To solve this problem, let’s preprocess the input first. We'll store the input cactus graph in an adjacency list. Let’s pick any random starting vertex \(s\) from which to perform a DFS traversal, storing the pre- and post-visiting times for each vertex \(v\) as \(tin[v]\) and \(tout[v]\) respectively.

After reversing the adjacency list for each vertex, let’s repeat the DFS from the vertex \(s\) (we'll consider all adjacent vertices in reverse order this time) and store the pre- and post-visiting times for each vertex \(v\) as \(tin'[v]\) and \(tout'[v]\) respectively. Using DFS and pre/post visiting times for every edge \(u \to v\), we can determine the following:

* Whether this edge is the first or the last in its cycle when traversing the cactus starting from \(s\)
* If \(tin[v] < tin[u]\) and \(tin'[v] < tin'[u]\), then this edge is the last of the cycle and vertex \(v\) is always the first vertex of the cycle visited on the way from vertex \(s\). Let’s denote such a vertex as a root of a cycle. Likewise, we can determine the first edge of the cycle.
* The index of the cycle this edge belongs to (or \(-1\) if it doesn’t belong to any cycle)
* This can be done in one DFS from vertex \(s\) with an additional stack to recover the path from the first cycle’s vertex to itself.

With an additional DFS traversal from vertex \(s\), we can determine the length of each cycle and the distance from each vertex to the root of the cycle it belongs to.

For our solution, we will consider all edges as directed, in order of traversal from the starting vertex \(s\). Note that for simplicity, we can consider different traversal orders of the same cycle as two different cycles (but we'll need to mark them as interchangeable). Let’s define \(dp(e, dist1, dist2, mandatory)\), where:

* \(e\) – edge ID that uniquely identifies the start \(u\) and end \(v\) of the edge;
* \(dist1\) – distance between vertex \(v\) and the closest already placed kiosk;
* \(dist2\) – distance between vertex \(v\) and the cycle root (if edge \(e\) belongs to a cycle);
* \(mandatory\) – flag for whether we must place a kiosk in this or one of the following DP states.

The number of edges \(M\) in a cactus graph of size \(N\) is at most \(3N / 2\). Any valid distance can be up to \((2K + 1) - K\) edges from the previous kiosk, plus \(1\) edge, plus \(K\) edges to the next kiosk. The flag can take on two values, so the total number of states is approximately \(3N * (2K + 1)^2\). In addition, we can force all states with \(dist1 > K + 1\) to have \(mandatory = 1\).

For convenience, we can create a dummy edge from dummy vertex \(d\) to \(s\) and use this edge as a starting edge \(e_0\). The answer to the problem is the minimum \(dp(e_0, d1, x, 1)\) across all \(d_1\) from \(K\) to \(2K + 1\). Here, \(x\) denotes any value representing edge belonging to no cycle.

The transitions that need to be considered are the following:

1. Build a kiosk at vertex \(v\).
2. If \(dist1 < 2K + 1\), keep transitioning to the next edge without building a kiosk at \(v\). For each edge, the value of \(dist'\) will simply be \(dist1 + 1\).
3. For exactly one adjacent edge, set a new strict value of \(dist1'\) that's greater than the current \(dist1 + 1\) (set \(mandatory = 1\) for this edge). This is equal to moving the next kiosk closer to the current edge. For other edges, recompute the value of \(dist1'\) considering the kiosk at the selected edge (\(mandatory = 0\) for all such edges). This can be done in \(\mathcal{O}(K)\) by iterating over all possible values of the new \(dist1\).

As previously mentioned, the total number of states is \(\mathcal{O}(NK^2)\). Each state has \(\mathcal{O}(K)\) transitions, so the total complexity of the solution is \(\mathcal{O}(NK^3)\).