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;
  vector<int> a[2];
  vector<int> dp[2];
  vector<int> trace[2];
  int main(int argc, char const *argv[]) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL), cout.tie(NULL);
    int n, t;
    cin >> t;
    while (t--) {
      cin >> n;
      a[0].resize(n + 1), a[1].resize(n + 1);
      for (int i = 1; i <= n; ++i) {
        cin >> a[0][i];
        a[1][i] = -a[0][i];
      }
      dp[0].resize(n + 1), dp[1].resize(n + 1);
      trace[0].resize(n + 1), trace[1].resize(n + 1);
      a[0][0] = a[1][0] = -n - 1;
      dp[0][0] = dp[1][0] = -n - 1;
      for (int i = 1; i <= n; ++i) {
        dp[0][i] = dp[1][i] = n + 1;
        for (int j = 0; j < 2; ++j) {
          if (dp[j][i - 1] > n) continue;
          for (int k = 0; k < 2; ++k) {
            if (dp[j][i - 1] < a[k][i] && a[j][i - 1] < dp[k][i]) {
              dp[k][i] = a[j][i - 1];
              trace[k][i] = j;
            }
            if (a[j][i - 1] < a[k][i] && dp[j][i - 1] < dp[k][i]) {
              dp[k][i] = dp[j][i - 1];
              trace[k][i] = j;
            }
          }
        }
      }
      if (dp[0][n] > n && dp[1][n] > n)
        cout << "NO\n";
      else {
        cout << "YES\n";
        int p = trace[0][n] > n;
        for (int i = n; i > 0; --i) {
          a[0][i] = a[p][i];
          p = trace[p][i];
        }
        for (int i = 1; i <= n; ++i) cout << a[0][i] << " \n"[i == n];
      }
    }
    return 0;
  }
  