CF 1550C

链接这里

题意：在区间内选取三个点，若两个点曼哈顿距离与XXX相等，则为Bad，否则为Good；
推一下可以发现，超过4个点，那么这个区间必为Bad，我们只需要枚举3个点与四个点是否为Bad就行。

#include<bits/stdc++.h>

using namespace std;

#define int long long
#define gcd(a, b) __gcd(a,b)
const long long mod = 1e9 + 7;
const int maxn = 1e5 + 5;

signed main() {
    cin.tie(nullptr)->sync_with_stdio(false);

    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<int> a(n);
        int num = 0;
        for (int i = 0; i < n; ++i) {
            cin >> a[i];
        }
        for (int i = 0; i < n - 2; ++i) {
            if (a[i + 1] <= max(a[i + 2], a[i]) && a[i + 1] >= min(a[i], a[i + 2]));
            else {
                num++;
            }

        }
        for (int i = 0; i < n - 3; ++i) {

            if ((a[i + 1] <= max(a[i], a[i + 3]) && a[i + 1] >= min(a[i], a[i + 3])) ||
                (a[i + 2] <= max(a[i], a[i + 3]) && a[i + 2] >= min(a[i], a[i + 3])) ||
                (a[i + 1] <= max(a[i + 2], a[i]) && a[i + 1] >= min(a[i + 2], a[i])) ||
                (a[i + 2] <= max(a[i + 1], a[i + 3]) && a[i + 2] >= min(a[i + 1], a[i + 3])));
                //  cout << a[i] << ' ' << a[i + 1] << ' ' << a[i + 2] << endl;
            else {
                //cout << a[i] << endl;
                num++;
            }
        }
        cout << num + 2 * n - 1 << endl;
    }

    return 0;
}