## 1) Abridged problem statement

Given an undirected weighted graph with `N` junctions and `M` roads. A criminal starts at junction `X` and will travel to junction `Y` using **some shortest path** from `X` to `Y` (not necessarily unique).

For each junction `K` (1..N), consider the moment when the criminal **could be** at junction `K` (i.e., after `dist(X,K)` time units). At that same time, depending on which shortest path the criminal chose, he might be in multiple possible points of the village: either exactly at some junction or at some point inside a road.

Let `A_K` be the number of distinct points in the graph that the criminal could occupy **at time `dist(X,K)`**, assuming he follows a shortest `X→Y` route. If junction `K` is not on any shortest `X→Y` route, then `A_K = 0`.

Output `A_1, A_2, ..., A_N`.

---

## 2) Detailed editorial (explaining the given solution)

### Key idea: Only shortest-path structure matters

Let `dist[u]` be the shortest distance from `X` to `u` (Dijkstra from `X`).

A path from `X` to `Y` is shortest iff it always moves along edges `(u,v,w)` where:
- `dist[u] + w = dist[v]` (in the direction of increasing distance)

These edges form a directed acyclic graph (DAG) when oriented from smaller `dist` to larger `dist`. Call it the **shortest-path DAG from X**.

But we only care about shortest paths that actually can reach `Y`. So we further restrict to nodes/edges that lie on **some** shortest path from `X` to `Y`.

---

### Step 1: Dijkstra from `X`

Compute all `dist[u]` in `O((N+M) log N)`.

---

### Step 2: Build reverse graph of shortest-path DAG and mark nodes that can reach `Y`

For every undirected road `(u,v,w)`, if `dist[u]+w==dist[v]`, then in the DAG we have `u -> v`.
To compute which nodes lie on some shortest path to `Y`, it's convenient to store **reverse edges**:

- If `u -> v` is a shortest edge, store `u` in `dag_rev[v]`.

Then run DFS/BFS starting from `Y` in `dag_rev`.
If `reachable[u] = true`, then there exists a shortest path from `X` to `Y` passing through `u`.

If `reachable[K]` is false, then `A_K = 0` immediately.

---

### Step 3: What are the possible positions at a given time `t`?

Fix a time value `t`. The criminal moves at constant speed 1 along roads.

Along any shortest-path-to-Y route:
- He can be at a vertex `u` exactly when `t = dist[u]`.
- He can be inside a DAG edge `u -> v` (length `w`) at time `t` iff:
  - he has reached `u` at time `dist[u]`,
  - then moves along the edge,
  - so possible times along that edge are in `(dist[u], dist[v])`.

Because we query only at times `t = dist[K]` (integers), and we must not count vertex points as "inside an edge", the edge contributes for integer times:

- `t ∈ [dist[u]+1, dist[v]-1]`

But only for edges that are actually usable on a shortest path to `Y`, i.e. where `reachable[v] = true` (since after entering `v`, you must still be able to get to `Y` via DAG).

Also, every reachable vertex `u` contributes **one possible point** at time `dist[u]` (being exactly at `u`), no matter what.

So for each reachable vertex `k`, `A_k` equals:

> (# of reachable vertices u with dist[u] == dist[k])
> + (# of shortest DAG edges u->v with reachable[v] where dist[k] is strictly between dist[u] and dist[v])

---

### Step 4: Reduce to offline "range add + point query"

Each usable edge `u -> v` contributes +1 to all integer times in:
- `[L, R] = [dist[u]+1, dist[v]-1]` (if `L <= R`)

Each reachable vertex `u` contributes +1 at the single time:
- `t = dist[u]`

Then for each reachable `k`, we query the total count at `t = dist[k]`.

This is a classic offline sweep-line with events:

- IN event at `L`: `balance += 1`
- OUT event at `R`: `balance -= 1` (done after queries at `R`)
- vertex point contribution can be handled as a tiny range `[dist[u], dist[u]]` via IN at `dist[u]` and OUT at `dist[u]`
- QUERY event at `dist[k]`: record current `balance`

Important: ordering at same coordinate must be:
1. IN
2. QUERY
3. OUT
so that ranges including the point are counted.

