G. X Aura
graphs; math; shortest paths
Mount ICPC can be represented as a grid of \(R\) rows (numbered from \(1\) to \(R\)) and \(C\) columns (numbered from \(1\) to \(C\)). The cell located at row \(r\) and column \(c\) is denoted as \((r, c)\) and has a height of \(H_{r, c}\). Two cells are adjacent to each other if they share a side. Formally, \((r, c)\) is adjacent to \((r-1, c)\), \((r+1, c)\), \((r, c-1)\), and \((r, c+1)\), if any exists.You can move only between adjacent cells, and each move comes with a penalty. With an aura of an odd positive integer \(X\), moving from a cell with height \(h_1\) to a cell with height \(h_2\) gives you a penalty of \((h_1 - h_2)^X\). Note that the penalty can be negative.You want to answer \(Q\) independent scenarios. In each scenario, you start at the starting cell \((R_s, C_s)\) and you want to go to the destination cell \((R_f, C_f)\) with minimum total penalty. In some scenarios, the total penalty might become arbitrarily small; such a scenario is called invalid. Find the minimum total penalty to move from the starting cell to the destination cell, or determine if the scenario is invalid.
The first line consists of three integers \(R\) \(C\) \(X\) (\(1 \leq R, C \leq 1000; 1 \leq X \leq 9; X\) is an odd integer).Each of the next \(R\) lines consists of a string \(H_r\) of length \(C\). Each character in \(H_r\) is a number from 0 to 9. The \(c\)-th character of \(H_r\) represents the height of cell \((r, c)\), or \(H_{r, c}\).The next line consists of an integer \(Q\) (\(1 \leq Q \leq 100\,000)\).Each of the next \(Q\) lines consists of four integers \(R_s\) \(C_s\) \(R_f\) \(C_f\) (\(1 \leq R_s, R_f \leq R; 1 \leq C_s, C_f \leq C\)).
For each scenario, output the following in a single line. If the scenario is invalid, output INVALID. Otherwise, output a single integer representing the minimum total penalty to move from the starting cell to the destination cell.
Explanation for the sample input/output #1For the first scenario, one of the solutions is to move as follows: \((1, 1) \rightarrow (2, 1) \rightarrow (3, 1) \rightarrow (3, 2) \rightarrow (3, 3) \rightarrow (3, 4)\). The total penalty of this solution is \((3 - 4)^1 + (4 - 3)^1 + (3 - 6)^1 + (6 - 8)^1 + (8 - 1)^1 = 2\).Explanation for the sample input/output #2For the first scenario, the cycle \((1, 1) \rightarrow (2, 1) \rightarrow (2, 2) \rightarrow (1, 2) \rightarrow (1, 1)\) has a penalty of \((1 - 2)^5 + (2 - 0)^5 + (0 - 9)^5 + (9 - 1)^5 = -26250\). You can keep repeating this cycle to make your total penalty arbitrarily small. Similarly, for the second scenario, you can move to \((1, 1)\) first, then repeat the same cycle.