maxn];int main(){    memset(s, 0, sizeof(s));    memset(tv, 0, sizeof(tv));    cin >> n;    for (int i = 1; i <= n; ++i) {        cin >> a[i];        if (!l(a[i])) l(a[i]) = i;        r(a[i]) = i;    }    dp[0] = 0;    for (int i = 1; i <= n; ++i) {        dp[i] = dp[i - 1];        if (r(a[i]) > i) continue;        int val = a[i];        int ml = l(a[i]);        memset(vis, 0, sizeof(vis));        vis[a[i]] = true;        int j;        for (j = i; j > ml; --j) {            if (vis[a[j]]) continue;            vis[a[j]] = true;            if (r(a[j]) > i) break;            if (l(a[j]) < ml) ml = l(a[j]);            val ^= a[j];        }        if (j == ml) dp[i] = max(dp[i - 1], val + dp[j - 1]);    }    cout << dp[n];    return 0;}