The code encodes event type as:
- `0` = IN
- `1` = QUERY
- `2` = OUT
and sorts tuples `(pos, type, data)` so this priority is automatic.

Complexity: sorting `O(E log E)` where `E = O(N + M)` events.

---

## 3) C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

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;
};

const int64_t inf = (int64_t)1e18 + 42;

int n, x, y, m;
vector<vector<pair<int, int>>> adj;

void read() {
    cin >> n >> x >> y;
    x--;
    y--;
    cin >> m;
    adj.assign(n, {});
    for(int i = 0; i < m; i++) {
        int u, v, w;
        cin >> u >> v >> w;
        u--;
        v--;
        adj[u].push_back({v, w});
        adj[v].push_back({u, w});
    }
}

void solve() {
    // We are only interested in shortest paths, so the first step is to build
    // the Dijkstra DAG of optimal edges. The Dijkstra will start from X, but we
    // also want to afterwards compute reachable(u) which will be true if Y is
    // reachable from u through the DAG (this can be done with one DFS).
    //
    // We are now interested in the number of points in the DAG that have
    // distance equal to dist(i) for each i. And we want to compute this for all
    // i quickly. Let's denote this with a function f(a) which will denote the
    // number of points with distance equal to a. We will maintain this in a
    // smart way, so then then we simply want to evaluate the function in N
    // points. We can first separately figure the contribution from points that
    // fall within a vertex - if u is a vertex with reachable(u) = true, then
    // f(dist(u)) should be increased by 1. Now we are left with the case of the
    // contribution coming from points on edges. Let's say we have an edge (u,
    // v), and reachable(v) = true. Then we can be on that edge at any point a,
    // such that dist(u) < a < dist(v). This is essentially a range of values.
    // We only care about integer query points, and we want to be careful about
    // not over-counting points that fall in vertices. Hence, we can increase
    // f(dist(u)+1), ..., f(dist(v)-1) by one, or one range per edge.
    //
    // This now ends up being a standard problem - we have N ranges, and N
    // queries, where for every query we want to figure out by how many ranges
    // it's covered. This can be done with splitting the ranges into a pair of
    // events - an IN event with "balance += 1" at L, and an OUT event with
    // "balance -= 1" at R. And we also add QUERY events that will look at the
    // balance. Priority should be IN, QUERY, OUT. Note that this approach is
    // offline, while online we could do segment trees but this is less pleasant
    // here as the edges have higher weights so we need to use a dynamic segment
    // tree / treap.
    //
    // All steps of the solution are in O((N + M) log M).

    vector<int64_t> dist(n, inf);
    priority_queue<pair<int64_t, int>, vector<pair<int64_t, int>>, greater<>>
        pq;
    dist[x] = 0;
    pq.push({0, x});
    while(!pq.empty()) {
        auto [d, u] = pq.top();
        pq.pop();
        if(d > dist[u]) {
            continue;
        }
        for(auto [v, w]: adj[u]) {
            if(dist[u] + w < dist[v]) {
                dist[v] = dist[u] + w;
                pq.push({dist[v], v});
            }
        }
    }

    vector<vector<int>> dag_rev(n);
    for(int u = 0; u < n; u++) {
        for(auto [v, w]: adj[u]) {
            if(dist[u] + w == dist[v]) {
                dag_rev[v].push_back(u);
            }
        }
    }

    vector<bool> reachable(n, false);
    function<void(int)> dfs = [&](int u) {
        if(reachable[u]) {
            return;
        }
        reachable[u] = true;
        for(int v: dag_rev[u]) {
            dfs(v);
        }
    };
    dfs(y);

    vector<tuple<int64_t, int, int>> events;
    for(int u = 0; u < n; u++) {
        if(reachable[u]) {
            events.push_back({dist[u], 0, 0});
            events.push_back({dist[u], 2, 0});
        }
        for(auto [v, w]: adj[u]) {
            if(dist[u] + w == dist[v] && reachable[v]) {
                int64_t l = dist[u] + 1;
                int64_t r = dist[v] - 1;
                if(l <= r) {
                    events.push_back({l, 0, 0});
                    events.push_back({r, 2, 0});
                }
            }
        }
    }
    for(int k = 0; k < n; k++) {
        if(reachable[k]) {
            events.push_back({dist[k], 1, k});
        }
    }

