Merge pull request #193 from SkqLiao/patch-1

### 动态树

> 本节**性质**部分内容翻译自[wcipeg](http://wcipeg.com/wiki/Lowest_common_ancestor)，并做过修改。

### 标准RMQ

时间复杂度 $O(N)-O(1)$，空间复杂度 $O(N)$ ，支持在线查询，常数较大，编程复杂度较高。

流程：

- 通过DFS序将树上LCA问题转为序列RMQ问题

- 通过单调栈将序列转为笛卡尔树

- 在笛卡尔树上求欧拉序，如此转化为 $\pm 1$ RMQ

- 对新序列分块，做分块ST表，块内通过二进制状压DP维护

每一步的复杂度都是 $O(N)$ 的，因此总复杂度依然是 $O(N)$ 。

提供RMQ转标准RMQ的代码，为洛谷上ST表的例题[**P3865** 【模板】ST表](https://www.luogu.org/problemnew/show/P3865)

```cpp

// Copyright (C) 2018 Skqliao. All rights served.

#include <bits/stdc++.h>

#define rep(i, l, r) for (int i = (l), _##i##_ = (r); i < _##i##_; ++i)

#define rof(i, l, r) for (int i = (l) - 1, _##i##_ = (r); i >= _##i##_; --i)

#define ALL(x) (x).begin(), (x).end()

#define SZ(x) static_cast<int>((x).size())

typedef long long ll;

typedef std::pair<int, int> pii;

template<typename T> inline bool chkMin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }

template<typename T> inline bool chkMax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }

const int MAXN = 1e5 + 5;

struct PlusMinusOneRMQ {

    const static int M = 8;

    int blocklen, block, Minv[MAXN], F[MAXN / M * 2 + 5][M << 1],

        T[MAXN], f[1 << M][M][M], S[MAXN];

    void init(int n) {

        blocklen = std::max(1, (int)(log(n * 1.0) / log(2.0)) / 2);

        block = n / blocklen + (n % blocklen > 0);

        int total = 1 << (blocklen - 1);

        for (int i = 0; i < total; i++) {

            for (int l = 0; l < blocklen; l++) {

                f[i][l][l] = l;

                int now = 0, minv = 0;

                for (int r = l + 1; r < blocklen; r++) {

                    f[i][l][r] = f[i][l][r - 1];

                    if ((1 << (r - 1)) & i) {

                        now++;

                    } else {

                        now--;

                        if (now < minv) {

                            minv = now;

                            f[i][l][r] = r;

                        }

                    }

                }

            }

        }

        T[1] = 0;

        for (int i = 2; i < MAXN; i++) {

            T[i] = T[i - 1];

            if (!(i & (i - 1))) {

                T[i]++;

            }

        }

    }

    void initmin(int a[], int n) {

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

            if (i % blocklen == 0) {

                Minv[i / blocklen] = i;

                S[i / blocklen] = 0;

            } else {

                if (a[i] < a[Minv[i / blocklen]]) {

                    Minv[i / blocklen] = i;

                }

                if (a[i] > a[i - 1]) {

                    S[i / blocklen] |= 1 << (i % blocklen - 1);

                }

            }

        }

        for (int i = 0; i < block; i++) {

            F[i][0] = Minv[i];

        }

        for (int j = 1; (1 << j) <= block; j++) {

            for (int i = 0; i + (1 << j) - 1 < block; i++) {

                int b1 = F[i][j - 1], b2 = F[i + (1 << (j - 1))][j - 1];

                F[i][j] = a[b1] < a[b2] ? b1 : b2;

            }

        }

    }

    int querymin(int a[], int L, int R) {

        int idl = L / blocklen, idr = R / blocklen;

        if (idl == idr)

            return idl * blocklen + f[S[idl]][L % blocklen][R % blocklen];

        else {

            int b1 = idl * blocklen + f[S[idl]][L % blocklen][blocklen - 1];

            int b2 = idr * blocklen + f[S[idr]][0][R % blocklen];

            int buf = a[b1] < a[b2] ? b1 : b2;

            int c = T[idr - idl - 1];

            if (idr - idl - 1) {

                int b1 = F[idl + 1][c];

                int b2 = F[idr - 1 - (1 << c) + 1][c];

                int b = a[b1] < a[b2] ? b1 : b2;

                return a[buf] < a[b] ? buf : b;

            }

            return buf;

        }

    }

};

struct CartesianTree {

private:

    struct Node {

        int key, value, l, r;

        Node(int key, int value) {

            this->key = key;

            this->value = value;

            l = r = 0;

        }

        Node() {}

    };

    Node tree[MAXN];

    int sz;

    int S[MAXN], top;

public:

    void build(int a[], int n) {

        top = 0;

        tree[0] = Node(-1, INT_MAX);

        S[top++] = 0;

        sz = 0;

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

            tree[++sz] = Node(i, a[i]);

            int last = 0;

            while (tree[S[top - 1]].value <= tree[sz].value) {

                last = S[top - 1];

                top--;

            }

            tree[sz].l = last;

            tree[S[top - 1]].r = sz;

            S[top++] = sz;

        }

    }

    Node &operator [] (const int x) {

        return tree[x];

    }

};

class stdRMQ {

public:

    void work(int a[], int n) {

        ct.build(a, n);

        dfs_clock = 0;

        dfs(0, 0);

        rmq.init(dfs_clock);

        rmq.initmin(depseq, dfs_clock);

    }

    int query(int L, int R) {

        int cl = clk[L], cr = clk[R];

        if (cl > cr) {

            std::swap(cl, cr);

        }

        return Val[rmq.querymin(depseq, cl, cr)];

    }

private:

    CartesianTree ct;

    PlusMinusOneRMQ rmq;

    int dfs_clock, clk[MAXN], Val[MAXN << 1], depseq[MAXN << 1];

    void dfs(int rt, int d) {

        clk[ct[rt].key] = dfs_clock;

        depseq[dfs_clock] = d;

        Val[dfs_clock++] = ct[rt].value;

        if (ct[rt].l) {

            dfs(ct[rt].l, d + 1);

            depseq[dfs_clock] = d;

            Val[dfs_clock++] = ct[rt].value;

        }

        if (ct[rt].r) {

            dfs(ct[rt].r, d + 1);

            depseq[dfs_clock] = d;

            Val[dfs_clock++] = ct[rt].value;

        }

    }

} doit;

int A[MAXN];

int main() {

    int n, m, l, r;

    scanf("%d%d", &n, &m);

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

        scanf("%d", &A[i]);

    }

    doit.work(A, n);

    while (m--) {

        scanf("%d%d", &l, &r);

        printf("%d\n", doit.query(l - 1, r - 1));

    }

    return 0;

}

```