AGC029C - Lexicographic constraints

I record the mentality of it, this question exposed a lot of small water channel my problems.

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Given \ (n-\) string length \ (a_i \) , find the smallest character set how to construct a comparison lexicographical ordering \ (S_1 <S_2 <\ DOTS <S_N \) .

When \ (a_i <a_ {i + 1} \) , the apparent full complement \ (0 \) on the line. Otherwise, it is a high-precision \ (+ 1 \) . Binary alphabet size, determined okay.

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Here is what I do question the process.

First, in addition to half of some, all, launch it. But the addition of written details rotten, no sentence variety.

Then no special character set is sentenced \ (1 \) case, almost literally ran 1e9 T. (The last change to change or T)

Then consider greedy process, wrote a wrong. The character is not enough time to open a new, in fact, lead to the front of the waste.

blog wrote last discovery is still a problem sb, when the water on a blog it.

#include <bits/stdc++.h>
const int MAXN = 500010;

int n, st[MAXN], col[MAXN], top;
void add() {
    if (st[top] > 1 && st[top - 1] != st[top] - 1) {
        st[top + 1] = st[top];
        col[top + 1] = col[top];
        --st[top]; ++top;
    }
    ++col[top];
}
int A[MAXN];
bool judge(int cnt) {
    int lst = 0;
    memset(st, 0, top + 1 << 2);
    memset(col, 0, top + 1 << 2);
    top = 0;
    for (int i = 1, t; i <= n; ++i) {
        t = A[i];
        if (lst < t) {
            if (!top || col[top] != 0) st[++top] = t, col[top] = 0;
            else st[top] = t;
        } else {
            int k = -1;
            while (st[top] > t) k = col[top], --top;
            if (st[top] != t) st[++top] = t, col[top] = k;
            if (cnt > 1) {
                add();
                while (top && col[top] >= cnt) {
                    int at = st[top]; --top;
                    if (st[top] + 1 != at)
                        st[++top] = at - 1;
                    add();
                }
            } else top = 0, col[0] = 1;
            if (col[0]) return false;
            if (st[top] != t) st[++top] = t, col[top] = 0;
        }
        lst = t;
    }
    return true;
}
int main() {
    std::ios_base::sync_with_stdio(false), std::cin.tie(0);
    std::cin >> n;
    for (int i = 1; i <= n; ++i) std::cin >> A[i];
    int l = 1, r = n, ans = 0;
    while (l <= r) {
        int mid = l + r >> 1;
        if (judge(mid)) ans = mid, r = mid - 1;
        else l = mid + 1;
    }
    std::cout << ans << std::endl;
    return 0;
}