codeforces 899e Segments Removal

Segments Removal

Vasya has an array of integers of length n.

Vasya performs the following operations on the array: on each step he finds the longest segment of consecutive equal integers (the leftmost, if there are several such segments) and removes it. For example, if Vasya’s array is [13, 13, 7, 7, 7, 2, 2, 2], then after one operation it becomes [13, 13, 2, 2, 2].

Compute the number of operations Vasya should make until the array becomes empty, i.e. Vasya removes all elements from it.

Input

The first line contains a single integer n (1 ≤ n ≤ 200 000) — the length of the array.

The second line contains a sequence a1, a2, …, an (1 ≤ ai ≤ 109) — Vasya’s array.

Output

Print the number of operations Vasya should make to remove all elements from the array.

Examples

input

42 5 5 2

output

2

input

56 3 4 1 5

output

5

input

84 4 4 2 2 100 100 100

output

3

input

610 10 50 10 50 50

output

4

题面不难，给你一个序列，每次把其中最左边连续相等的数字去除，问最多需要多少次可以将整个序列全部除去。开始一直都是用set存每个线段的起点，终点和长度，但是每次在删除和合并的时候不知道如何处理线段的长度问题。看了题解之后，才发现，我们可以有两种方式储存线段，一是起点加终点，二是起点加长度。在这道题中，我们在删除一个线段后，就不是很容易算出合并之后的线段长度，用起点加长度就可以避免这个问题。
代码如下：

#include<bits/stdc++.h>using namespace std;const int maxn = 2e5 + 6;int n;int a[maxn];set<pair<int, int> >s1, s2;int main(){    ios::sync_with_stdio(false);    cin >> n;    int l, r;    for(l = 1, r = 1; r <= n; r++) {        cin >> a[r];        if(a[r] != a[l]) {            s1.insert({l - r, l});            s2.insert({l, r - l});            l = r;        }    }    s1.insert({l - r, l});    s2.insert({l, r - l});    int cnt = 0;    while(s1.size() > 1) {        cnt++;        pair<int, int> temp = *s1.begin();        s1.erase(temp);        int st = temp.second;        int len = -temp.first;        auto lpos = s2.lower_bound({st, len});        auto rpos = s2.upper_bound({st, len});        if(lpos == s2.begin() || rpos == s2.end()) {            s2.erase({st, len});            continue;        }        lpos--;        if(a[lpos->first] == a[rpos->first]) {            int st1 = lpos->first, st2 = rpos->first;            int len1 = lpos->second, len2 = rpos->second;            s1.erase({ -len1, st1});            s1.erase({ -len2, st2});            s2.erase({st1, len1}), s2.erase({st2, len2});            s1.insert({ -len1 - len2, st1});            s2.insert({st1, len1 + len2});        }        s2.erase({st, len});    }    cnt++;    cout << cnt << endl;    return 0;}