维护好每个人左右两边的最长单调子序列长度即可

时间复杂度是 n^2

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 2e5 + 5;
int __t = 1, n;
void solve() {
    cin >> n;
    vector<int> h(n);
    for (int i = 0; i < n; i++)
        cin >> h[i];
    vector<int> inc(n, 1);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (h[j] < h[i] && inc[j] + 1 > inc[i]) {
                inc[i] = inc[j] + 1;
            }
        }
    }
    vector<int> dec(n, 1);
    for (int i = n - 1; i >= 0; i--) {
        for (int j = n - 1; j > i; j--) {
            if (h[j] < h[i] && dec[j] + 1 > dec[i]) {
                dec[i] = dec[j] + 1;
            }
        }
    }
    int ma = 0;
    for (int i = 0; i < n; i++) {
        ma = max(ma, inc[i] + dec[i] - 1);
    }
    cout << n - ma << "\n";
}
int32_t main() {
#ifdef ONLINE_JUDGE
    ios::sync_with_stdio(false);
    cin.tie(0);
#endif
    // cin >> __t;
    while (__t--)
        solve();
    return 0;
}

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