description: You are given a permutation p consisting of n integers 1, 2, ..., n (a permutation is an array where each element from 1 to n occurs exactly once).
  
  Let's call an array a bipartite if the following undirected graph is bipartite:
  
    * the graph consists of n vertices; 
    * two vertices i and j are connected by an edge if i < j and a_i > a_j. 
  
  
  
  Your task is to find a bipartite array of integers a of size n, such that a_i = p_i or a_i = -p_i, or report that no such array exists. If there are multiple answers, print any of them.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^5) — the number of test cases.
  
  The first line of each test case contains a single integer n (1 ≤ n ≤ 10^6) — the size of the permutation.
  
  The second line contains n integers p_1, p_2, ..., p_n.
  
  The sum of n over all test cases doesn't exceed 10^6.
  
  Output
  
  For each test case, print the answer in the following format. If such an array a does not exist, print "NO" in a single line. Otherwise, print "YES" in the first line and n integers — array a in the second line.
  
  Example
  
  Input
  
  
  4
  3
  1 2 3
  6
  1 3 2 6 5 4
  4
  4 1 3 2
  8
  3 2 1 6 7 8 5 4
  
  
  Output
  
  
  YES
  1 2 3
  NO
  YES
  -4 -1 -3 -2
  YES
  -3 -2 1 6 7 -8 -5 -4
solution: #include <bits/stdc++.h>
  using namespace std;
  const int N = 1000001;
  map<pair<int, int>, pair<int, int>> parent_state[N];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      vector<int> p(n);
      for (auto &i : p) cin >> i;
      fill(parent_state, parent_state + n + 1,
           map<pair<int, int>, pair<int, int>>());
      parent_state[0][{INT_MIN, INT_MIN}];
      for (int i = 0; i < n; ++i) {
        auto it = parent_state[i].begin();
        while (it != parent_state[i].end()) {
          auto cur = it++;
          auto [x, y] = cur->first;
          bool redundant = false;
          for (auto oth = parent_state[i].begin(); oth != cur; ++oth) {
            auto [x_oth, y_oth] = oth->first;
            redundant |= x >= x_oth && y >= y_oth;
          }
          if (redundant) {
            parent_state[i].erase(cur);
            continue;
          }
          for (auto p_nxt : {-p[i], p[i]}) {
            if (p_nxt > x) {
              parent_state[i + 1][{p_nxt, y}] = {x, y};
            } else if (p_nxt > y) {
              parent_state[i + 1][{x, p_nxt}] = {x, y};
            }
          }
        }
      }
      if (parent_state[n].empty()) {
        cout << "NO\n";
        continue;
      }
      auto [x, y] = parent_state[n].begin()->first;
      for (int i = n - 1; i >= 0; --i) {
        auto [x_parent, y_parent] = parent_state[i + 1][{x, y}];
        p[i] = x != x_parent ? x : y;
        x = x_parent, y = y_parent;
      }
      cout << "YES\n";
      for (auto i : p) {
        cout << i << " ";
      }
      cout << "\n";
    }
  }
  