(Analysis by Brian Dean)

There are a few ways to approach this problem. Probably the most straightforward is a binary search on the answer (say, $x$), where in each step we preserve only the edges of cost $<x$ in our graph and count the number of connected components. If the number is less than $K$, then our guess for $x$ was too high. Unfortunately, the running time here is $O(N^2)$ to find connected components, times the number of steps in the binary search, which is the log of 2019201997 --- about 33. This will likely time out on large input cases.

The key structural insight is that the answer to this problem comes from a minimum spanning tree (MST) of our graph. If you want to split into $K=2$ clusters, you can do this by finding an MST and removing its highest-weight edge, thereby splitting it into two pieces that represent two sides of our partition. More generally, to split into $K$ clusters, you just remove the $K-1$ highest-weight edges from the MST. This makes sense if you think of the operation of Kruskal's MST algorithm: consider the state of the algorithm when it has built up $K$ connected components and is about to add an edge of weight $x$ to join the next pair of components together --- at this point, each component is internally connected by a spanning tree containing edges of weight at most $x$, and running between these components are only edges of weight at least $x$ (otherwise components would have been joined earlier in the process).

So we see that we can answer this question by taking the $(K-1)$th largest edge weight in an MST, which means all we need to do is compute an MST. There are many MST algorithms out there, but unfortunately many of them (e.g., Kruskal, Boruvka) run in $O(N^2 \log N)$ time for this problem since it involves a dense graph. Kruskal's algorithm can be modified to run in a dense graph in $O(N^2)$ time (left as an exercise to the reader), or we can note that the sorting step is really the bottleneck with Kruskal, so we can sort the edges quickly (e.g., with something like a two-pass radix sort in linear time), after which Kruskal's algorithm runs in nearly linear time. Probably even better, however, would be to instead use Jarnik's algorithm (perhaps more commonly known as Prim's algorithm, which is also pretty much equivalent to Dijkstra's shortest path algorithm --- in the MST context, I prefer to call it Jarnik's algorithm since Jarnik's publication pre-dated Prim's by a substantial amount). This gives a simple way to build the MST in just $O(N^2)$ time.

Some mathematically enterprising students even managed to reverse-engineer our "random" edge weight function to solve the problem without using algorithmic methods --- although this was not the intent of the problem (and indeed, such solutions would break if a different edge weight function was used), these solutions were still awarded full marks.

Here is my C++ code. It basically runs Jarnik's algorithm to find an MST, then sorts the MST edges to find the right one.

#include <iostream>
#include <fstream>
#include <algorithm>
using namespace std;
#define MAX_N 7500

int N, K, visited[MAX_N+1];
long long D[MAX_N+1];

// Prim/Jarnik MST algorithm
void jarnik(int source)
{
for (int i=1; i<=N; i++) D[i] = 2019201997;
for (int iter=0; iter<N; iter++) {
int min_i = 0;
for (int i=1; i<=N; i++)
if (!visited[i] && (min_i==0 || D[i] < D[min_i])) min_i = i;
visited[min_i] = 1;
for (int j=1; j<=N; j++)
if (!visited[j])
D[j] = min(D[j], (2019201913LL*min(min_i,j)+2019201949LL*max(min_i,j)) % 2019201997LL);
}
}

int main(void)
{
ifstream fin ("walk.in");
fin >> N >> K;
ofstream fout ("walk.out");
jarnik(1);
sort (D+1,D+N+1);
fout << D[N+1-K] << "\n";
return 0;
}