牛客第五场D drop Voicing - LIS乱搞

牛客第五场D drop Voicing - LIS乱搞

Face

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题意

• 给定一个排列支持两种操作
  • 把倒数第二个拿到第一个去
  • 把第一个拿到最后一个去
• 求吧整个序列调整成升序 最少需要连续操作1的操作次数

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数据范围: $1\le n，m\le 10^5,a[i]\le n$

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前置技能

• LIS

Tutorial:

• 原理

插入查找复杂度:$O(nlogn)$

总复杂度:$O(n^{2}log(n))$

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code:

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#include 

using namespace std;

#define local
#ifdef local

template<class T>
void _E(T x) { cerr << x; }

void _E(double x) { cerr << fixed << setprecision(6) << x; }

void _E(string s) { cerr << "\"" << s << "\""; }

template<class A, class B>
void _E(pair<A, B> x) {
    cerr << '(';
    _E(x.first);
    cerr << ", ";
    _E(x.second);
    cerr << ")";
}

template<class T>
void _E(vector<T> x) {
    cerr << "[";
    for (auto it = x.begin(); it != x.end(); ++it) {
        if (it != x.begin()) cerr << ", ";
        _E(*it);
    }
    cerr << "]";
}

void ERR() {}

template<class A, class... B>
void ERR(A x, B... y) {
    _E(x);
    cerr << (sizeof...(y) ? ", " : " ");
    ERR(y...);
}

#define debug(x...) do { cerr << "{ "#x" } -> { "; ERR(x); cerr << "}" << endl; } while(false)
#else
#define debug(...) 114514.1919810
#endif
#define _rep(n, a, b) for (ll n = (a); n <= (b); ++n)
#define _rev(n, a, b) for (ll n = (a); n >= (b); --n)
#define _for(n, a, b) for (ll n = (a); n < (b); ++n)
#define _rof(n, a, b) for (ll n = (a); n > (b); --n)
#define oo 0x3f3f3f3f3f3f
#define ll long long
#define db long double
#define eps 1e-3
#define bin(x) cout << bitset<10>(x) << endl;
#define what_is(x) cerr << #x << " is " << x << endl
#define met(a, b) memset(a, b, sizeof(a))
#define all(x) x.begin(), x.end()
#define pii pair
#define pdd pair
#define endl "\n"
#define ls ch[now][0]
#define rs ch[now][1]
const ll mod = 998244353;
const ll maxn = 2e5 + 10;


ll qpow(ll a, ll b) {
    ll ret = 1;
    for (; b; a = a * a % mod, b >>= 1) {
        if (b & 1) {
            ret = ret * a % mod;
        }
    }
    return ret;
}

vector<ll> f(maxn), invf(maxn);

ll inv(ll a) {
    return qpow(a, mod - 2);
}

void prework() {
    f[0] = 1;
    _rep(i, 1, maxn - 1) {
        f[i] = f[i - 1] * i % mod;
    }
    invf[maxn - 1] = qpow(f[maxn - 1], mod - 2);
    for (ll i = maxn - 2; i >= 0; i--) {
        invf[i] = invf[i + 1] * (i + 1) % mod;
    }

}

ll C(ll n, ll m) {
    if (n > m || m < 0)return 0;
    if (n == 0 || m == n) return 1;
    ll res = (f[m] * invf[m - n] % mod * invf[n]) % mod;
    return res;
}

vector<pii > G[maxn];
vector<ll> a;
int n;

ll lis(int l, int r) {
    vector<ll> dp(n + 1, oo);
    _rep(i, l, r) {
        *lower_bound(all(dp), a[i]) = a[i];
    }
    return lower_bound(all(dp), oo) - dp.begin();
}

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    a.resize(n * 2 + 1);
    _rep(i, 1, n) {
        cin >> a[i];
        a[i + n] = a[i];
    }
    ll res = 0;

    _rep(i, 1, n) {
        res = max(res, lis(i, i + n - 1));
    }
    cout << n-res << endl;
}