C2. Adjust The Presentation (Hard Version)

onlyblues發表於2024-10-06

C2. Adjust The Presentation (Hard Version)

This is the hard version of the problem. In the two versions, the constraints on $q$ and the time limit are different. In this version, $0 \leq q \leq 2 \cdot 10^5$. You can make hacks only if all the versions of the problem are solved.

A team consisting of $n$ members, numbered from $1$ to $n$, is set to present a slide show at a large meeting. The slide show contains $m$ slides.

There is an array $a$ of length $n$. Initially, the members are standing in a line in the order of $a_1, a_2, \ldots, a_n$ from front to back. The slide show will be presented in order from slide $1$ to slide $m$. Each section will be presented by the member at the front of the line. After each slide is presented, you can move the member at the front of the line to any position in the lineup (without changing the order of the rest of the members). For example, suppose the line of members is $[\color{red}{3},1,2,4]$. After member $3$ presents the current slide, you can change the line of members into either $[\color{red}{3},1,2,4]$, $[1,\color{red}{3},2,4]$, $[1,2,\color{red}{3},4]$ or $[1,2,4,\color{red}{3}]$.

There is also an array $b$ of length $m$. The slide show is considered good if it is possible to make member $b_i$ present slide $i$ for all $i$ from $1$ to $m$ under these constraints.

However, your annoying boss wants to make $q$ updates to the array $b$. In the $i$-th update, he will choose a slide $s_i$ and a member $t_i$ and set $b_{s_i} := t_i$. Note that these updates are persistent, that is changes made to the array $b$ will apply when processing future updates.

For each of the $q+1$ states of array $b$, the initial state and after each of the $q$ updates, determine if the slideshow is good.

Input

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$ and $q$ ($1 \le n, m \le 2 \cdot 10^5$; $0 \leq q \leq 2 \cdot 10^5$) — the number of members and the number of sections.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1 \le a_i \le n$) — the initial order of the members from front to back. It is guaranteed that each integer from $1$ to $n$ appears exactly once in $a$.

The third line of each test case contains $m$ integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the members who should present each section.

Each of the next $q$ lines contains two integers $s_i$ and $t_i$ ($1 \le s_i \le m$, $1 \le t_i \le n$) — parameters of an update.

It is guaranteed that the sum of $n$, the sum of $m$ and the sum of $q$ over all test cases do not exceed $2 \cdot 10^5$ respectively.

Output

For each test case, output $q+1$ lines corresponding to the $q+1$ states of the array $b$. Output "YA" if the slide show is good, and "TIDAK" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yA", "Ya", "ya", and "YA" will be recognized as positive responses.

Example

Input

3
4 2 2
1 2 3 4
1 1
1 2
1 1
3 6 2
1 2 3
1 1 2 3 3 2
3 3
2 2
4 6 2
3 1 4 2
3 1 1 2 3 4
3 4
4 2

Output

YA
TIDAK
YA
YA
TIDAK
YA
TIDAK
YA
YA

Note

For the first test case, you do not need to move the members as both slides are presented by member $1$, who is already at the front of the line. After that, set $b_1 := 2$, now slide $1$ must be presented by member $2$ which is impossible as member $1$ will present slide $1$ first. Then, set $b_1 = 1$, the $b$ is the same as the initial $b$, making a good presentation possible.

解題思路

  當 $b$ 固定後，要判斷其是否是好的，只需從 $b$ 中依次把第一次出現的數選出來，並判斷所構成的序列 $c$ 是否等於 $a$ 的字首即可。例如有 $a = [2, 4, 1, 3]$，$b = [2, 2, 4, 2, 4, 1, 4]$，依次選出第一次出現的數就是 $c = [2, 4, 1]$，等於 $a$ 的字首，說明是好的。如果 $c$ 不是 $a$ 的字首，說明必定存在最小的位置 $i$ 使得 $a_i \ne c_i$，說明在 $a$ 中 $a_i$ 比 $c_i$ 先出現。又因為 $c_1 \sim c_{i-1}$ 中不存在 $a_i$，因此 $a_i$ 必定會與 $b$ 中某個位置發生失配。否則 $c$ 是 $a$ 的字首，容易知道存在構造方案使得 $b$ 是好的。

  由於涉及到修改，容易想到維護 $c$，每次修改相當於令 $c$ 中某一段迴圈左移或右移一個單位，非常難維護而且還要判斷是否為 $a$ 的字首，一開始死磕這種方法做不出來。參考了一下別人的程式碼，發現實際上維護的是每個值第一次出現的位置。定義 $f(i)$ 表示在 $b$ 中值 $a_i$ 第一次出現的位置（不存在則設為 $+\infty$），如果 $c$ 是 $a$ 的字首，說明在 $b$ 中一定是先出現 $a_1$，再出現 $a_2$，再出現 $a_3$ 以此類推，這等價於 $f(1) < f(2) < \cdots < f(n)$。所以可以用一個變數 $\text{cnt}$ 統計不滿足 $f(i) < f(i+1)$ 的數目，如果 $\text{cnt} = 0$ 說明 $b$ 是好的，否則不好。

  給 $1 \sim n$ 每個值開一個 std::set 存 $b$ 中出現的位置。當將 $b_x$ 修改成 $t$ 時，因為 $f(p_{b_x})$ 和 $f(p_{t})$ 會改變，其中 $p_x$ 表示 $a$ 中值 $x$ 的下標，因此受影響的只有 $f(p_{b_x} - 1)$，$f(p_{b_x})$，$f(p_{b_x} + 1)$ 之間的關係，以及 $f(p_{t} - 1)$，$f(p_{t})$，$f(p_{t} + 1)$ 之間的關係。在將 $b_x$ 改成 $t$ 的過程中相應地更新 $\text{cnt}$ 即可，同時還要維護 $b_x$ 和 $t$ 對應的 std::set。

  AC 程式碼如下，時間複雜度為 $O(q\log{n})$：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N = 2e5 + 5;

int n, m, k;
int a[N], b[N], p[N];
set<int> st[N];
int f[N], cnt;

void upd(int x, int op) {
    if (p[b[x]] - 1 > 0) cnt -= f[p[b[x]] - 1] > f[p[b[x]]];
    if (p[b[x]] + 1 <= n) cnt -= f[p[b[x]]] > f[p[b[x]] + 1];
    if (op) st[b[x]].insert(x);
    else st[b[x]].erase(x);
    f[p[b[x]]] = *st[b[x]].begin();
    if (p[b[x]] - 1 > 0) cnt += f[p[b[x]] - 1] > f[p[b[x]]];
    if (p[b[x]] + 1 <= n) cnt += f[p[b[x]]] > f[p[b[x]] + 1];
}

void solve() {
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        p[a[i]] = i;
    }
    for (int i = 1; i <= n; i++) {
        st[i].clear();
        st[i].insert(m + 1);
    }
    for (int i = 1; i <= m; i++) {
        cin >> b[i];
        st[b[i]].insert(i);
    }
    for (int i = 1; i <= n; i++) {
        f[i] = *st[a[i]].begin();
    }
    cnt = 0;
    for (int i = 1; i < n; i++) {
        cnt += f[i] > f[i + 1];
    }
    cout << (cnt ? "TIDAK" : "YA") << '\n';
    while (k--) {
        int x, y;
        cin >> x >> y;
        upd(x, 0);
        b[x] = y;
        upd(x, 1);
        cout << (cnt ? "TIDAK" : "YA") << '\n';
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
    
    return 0;
}

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