1. Abridged Problem Statement  
Given a directed graph with n vertices and m positive‐weight edges, a source s and target t, report the lengths of the k shortest s→t paths (paths may revisit vertices or edges). Output them in non‐decreasing order; if fewer than k distinct paths exist, print “NO” for each missing one.

2. Detailed Editorial  
The classic way to enumerate k shortest paths (allowing cycles) in near-optimal time is Eppstein’s algorithm. It runs Dijkstra once in reverse from the target t to compute dist[v] = shortest‐path‐distance from v to t, and it builds a shortest‐path tree (actually a directed arborescence) T over those distances. Any other edge (u→v) not in T “deviates” from the tree path at u, adding extra cost = w(u→v) + dist[v] – dist[u]. Such deviations represent one‐edge detours off the shortest‐path tree, each of which yields a candidate path.  

Key steps:  
1. Reverse‐Dijkstra from t to get dist[] and pick for each u the outgoing edge in the shortest‐path tree T.  
2. For each u, collect all deviations: for each edge u→v not used in T, compute penalty cost c = w + dist[v] – dist[u], and insert (c,v) into a min‐heap Hu of deviations at u.  
3. Topologically sort the tree T from leaves up, and in that order merge each node’s heap into its parent’s heap. After this, Hs (at source s) contains all possible single‐deviation costs from the unique shortest s→t path.  
4. The first shortest path length is dist[s]. To extract the next k–1 shortest, maintain a global min‐queue of states (total_cost, heap_state). Initially push (dist[s] + top(Hs).cost, copy_of_Hs). Each time you pop (d, H), you record d, then split H into (head, left, right) via the trio operation: head = cheapest deviation at this heap node, and left/right are the subheaps representing alternative deviations. Push states reflecting continuing along left and right, and also, if the deviation leads to node v, push (d + top(Hv).cost, Hv). Repeat until k are output or queue exhausts.  

This runs in O((n + m) log n + k log k) time roughly, and memory O(n + m + k). The critical data structure is a *meldable* heap (e.g. randomized / leftist / pairing heap) supporting merge in O(log N).

3. The Provided C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Convenience operators for reading/writing pairs and vectors
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) in >> x;
    return in;
}
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto& x: a) out << x << ' ';
    return out;
}

// A randomized meldable min-heap storing keys of type T
template<class T = int>
class MeldableHeap {
  private:
    // Pseudo-random bit generator
    static uint32_t rng() {
        static mt19937 gen(random_device{}());
        return gen();
    }

    // Nodes in the heap
    struct Node {
        T key;
        Node *left, *right;
        Node(T k): key(k), left(nullptr), right(nullptr) {}
    };

    // Merge two heap roots a and b, return merged root
    Node* merge(Node* a, Node* b) {
        if(!a) return b;
        if(!b) return a;
        // Ensure a->key <= b->key
        if(a->key > b->key) swap(a, b);
        // Copy node a to avoid destroying original (for persistence)
        Node* r = new Node(a->key);
        // Randomly choose which subtree to merge into
        if(rng() & 1) {
            r->left  = merge(a->left,  b);
            r->right = a->right;
        } else {
            r->left  = a->left;
            r->right = merge(a->right, b);
        }
        return r;
    }

    // Pop top key from heap rooted at a: returns (headNode, mergedTail)
    pair<Node*, Node*> pop(Node* a) {
        Node* head = new Node(a->key);
        Node* tail = merge(a->left, a->right);
        return {head, tail};
    }

  public:
    Node* root;
    MeldableHeap(): root(nullptr) {}
    MeldableHeap(Node* r): root(r) {}
    bool empty() const { return root == nullptr; }
    T top()    const { return root->key; }

    // Copy-on-write: share the same root pointer
    MeldableHeap copy() const { return MeldableHeap(root); }

    // Push a new key by merging a singleton node
    void push(T key) {
        Node* node = new Node(key);
        root = merge(root, node);
    }

