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 MX = 1e6 + 4;
  int p[MX], dp[MX][2][2];
  pair<int, int> from[MX][2][2];
  void upd(int i, int j, int k, int val, int pj, int pk) {
    if (val < dp[i][j][k]) {
      dp[i][j][k] = val;
      from[i][j][k] = make_pair(pj, pk);
    }
  }
  bool ans(int i, int j, int k) {
    if (dp[i][j][k] == INT_MAX) return false;
    int n = i + 1;
    int o[n];
    while (i >= 0) {
      o[i] = j ? p[i] : -p[i];
      pair<int, int> p = from[i][j][k];
      j = p.first;
      k = p.second;
      i--;
    }
    cout << "YES\n";
    for (int i = 0; i < n; i++) cout << o[i] << ' ';
    cout << '\n';
    return true;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      for (int i = 0; i < n; i++) cin >> p[i];
      for (int i = 0; i < n; i++)
        for (int j = 0; j < 2; j++)
          for (int k = 0; k < 2; k++) dp[i][j][k] = INT_MAX;
      dp[0][0][0] = INT_MIN;
      for (int i = 1; i < n; i++) {
        for (int j = 0; j < 2; j++)
          for (int k = 0; k < 2; k++)
            for (int w = 0; w < 2; w++) {
              if (dp[i - 1][j][k] == INT_MAX) continue;
              int pmax = p[i - 1];
              int pmyx = dp[i - 1][j][k];
              if (!j) pmax *= -1;
              if (k) swap(pmax, pmyx);
              int curr = p[i];
              if (!w) curr *= -1;
              if (curr < pmyx) continue;
              if (curr < pmax)
                upd(i, w, 1, pmax, j, k);
              else
                upd(i, w, 0, pmyx, j, k);
            }
      }
      bool ok;
      ok = ans(n - 1, 1, 0);
      if (!ok) ok = ans(n - 1, 1, 1);
      if (!ok) ok = ans(n - 1, 0, 0);
      if (!ok) cout << "NO\n";
    }
  }
  