You are presented with a mystery function $f:\mathbb N^2\to\mathbb N$ which is known to satisfy \[f(x+1,y)>f(x,y)\quad\text{and}\quad f(x,y+1)>f(x,y)\] for all $(x,y)\in\mathbb N^2$. I will tell you the value of $f(x,y)$ for \$1. What's the minimum cost, in dollars, that it takes to compute the $19$th smallest element of $\{f(x,y)\mid(x,y)\in\mathbb N^2\}$? Here, $\mathbb N=\{1,2,3,\dots\}$ denotes the set of positive integers.
To solve this problem, we need to determine the 19th smallest element of the set \(\{f(x,y) \mid (x,y) \in \mathbb{N}^2\}\) given the properties of the function \(f\). The properties are:
1. \(f(x+1, y) > f(x, y)\)
2. \(f(x, y+1) > f(x, y)\)

These properties imply that \(f\) is strictly increasing in both arguments. Therefore, the values of \(f(x, y)\) can be visualized as a strictly increasing sequence when traversing the grid \((x, y)\) in a specific order.

To find the 19th smallest element, we can use a min-heap (priority queue) to efficiently track the smallest elements. Here is a step-by-step approach:

1. **Initialize the Min-Heap:**
   - Start with the smallest element \(f(1,1)\) and insert it into the min-heap.
   - Keep track of the elements that have been visited to avoid duplicates.

2. **Extract the Minimum Element:**
   - Extract the smallest element from the min-heap. This element is the next smallest element in the sequence.
   - Insert the next possible elements \(f(x+1, y)\) and \(f(x, y+1)\) into the min-heap if they have not been visited.

3. **Repeat the Process:**
   - Repeat the extraction and insertion process until the 19th smallest element is extracted.

Let's detail the steps with an example:

1. **Initialization:**
   - Insert \(f(1,1)\) into the min-heap.
   - Mark \((1,1)\) as visited.

2. **First Extraction:**
   - Extract \(f(1,1)\) from the min-heap.
   - Insert \(f(2,1)\) and \(f(1,2)\) into the min-heap.
   - Mark \((2,1)\) and \((1,2)\) as visited.

3. **Subsequent Extractions:**
   - Continue extracting the smallest element and inserting the next possible elements until the 19th extraction.

The cost of each extraction and insertion operation in a min-heap is \(O(\log k)\), where \(k\) is the number of elements in the heap. Since we need to perform 19 extractions, the total cost is \(O(19 \log 19)\).

The final answer is \( \boxed{ O(19 \log 19) } \) dollars.