We'll process the agents in increasing order of authorization level, computing the \(D\) value for each one in turn. Let \(G_{i,j} = 1\) if agents with the current authorization level may enter room \((i, j)\), which is the only information required to process an agent. All \(G\) values are initially \(0\), and before processing each next agent with an authorization level of \(x\), we'll consider each room \((i, j)\) such that \(A_{i,j}\) is greater than the previous agent's authorization level and no greater than \(x\), and set \(G_{i,j}\) to \(1\) for each one. Overall, this process boils down to processing a list of \(3N + M\) events sorted by authorization level.

When processing agent \(i\), assuming \(R1_i \le R2_i\) w.l.o.g., we can consider any potentially optimal route to consist of the following sequence of segments:

1. Move from room \((R1_i, C1_i)\) to some room \((R1_i, a)\) in the initial row such that \(G_{R1_i,a} = 1\).
a) If \(a = C1_i\), this requires \(0\) moves.
b) If \(|C1_i - a| = 1\), this requires \(1\) horizontal move.
c) Otherwise, if \(|C1_i - a| = 2\), this can be done by first moving up to cell \((x, C1_i)\) (in some row \(x\) such that \(x \le R1_i\)), then across to cell \((x, a)\), and finally back down to cell \((R1_i, a)\), assuming all involved cells have \(G\) values of \(1\). This requires \(2(R1_i - x) + 2\) moves, though there might be no valid choice of \(x\). If any valid rows \(x\) do exist, the best one will be the largest \(x\) such that \(x \le R1_i\) and \(G_{x,2} = 1\).
2. Move from room \((R1_i, a)\) down to some room \((R2_i, b)\) in the final row such that \(G_{R2_i,b} = 1\), using only downward and horizontal moves (upward moves would be of no use).
3. Move from room \((R2_i, b)\) to the final room \((R2_i, C2_i)\). This segment is nearly identical to segment #1, just mirrored vertically — for example, if \(|C2_i - b| = 2\), this can be done by moving down to some row \(y\) such that \(y \ge R2_i\).

For each agent \(i\), we'll consider all \(9\) possible combinations of \(a\) and \(b\), and evaluate the number of moves involved in the route dictated by each (if a valid path exists).

For segment #1, only the \(|C1_i - a| = 2\) case is tricky. To handle it, we should start by computing the best candidate value of \(x\), which can be done in \(O(\log(N))\) time if we maintain an ordered set of row indices \(j\) such that \(G_{j,2} = 1\) (which we can in a total of \(O(N\log(N))\) time). To determine whether \(x\) is valid, we can compare it against the largest row \(x'\) such that \(x' \le R1_i\) and \(G_{x',0} = 0\) or \(G_{x',2} = 0\). A set of such rows \(x'\) can similarly be maintained in a total of \(O(N \log(N))\) time and queried in \(O(\log(N))\) time per route. Finally, a near-identical process may be repeated for segment #3.

To help with segment #2, we'll maintain a segment tree over the \(N\) rows. Each node \(i\) representing an interval of rows \(x_i..y_i\) will store a \(3 \times 3\) matrix \(M_i\) such that \(M_{i,a,b}\) is the minimum number of moves currently required to move from room \((x_i, a)\) to room \((y_i, b)\) using only downward and horizontal moves.

The matrices from two adjacent intervals' nodes \(i\) and \(j\) (such that \(y_i + 1 = x_j)\) can be combined to obtain a matrix \(M'\) for rows \(x_i..y_j\) by considering each possible transitionary column \(c\) for moving from row \(y_i\) to row \(x_j\) — in particular, \(\displaystyle M'_{a,b} = \min_{c=1..3} \{{M_i}_{a,c} + 1 + {M_j}_{c,b}\}\). This can be applied when recomputing a node's matrix based on those of its children (during segment tree updates), and when combining multiple nodes' matrices (during segment tree queries). A matrix computed from querying the segment tree provides exactly the information required to evaluate segment #2 of a route.

Setting each \(G\) value to \(1\) corresponds to a segment tree update, while processing each agent \(i\) requires querying the segment tree for the interval of rows \(R1_i..R2_i\) (assuming \(R1_i \le R2_i)\), each of which can be done in \(O(\log(N))\) time (with a factor of \(3^3\) for dealing with the matrices). This brings up the total time complexity to \(O((N + M)\log(N))\).

[See David Harmeyer (SecondThread)'s solution video here.](https://youtu.be/5sNUbjuRlBA)