CF1707E Replace

由题意可以发现一个性质：

\[f[(l, r)] = \bigcup_{i = l}^{r - 1} f[(i, i + 1)] \]

进而可以推广至：

\[f^k[(l, r)] = \bigcup_{i = l}^{r - 1} f^k[(i, i + 1)] \]

证明显然，即若 \([l_1, r_1] \cap [l_2, r_2]\)，则 \(f([l_1, r_1] \cup [l_2, r_2]) = f((l_1, r_1)) \cup f((l_2, r_2))\)。

然后可以倍增维护。

设 \(g(i, j, k)\) 表示 \(f^{2^k}[(i, i + 2^j)]\)，正常做就好。时间复杂度 \(\mathcal O(n \log^2 n + q \log n)\)。

代码：

#include <bits/stdc++.h>

using namespace std;

constexpr int N = 1e5 + 10, K = 18;

int n, q, lg[N], a[N];

struct Node {
    int l, r;

    Node operator+(const Node &rhs) const {return {min(l, rhs.l), max(r, rhs.r)};}
} g[N][K][K];

inline Node f(int l, int r, int k) {
    int ln = lg[r - l];
    return g[l][ln][k] + g[r - (1 << ln)][ln][k];
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(nullptr), cout.tie(nullptr);
    cin >> n >> q; for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 2; i <= n; i++) lg[i] = lg[i >> 1] + 1;
    for (int i = 1; i < n; i++) g[i][0][0] = {min(a[i], a[i + 1]), max(a[i], a[i + 1])};
    for (int j = 1; j < K; j++) {
        for (int i = 1; i + (1 << j) <= n; i++) {
            g[i][j][0] = g[i][j - 1][0] + g[i + (1 << (j - 1))][j - 1][0];
        }
    }
    for (int k = 1; k < K; k++) {
        for (int j = 0; j < K; j++) {
            for (int i = 1; i + (1 << j) <= n; i++) {
                g[i][j][k] = f(g[i][j][k - 1].l, g[i][j][k - 1].r, k - 1);
            }
        }
    }
    while (q--) {
        int l, r, ans = 1; cin >> l >> r;
        if (r - l + 1 == n) {cout << "0\n"; continue;}
        if (l == r) {cout << "-1\n"; continue;}
        for (int k = K - 1; k >= 0; k--) {
            Node to = f(l, r, k);
            if (1 < to.l || to.r < n) l = to.l, r = to.r, ans += (1 << k);
        }
        Node now = f(l, r, 1);
        if (now.l == 1 && now.r == n) cout << ans << '\n';
        else cout << "-1\n";
    }
    return 0;
}