    sort(events.begin(), events.end());

    vector<int> answer(n, 0);
    int balance = 0;
    for(auto [pos, type, data]: events) {
        if(type == 0) {
            balance++;
        } else if(type == 1) {
            answer[data] = balance;
        } else {
            balance--;
        }
    }

    cout << answer << '\n';
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (same approach) with detailed comments

```python
import sys
import heapq

INF = 10**30

def solve() -> None:
    data = list(map(int, sys.stdin.buffer.read().split()))
    it = iter(data)

    n = next(it)
    x = next(it) - 1   # 0-based
    y = next(it) - 1

    m = next(it)

    # Build adjacency list for undirected weighted graph
    adj = [[] for _ in range(n)]
    for _ in range(m):
        u = next(it) - 1
        v = next(it) - 1
        w = next(it)
        adj[u].append((v, w))
        adj[v].append((u, w))

    # 1) Dijkstra from x to get shortest distances dist[]
    dist = [INF] * n
    dist[x] = 0
    pq = [(0, x)]  # (distance, node)

    while pq:
        d, u = heapq.heappop(pq)
        if d != dist[u]:
            continue  # stale entry
        for v, w in adj[u]:
            nd = d + w
            if nd < dist[v]:
                dist[v] = nd
                heapq.heappush(pq, (nd, v))

    # 2) Build reverse edges of shortest-path DAG:
    # if dist[u] + w == dist[v], then u -> v is in DAG, so store u in dag_rev[v]
    dag_rev = [[] for _ in range(n)]
    for u in range(n):
        du = dist[u]
        for v, w in adj[u]:
            if du + w == dist[v]:
                dag_rev[v].append(u)

    # 3) Mark all nodes that can reach y using only DAG edges (reverse DFS/BFS from y)
    reachable = [False] * n
    stack = [y]
    while stack:
        u = stack.pop()
        if reachable[u]:
            continue
        reachable[u] = True
        for p in dag_rev[u]:   # predecessors in DAG
            if not reachable[p]:
                stack.append(p)

    # 4) Build offline sweep-line events:
    # event = (pos, type, data), type: 0=IN, 1=QUERY, 2=OUT
    events = []

    for u in range(n):
        if reachable[u]:
            # vertex point at time dist[u] (as IN and OUT at same position)
            events.append((dist[u], 0, -1))
            events.append((dist[u], 2, -1))

        # for each DAG edge u->v with reachable[v], add range [dist[u]+1, dist[v]-1]
        du = dist[u]
        for v, w in adj[u]:
            if du + w == dist[v] and reachable[v]:
                l = du + 1
                r = dist[v] - 1
                if l <= r:
                    events.append((l, 0, -1))
                    events.append((r, 2, -1))

    # queries: answer[k] at time dist[k] for reachable nodes
    for k in range(n):
        if reachable[k]:
            events.append((dist[k], 1, k))

    # Sort so that at same pos: IN (0) before QUERY (1) before OUT (2)
    events.sort()

    ans = [0] * n
    balance = 0
    for pos, typ, dat in events:
        if typ == 0:
            balance += 1
        elif typ == 1:
            ans[dat] = balance
        else:
            balance -= 1

    # Print all answers space-separated (matching the C++ behavior)
    sys.stdout.write(" ".join(map(str, ans)) + "\n")


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

---

## 5) Compressed editorial

1. Run Dijkstra from `X` to get `dist[u]`.
2. Build shortest-path DAG edges by condition `dist[u]+w==dist[v]` (directed increasing `dist`). Store them reversed.
3. DFS from `Y` on reversed DAG to mark `reachable[u]` = node lies on some shortest `X→Y` path.
4. For each reachable vertex `u`, add a point contribution at time `dist[u]`.
5. For each DAG edge `u→v` with `reachable[v]`, add a +1 range on integer times `[dist[u]+1, dist[v]-1]`.
6. Offline sweep-line over all range endpoints and queries at `dist[k]`:
   - IN before QUERY before OUT at equal coordinate.
   - Current `balance` gives `A_k`. Unreachable nodes output 0.
