$Splay + 树状数组$

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>

using namespace std;

const int N = 100010, INF = 2e9;

struct node {
    int s[2], p;
    int val;
    int size;
    void init(int _val, int _p) { s[0] = s[1] = 0, p = _p, val = _val, size = 1; }
} tr[N];
int root, idx;

int n;
int f[N];
int a[N], k;
int td[N];
int res[N];

void pushup(int u)
{
    tr[u].size = tr[tr[u].s[0]].size + tr[tr[u].s[1]].size + 1;
}

void rotate(int x)
{
    int y = tr[x].p, z = tr[y].p;
    int k = tr[y].s[1] == x;

    tr[z].s[tr[z].s[1] == y] = x, tr[x].p = z;
    tr[y].s[k] = tr[x].s[k ^ 1], tr[tr[x].s[k ^ 1]].p = y;
    tr[x].s[k ^ 1] = y, tr[y].p = x;

    pushup(y), pushup(x);
}

void splay(int x, int k)
{
    while(tr[x].p != k)
    {
        int y = tr[x].p, z = tr[y].p;
        if(z != k)
            if((tr[z].s[1] == y) == (tr[y].s[1] == x)) rotate(y);
            else rotate(x);
        rotate(x);
    }
    if(!k) root = x;
}

void insert(int val)
{
    int u = root, p = 0;
    while(u) p = u, u = tr[u].s[val > tr[u].val];
    u = ++ idx;
    if(p) tr[p].s[val > tr[p].val] = u;
    tr[u].init(val, p);
    splay(u, 0);
}

int get_k(int k)
{
    int u = root;
    while(u)
    {
        if(tr[tr[u].s[0]].size >= k) u = tr[u].s[0];
        else if(tr[tr[u].s[0]].size + 1 == k) return u;
        else k -= tr[tr[u].s[0]].size + 1, u = tr[u].s[1];
    }
    return -1;
}

void dfs(int u)
{
    if(tr[u].s[0]) dfs(tr[u].s[0]);
    if(tr[u].val > -INF && tr[u].val < INF) a[++ k] = tr[u].val;
    if(tr[u].s[1]) dfs(tr[u].s[1]);
}

int lowbit(int x)
{
    return x & -x;
}

void add(int x, int val)
{
    for(int i = x; i <= n; i += lowbit(i)) td[i] = max(td[i], val);
}

int query(int x)
{
    int res = 0;
    for(int i = x; i; i -= lowbit(i)) res = max(res, td[i]);
    return res;
}

int main()
{
    scanf("%d", &n);
    insert(-INF), insert(INF);
    for(int i = 1; i <= n; i ++)
    {
        // splay构造序列
        int x;
        scanf("%d", &x);
        // 按插入要求构造
        int l = get_k(x + 1), r = get_k(x + 2);
        splay(l, 0), splay(r, l);
        int u = ++ idx;
        tr[u].init(i, r);
        tr[r].s[0] = u;
        pushup(r), pushup(l);
        splay(u, 0);
    }

    dfs(root);  // 获取序列

    for(int i = 1; i <= k; i ++) add(a[i], f[a[i]] = query(a[i]) + 1);
    // 每次插入一个元素，如果该元素对当前答案有影响，则取f[i], 否则就取f[i - 1]
    for(int i = 1; i <= n; i ++) f[i] = max(f[i], f[i - 1]);
    for(int i = 1; i <= n; i ++) printf("%d\n", f[i]);
    return 0;
}