    // Pop the minimum key and return it
    T pop() {
        assert(root);
        auto pr = pop(root);
        root = pr.second;
        return pr.first->key;
    }

    // Extract trio: (root key, left-heap, right-heap)
    tuple<T, MeldableHeap, MeldableHeap> trio() const {
        return { root->key,
                 MeldableHeap(root->left),
                 MeldableHeap(root->right) };
    }

    // Merge with another heap, returning new heap (persistent)
    MeldableHeap merge(const MeldableHeap& other) const {
        MeldableHeap H;
        H.root = merge(root, other.root);
        return H;
    }
    friend MeldableHeap merge(const MeldableHeap& a, const MeldableHeap& b) {
        return a.merge(b);
    }

    // For priority_queue comparison
    bool operator<(const MeldableHeap& other) const {
        return top() < other.top();
    }
};

// Implements Eppstein's k-shortest-paths algorithm
template<class T = int64_t>
class EppsteinShortestPaths {
  private:
    static constexpr T INF = numeric_limits<T>::max() / 2;

    // Graph edge representation
    struct Edge { int u, v; T w; };

    int n;
    vector<Edge> edges;
    vector<vector<pair<int,int>>> adj, rev_adj;

    // Run Dijkstra from t on the reversed graph, return (dist, parent_edge)
    pair<vector<T>,vector<int>> build_dijkstra_tree(int t) {
        vector<T> dist(n, INF);
        dist[t] = 0;
        priority_queue<pair<T,int>, vector<pair<T,int>>, greater<>> pq;
        pq.emplace(0, t);

        while(!pq.empty()) {
            auto [d,u] = pq.top(); pq.pop();
            if(d != dist[u]) continue;
            // Relax incoming edges u<-v in original graph → outgoing in rev_adj
            for(auto [v, idx]: rev_adj[u]) {
                T nd = d + edges[idx].w;
                if(nd < dist[v]) {
                    dist[v] = nd;
                    pq.emplace(nd, v);
                }
            }
        }
        // Build shortest-path tree: for each u, pick one outgoing edge
        vector<int> tree(n, -1);
        for(int u = 0; u < n; u++) {
            for(auto [v, idx]: adj[u]) {
                if(dist[u] == dist[v] + edges[idx].w) {
                    tree[u] = idx;
                    break;
                }
            }
        }
        return {dist, tree};
    }

    // Topological sort of the tree edges to process merges bottom-up
    vector<int> topsort(const vector<int>& tree) {
        vector<int> indeg(n, 0);
        for(int u=0; u<n; u++) {
            if(tree[u] != -1) {
                int v = edges[tree[u]].v;
                indeg[v]++;
            }
        }
        queue<int> q;
        for(int u=0; u<n; u++) if(indeg[u]==0) q.push(u);
        vector<int> order;
        while(!q.empty()) {
            int u = q.front(); q.pop();
            order.push_back(u);
            if(tree[u] != -1) {
                int v = edges[tree[u]].v;
                if(--indeg[v] == 0) q.push(v);
            }
        }
        return order;
    }

  public:
    EppsteinShortestPaths(int _n=0){ init(_n); }

    void init(int _n) {
        n = _n;
        edges.clear();
        adj.assign(n, {});
        rev_adj.assign(n, {});
    }

    // Add directed edge u->v with weight w
    int add_edge(int u, int v, T w) {
        int idx = edges.size();
        edges.emplace_back(u,v,w);
        adj[u].emplace_back(v,idx);
        rev_adj[v].emplace_back(u,idx);
        return idx;
    }

    // Return up to k shortest path lengths from s to t
    vector<T> get_k_shortest_paths(int s, int t, int k) {
        // 1) Build dist[] and tree from t
        auto [dist, tree] = build_dijkstra_tree(t);
        if(dist[s] == INF || k<=0) return {};

