笔记

思路：

先开一个存有动态数组的数组

枚举数列中的每个点充当x，用数组记录下与改元素相邻的元素，用来计算对答案的贡献。下面的问题就是计算x改变后对答案的贡献了，具体的，我们想知道把x改变成什么值才能对答案的贡献最大，自然是该数组中位数了，然后计算ans取小就是最终答案。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <ios>
#include <vector>
using namespace std;
typedef long long ll;
const int N = (int)1e5;
int n, m, a[N + 1];
vector<int> b[N + 1];

int main(int argc, char *argv[]) {
    freopen("note.in", "r", stdin);
    freopen("note.out", "w", stdout);

    ios :: sync_with_stdio(false);
    cin >> n >> m;
    for (int i = 1; i <= m; ++i) cin >> a[i];
    for (int i = 1; i <= m; ++i) {
        if (i > 1 && a[i - 1] != a[i]) b[a[i - 1]].push_back(a[i]);
        if (i < m && a[i + 1] != a[i]) b[a[i + 1]].push_back(a[i]);
    }

    ll ans = 0LL, sum = 0LL;
    for (int i = 1; i <= n; ++i) {
        if (!b[i].size()) continue;
        sort(b[i].begin(), b[i].end());//把这一串数按从小到大的顺序排起来 
        int y = b[i][b[i].size() >> 1];
        ll before = 0LL, after = 0LL;
        for (int j = 0; j < b[i].size(); ++j) {
            before += abs(i - b[i][j]);
            after += abs(y - b[i][j]);
        }
        ans = max(ans, before - after), sum += before;
    }

    cout << sum / 2 - ans << endl;

    fclose(stdin);
    fclose(stdout);
    return 0;
}