可以推出
min[i]要么是i要么是1，当a序列中存在这个数是1
max[i]的话就比较麻烦了
首先对于i来说，如果还没有被提到第一位的话，他的max可由他后面的这部分序列中 j>=i 的不同数多少所决定，这个可以用树状数组解决
其次就是两次被提到第一位的中间的空当，这个空当中不同的数的大小，也会决定max，这里的解法比较多样，我用的是主席树

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <complex>
#include <cstring>
#include <ctime>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <vector>
#define MP make_pair
#define ll long long
#define ld long double
#define null NULL
#define all(a) a.begin(), a.end()
#define forn(i, n) for (int i = 0; i < n; ++i)
#define sz(a) (int)a.size()
// #define lson l , m , rt << 1
// #define rson m + 1 , r , rt << 1 | 1
#define bitCount(a)  __builtin_popcount(a)
template<class T> int gmax(T &a, T b) { if (b > a) { a = b; return 1; } return 0; }
template<class T> int gmin(T &a, T b) { if (b < a) { a = b; return 1; } return 0; }
using namespace std;
const int INF = 0x3f3f3f3f;
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string) s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) { return "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; }
template <typename A>
string to_string(A v) { bool first = true; string res = "{"; for (const auto &x : v) { if (!first) { res += ", "; } first = false; res += to_string(x); } res += "}"; return res; }
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) { cerr << " " << to_string(H); debug_out(T...); }
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

// const int MAXN = 3e5 + 5;
// const int M = MAXN * 100;
int n, m;
vector<int> T, lson, rson, c;
int tot = 0;

int newNode() {
    lson.push_back(0); rson.push_back(0); c.push_back(0); 
    return tot ++;
}
int build(int l, int r) {
    int root = newNode();
    c[root] = 0;
    if(l != r) {
        int mid = (l + r) >> 1;
        lson[root] = build(l, mid);
        rson[root] = build(mid + 1, r);
    }
    return root;
}

int update(int root, int pos, int val) {
    int newroot = newNode(), tmp = newroot;
    c[newroot] = c[root] + val;
    int l = 1, r = n;
    while(l < r) {
        int mid = (l + r) >> 1;
        if(pos <= mid) {
            lson[newroot] = newNode(); rson[newroot] = rson[root];
            newroot = lson[newroot]; root = lson[root];
            r = mid;
        } else {
            rson[newroot] = newNode(); lson[newroot] = lson[root];
            newroot = rson[newroot]; root = rson[root];
            l = mid + 1;
        }
        c[newroot] = c[root] + val;
    }
    return tmp;
}

int query(int root, int pos) {
    int ret = 0;
    int l = 1, r = n;
    while(pos < r) {
        int mid = (l + r) >> 1;
        if(pos <= mid) {
            r = mid;
            root = lson[root];
        } else {
            ret += c[lson[root]];
            root = rson[root];
            l = mid + 1;
        }
    }
    return ret + c[root];
}


class BIT {
private:
    vector<int> tree;
    int treesize;
public:
    BIT(int x) {
        tree.resize(x + 5, 0);
        treesize = x + 5;
    }

    int Sum(int x) {
        if(x <= 0) return 0;
        if(x > treesize) x = treesize;
        int ans = 0;
        while(x > 0) {
            ans += tree[x];
            x -= x & -x;
        }
        return ans;
    }
    void Add(int x, int d) {
        // debug(x);
        while(x <= treesize) {
            tree[x] += d;
            x += x & -x;
        }
    }
};


int main() {
    
    while(~scanf("%d %d", &n, &m)) {
        vector<int> vc;
        vector<int> mp(max(n,m) + 5, 0);
        vector<int> maxx(n + 5, 0);
        vector<int> minn(n + 5, 0);
        vector<vector<int> > E(n + 5, vector<int>());
        for(int i = 1; i <= n; ++i) {
            maxx[i] = minn[i] = i;
        }   

        for(int i = 0; i < m; ++i) {
            int t; scanf("%d", &t);
            vc.push_back(t);
            E[t].push_back(i + 1);
            minn[t] = 1;
        }