        // 2) For each u, collect all non-tree outgoing edges as deviations
        vector<MeldableHeap<pair<T,int>>> H(n);
        for(int u=0; u<n; u++){
            for(auto [v, idx]: adj[u]){
                if(tree[u] == idx) continue;
                T cost = edges[idx].w + dist[v] - dist[u];
                H[u].push({cost, v});
            }
        }

        // 3) Merge heaps bottom-up along tree
        auto order = topsort(tree);
        reverse(order.begin(), order.end());
        for(int u: order){
            if(tree[u] != -1){
                int idx = tree[u];
                int parent = edges[idx].v ^ edges[idx].u ^ u;
                H[u] = merge(H[u], H[parent]);
            }
        }

        // 4) Extract answers
        vector<T> ans;
        ans.push_back(dist[s]); // the shortest path
        // If no deviations at s, we're done
        if((int)ans.size() >= k || H[s].empty()) return ans;

        // Min-heap of (candidate_total_cost, heap_state)
        priority_queue<
            pair<T,MeldableHeap<pair<T,int>>>,
            vector<pair<T,MeldableHeap<pair<T,int>>>>,
            greater<>
        > pq;
        // Push first deviation from s
        pq.emplace(dist[s] + H[s].top().first, H[s].copy());

        while(!pq.empty() && (int)ans.size() < k){
            auto [cur_cost, heap_state] = pq.top(); pq.pop();
            ans.push_back(cur_cost);

            // Split heap_state into head, left, right
            auto [hd, left_h, right_h] = heap_state.trio();
            T head_cost = hd.first;    // cost of this deviation
            int v = hd.second;          // node where deviation goes

            // Continue along left subheap
            if(!left_h.empty()){
                T nc = cur_cost - head_cost + left_h.top().first;
                pq.emplace(nc, left_h);
            }
            // Continue along right subheap
            if(!right_h.empty()){
                T nc = cur_cost - head_cost + right_h.top().first;
                pq.emplace(nc, right_h);
            }
            // Also consider deviations from node v
            if(!H[v].empty()){
                pq.emplace(cur_cost + H[v].top().first, H[v].copy());
            }
        }
        return ans;
    }
};

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, m, k, s, t;
    cin >> n >> m >> k;
    cin >> s >> t;
    --s; --t;

    EppsteinShortestPaths<int64_t> solver(n);
    for(int i=0,u,v,w; i<m; i++){
        cin >> u >> v >> w;
        solver.add_edge(u-1, v-1, w);
    }

    auto ans = solver.get_k_shortest_paths(s, t, k);
    // Output exactly k lines
    for(int i=0; i<k; i++){
        if(i < (int)ans.size())
            cout << ans[i] << "\n";
        else
            cout << "NO\n";
    }
    return 0;
}
```

4. Python Implementation with Detailed Comments  
```python
import sys
import threading
import heapq
def main():
    sys.setrecursionlimit(10**7)
    data = sys.stdin.read().split()
    it = iter(data)
    n = int(next(it))
    m = int(next(it))
    K = int(next(it))
    s = int(next(it)) - 1
    t = int(next(it)) - 1

    # Read graph
    adj  = [[] for _ in range(n)]
    radj = [[] for _ in range(n)]
    edges = []
    for _ in range(m):
        u = int(next(it)) - 1
        v = int(next(it)) - 1
        w = int(next(it))
        idx = len(edges)
        edges.append((u,v,w))
        adj[u].append((v, idx))
        radj[v].append((u, idx))

    INF = 10**30

    # 1) Reverse Dijkstra from t to get dist[v] = min distance v->...->t
    dist = [INF]*n
    dist[t] = 0
    hq = [(0, t)]
    while hq:
        d, u = heapq.heappop(hq)
        if d>dist[u]: continue
        for v, ei in radj[u]:
            w = edges[ei][2]
            nd = d + w
            if nd < dist[v]:
                dist[v] = nd
                heapq.heappush(hq, (nd, v))

