We can represent the given network of chargers and destinations as a weighted, undirected graph $G$, whose vertices correspond to chargers (vertex IDs $1$ through $C$) and destinations (vertex IDs $C+1$ through $N$) and whose edges correspond to roads.

Subtask 1

For the first subtask, we can run Dijkstra's algorithm once using each destination as a source to check whether that destination can reach $K$ distinct charging stations. This runs in $O(NM\log N)$ time.

Subtask 2

For the second subtask, we solve the problem in $O(M\log^2 N)$ for the case $K = 2$.

Given a partition of the chargers into two groups, we first describe how to find all destinations reachable from at least one charger in each group in $O(M\log N)$. For any group of chargers, we can find all destinations reachable by some charger in the group with a "multi-source" version of Dijkstra's algorithm by initializing all chargers in the group to distance $0$. Running this once for each group lets us identify all well-connected destinations "verified" by a given partition of the chargers into two groups.

To turn this into a full solution for $K=2$, we construct a set of partitions such that each pair of chargers ends up in distinct groups at least once. Specifically, partition the chargers by the $i$th binary digit of their vertex IDs, for each $i$ between $0$ and $\log N$. Each pair of vertex IDs differs in at least one bit, so this set of partitions will separate each pair of chargers at least once. Then, running the above subroutine for each partition will "verify" every well-connected vertex at least once. Since we have $O(\log N)$ partitions, this algorithm runs in $O(M\log^2 N)$.

Subtask 3

To solve the last subtask, one can directly modify Dijkstra's algorithm. Rather than running "multi-source" Dijkstra's as in Subtask 2, we simulate running $C$ parallel copies of Dijkstra's algorithm, one from each charger. In our simulation, we process all vertices at distance $d$ from each source before processing vertices at distance $d+1$. We also break after processing after processing all vertices at distance $>R$. Exactly those vertices that are visited from at least $K$ distinct chargers are well-connected.

However, this approach as described is inefficient, because each vertex can be visited up to $C$ times. The key observation to make this approach run in time is to note that we only need to visit each vertex $v$ from $K$ distinct chargers. We can skip further visits because the first $K$ visits to $v$ will cover any well-connected vertex $u$ that is reachable through $v$.

To implement these runs of Dijkstra from $C$ sources in parallel, we can have a heap consisting of tuples (distance, vertex, root), with root denoting the Dijkstra source of that heap element. And rather than having a boolean "visited" array (which would use $O(N^2)$ memory), we track visited pairs (vertex, root) in a hash map for $O(1)$ lookups and $O(1)$ memory per entry. When visiting vertices, we skip any vertex that has been visited $K$ times and any vertex with distance $>R$.

All together, this leads to an algorithm that runs in $O(KM\log N)$. Since each vertex is visited at most $K$ times, each edge is processed at most $2K$ times. Hence, at most $O(KM)$ elements are ever inserted into the heap.

#include <iostream>
#include <queue>
#include <tuple>
#include <unordered_set>
#include <vector>
using namespace std;
const int MAXN = 50000;
int N, M, C, R, K;
vector<pair<int, int>> adj[MAXN];
int visits[MAXN];
unordered_set<int> visitors[MAXN];
int main() {
  cin >> N >> M >> C >> R >> K;
  for (int i = 0; i < M; i++) {
    int u, v, l;
    cin >> u >> v >> l;
    u -= 1;
    v -= 1;
    adj[u].push_back({v, l});
    adj[v].push_back({u, l});
  priority_queue<tuple<int, int, int>, vector<tuple<int, int, int>>, greater<tuple<int, int, int>>> pq;
  // Initialize with all charging stations in the queue
  for (int i = 0; i < C; i++) {
    pq.push({0, i, i});
  // Dijkstra's algorithm, except we (i) visit each node up to K times and (ii) only visit nodes
  // along paths of length at most R
  int well_connected = 0;
  while (!pq.empty()) {
    const auto [d, u, r] = pq.top();
    if (visits[u] == K || visitors[u].find(r) != visitors[u].end()) {
    // Update the number of visits and our count of well-connected destinations
    visits[u] += 1;
    if (u >= C && visits[u] == K) {
      well_connected += 1;
    // Add all neighbors within the range that have not been visited K times to the queue
    for (const auto& [v, l] : adj[u]) {
      if (d + l <= R && visits[v] < K) {
        pq.push({d + l, v, r});
  cout << well_connected << "\n";
  for (int i = C; i < N; i++) {
    if (visits[i] == K) {
      cout << i + 1 << "\n";