        BIT bit = BIT(n + 5);
        // set<int> st;
        for(int i = 0; i < m; ++i) {
            if(mp[vc[i]] == 0) {
                bit.Add(vc[i], 1);
                maxx[vc[i]] = max(maxx[vc[i]], vc[i] + bit.Sum(n) - bit.Sum(vc[i]));
                debug(i, vc[i], maxx[vc[i]]);
            }
            mp[vc[i]] = 1;
        }
        
        for(int i = 1; i <= n; ++i) {
            if(mp[i] == 0) {
                maxx[i] = max(maxx[i], i + bit.Sum(n) - bit.Sum(i));
                // debug(i, maxx[i]);
            }
        }


        tot = 0;
        T.clear();
        for(int i = 0; i < m + 5; ++i) T.push_back(i);
        T[m + 1] = build(1, m);
        for(int i = 1; i <= n; ++i) mp[i] = 0;
        for(int i = m; i >= 1; -- i) {
            int target = vc[i-1];
            if(mp[target] == 0) {
                T[i] = update(T[i + 1], i,  1);
            } else {
                int tmp = update(T[i + 1], mp[target], -1);
                T[i] = update(tmp, i, 1);
            }
            mp[target] = i;
        }

        debug(query(T[m + 1], m + 1), query(T[m + 1], m));

        for(int i = 1; i <= n; ++i) {
            int pre = m + 1;
            for(int j = E[i].size() - 1; j >= 0; --j) {
                int tmp = query(T[E[i][j]], pre - 1);
                maxx[i] = max(maxx[i], tmp);
                // debug(i, E[i][j] + 1, pre - 1, tmp);
                pre = E[i][j];
            }
        }

        

        for(int i = 1; i <= n; ++i) {
            printf("%d %d\n", minn[i], maxx[i]);
        }
    }
    return 0;
}

官方给出了一种比较新颖的直接线段树的做法，我觉得写的非常有趣，

#include <bits/stdc++.h>

#define forn(i, n) for (int i = 0; i < int(n); i++)
#define x first 
#define y second

using namespace std;

const int N = 300 * 1000 + 13;

typedef pair<int, int> pt;

int n;
int a[N];
vector<int> pos[N];
pt ans[N];
int prv[N];

vector<int> t[4 * N];

void build(int v, int l, int r){
    if (l == r - 1){
        t[v].push_back(prv[l]);
        return;
    }
    int m = (l + r) / 2;
    build(v * 2, l, m);
    build(v * 2 + 1, m, r);
    t[v].resize(r - l);
    merge(t[v * 2].begin(), t[v * 2].end(), t[v * 2 + 1].begin(), t[v * 2 + 1].end(), t[v].begin());
}

int get(int v, int l, int r, int L, int R, int val){
    if (L >= R)
        return 0;
    if (l == L && r == R)
        return lower_bound(t[v].begin(), t[v].end(), val) - t[v].begin();
    int m = (l + r) / 2;
    return get(v * 2, l, m, L, min(m, R), val) + get(v * 2 + 1, m, r, max(m, L), R, val);
}

int f[N];

void upd(int x){
    for (int i = x; i >= 0; i = (i & (i + 1)) - 1)
        ++f[i];
}

int get(int x){
    int res = 0;
    for (int i = x; i < N; i |= i + 1)
        res += f[i];
    return res;
}

int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    forn(i, m){
        scanf("%d", &a[i]);
        --a[i];
    }
    forn(i, m){
        pos[a[i]].push_back(i);
    }
    
    vector<pt> qr;
    forn(i, n){
        for (int j = 1; j < int(pos[i].size()); ++j)
            qr.push_back(make_pair(pos[i][j - 1] + 1, pos[i][j] - 1));
        if (!pos[i].empty())
            qr.push_back(make_pair(pos[i].back() + 1, m - 1));
    }
    
    forn(i, n) ans[i] = {i, i};
    forn(i, m) ans[a[i]].x = 0;
    
    forn(i, n){
        int cur = -1;
        for (auto it : pos[i]){
            prv[it] = cur;
            cur = it;
        }
    }
    build(1, 0, m);
    
    forn(i, qr.size()){
        int l = qr[i].x;
        int r = qr[i].y;
        if (r < l) continue;
        int x = a[qr[i].x - 1];
        int cnt = get(1, 0, m, l, r + 1, l);
        ans[x].y = max(ans[x].y, cnt);
    }
    
    forn(i, m){
        if (i == pos[a[i]][0]){
            ans[a[i]].y = max(ans[a[i]].y, a[i] + get(a[i]));
            upd(a[i]);
        }
    }
    forn(i, n) if (pos[i].empty()){
        ans[i].y = max(ans[i].y, i + get(i));
    }
    
    forn(i, n) printf("%d %d\n", ans[i].x + 1, ans[i].y + 1);
    return 0;
}