    # 2) Build shortest-path tree: tree_edge[u] = edge idx in the shortest outgoing
    tree_edge = [-1]*n
    for u in range(n):
        for v, ei in adj[u]:
            w = edges[ei][2]
            # if taking u->v plus dist[v] equals dist[u], it's on a shortest path
            if dist[u] == w + dist[v]:
                tree_edge[u] = ei
                break

    # 3) For each node u, collect deviation edges into a min-heap H[u].
    #    Deviation cost = w + dist[v] - dist[u].
    H = [[] for _ in range(n)]
    for u in range(n):
        for v, ei in adj[u]:
            if ei == tree_edge[u]: continue
            w = edges[ei][2]
            penalty = w + dist[v] - dist[u]
            heapq.heappush(H[u], (penalty, v))

    # 4) Toposort the tree to merge heaps bottom-up
    indeg = [0]*n
    for u in range(n):
        ei = tree_edge[u]
        if ei != -1:
            vv = edges[ei][1]
            indeg[vv] += 1
    q = [u for u in range(n) if indeg[u]==0]
    topo = []
    for u in q:
        topo.append(u)
        ei = tree_edge[u]
        if ei != -1:
            vv = edges[ei][1]
            indeg[vv] -= 1
            if indeg[vv]==0:
                q.append(vv)
    # Merge in reverse topo: child into parent
    for u in reversed(topo):
        ei = tree_edge[u]
        if ei == -1: continue
        # merge H[u] and H[parent]
        p = edges[ei][1]  # parent of u
        if len(H[u]) > len(H[p]):
            H[u], H[p] = H[p], H[u]
        # now H[p] >= H[u]
        for item in H[u]:
            heapq.heappush(H[p], item)
        H[u].clear()

    # 5) Extract k shortest paths
    ans = []
    if dist[s] < INF:
        ans.append(dist[s])

    # A global heap of (total_cost, node_heap, at_node v, index in H[v])
    # But to avoid persistence, we store: (current_cost, u, idx, local_heap_snapshot)
    # Instead, we emulate Eppstein: each candidate is (d, heap_u, u)
    global_pq = []
    if H[s]:
        pen, v = H[s][0]
        # store (total_cost, node, base_heap_slice)
        heapq.heappush(global_pq, (dist[s] + pen, s, tuple(H[s])))

    while global_pq and len(ans) < K:
        d, u, heap_snapshot = heapq.heappop(global_pq)
        ans.append(d)
        # Convert snapshot back to a heap
        local = list(heap_snapshot)
        # pop the head deviation
        pen, v = local[0]
        # separate left/right is tricky in Python, so we simply remove first
        # and push the rest as new candidate
        rest = local[1:]
        if rest:
            # cheapest of rest
            pen2, v2 = rest[0]
            heapq.heappush(global_pq, (d - pen + pen2, u, tuple(rest)))
        # also consider deviations from node v
        if H[v]:
            pen3, _ = H[v][0]
            heapq.heappush(global_pq, (d + pen3, v, tuple(H[v])))

    # Output exactly K lines
    out = []
    for i in range(K):
        if i < len(ans):
            out.append(str(ans[i]))
        else:
            out.append("NO")
    sys.stdout.write("\n".join(out))

if __name__ == "__main__":
    main()
```

5. Compressed Editorial  
- Run one reverse Dijkstra from t to get dist[v] and build the shortest-path arborescence T.  
- For each non-tree edge u→v compute penalty = w + dist[v] – dist[u] and store it in a min-heap at u.  
- Merge these heaps up the tree so that at s you have all single-edge deviations.  
- The best path is dist[s]. Then repeatedly extract the next best deviation via a global min-queue of heap states, splitting each heap into subheaps and also spawning new candidates at the deviation’s endpoint v. This yields the k shortest (possibly non-simple) paths in O((n+m)log n + k log k).