Give solution in cpp. There are n variables; let...

Below is a completely stand-alone C++17 implementation that works in O(n3 + q·n2) time (≈ 125 M operations for the preprocessing and ≤ 2.5 · 10^8 very light operations for the worst-case 1000 queries, well inside the limits).

1. Main ideas

• Compress the m ≤ 4 ⋅ 10^5 operations into a dense weight-matrix
  w[u][v] = min z for the edge u → v.
  (only the smallest weight of parallel edges can ever matter).

• With n ≤ 500 we can afford a Floyd-Warshall on that matrix once, giving dist[u][v] = shortest-path weight.

• For every query we have an initial vector a[1…n] and may lower any entries (total budget k).

  For a fixed node i

  • an upper bound on the value it can end with in any order is

    text
    hi(i) = min(    a[i],
                    min over in-edges u→i  ( a[u] + w[u][i] ) )

    because each incoming edge must be executed once, and it will lower i at least to the minimum of the sources’ current values. Any source that already starts small enough will certainly do that, no matter when its edge is scheduled.
  • a lower bound reachable in a suitable order is

    lo(i) = min over all nodes s ( a[s] + dist[s][i] )
    

    – execute the edges of a shortest path backwards, so that each vertex on the path is relaxed after its source.

  If we can make lo(i) strictly smaller than hi(i) by spending at most k decrements, i is “i-unstable”.

  Only the first node s of the path needs to be decreased: lowering anything downstream would also lower hi(i) again. The cheapest way is therefore

  text
  need(i) = min over s that do **not** attain hi(i)
            max( 0 , a[s] + dist[s][i] - hi(i) + 1 )

  (+1 because we need a strict inequality). i is unstable iff need(i) ≤ k.

2. Complexity

• Pre-processing
  w-matrix filling: O(m)
  Floyd-Warshall: O(n³) ( ≤ 125 M primitive ops )
• Per query
  build hi ( n² ), build the “attains-hi” table ( n² ), compute every need(i) ( n² ) ⇒ O(n²)
• Space
  two n × n matrices of 64-bit integers (≈ 2 MB each) plus a boolean matrix (≈ 0.25 MB).

3. Implementation

cpp
#include <bits/stdc++.h>
using namespace std;

const long long INF = 4e18;      // enough for all path sums

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

    int n, m;                    // variables, operations
    if (!(cin >> n >> m)) return 0;

    /* -------- compress the edges ---------- */
    vector<vector<long long>> w(n, vector<long long>(n, INF));
    for (int i = 0; i < n; ++i) w[i][i] = 0;

    for (int i = 0; i < m; ++i) {
        int x, y; long long z;
        cin >> x >> y >> z;
        --x; --y;                      // 0-based
        w[y][x] = min(w[y][x], z);     // edge y → x with weight z
    }

    /* -------- all-pairs shortest paths ---------- */
    vector<vector<long long>> dist = w;          // copy
    for (int k = 0; k < n; ++k)
        for (int i = 0; i < n; ++i)
            if (dist[i][k] < INF)
                for (int j = 0; j < n; ++j)
                    if (dist[k][j] < INF &&
                        dist[i][j] > dist[i][k] + dist[k][j])
                        dist[i][j] = dist[i][k] + dist[k][j];

    /* -------- queries ---------- */
    int q;  cin >> q;
    vector<long long> a(n), hi(n);
    vector<vector<char>> isMin(n, vector<char>(n));

    string answer; answer.reserve(n);

    while (q--) {
        long long K;                   // budget
        cin >> K;
        for (int i = 0; i < n; ++i) cin >> a[i];

        /* build hi(i) and remember which nodes attain it  */
        fill(hi.begin(), hi.end(), 0);
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < n; ++j) isMin[i][j] = 0;

        for (int v = 0; v < n; ++v) {
            long long best = a[v];
            for (int u = 0; u < n; ++u)
                if (w[u][v] < INF)
                    best = min(best, a[u] + w[u][v]);
            hi[v] = best;

            /* mark minimisers */
            if (a[v] == best) isMin[v][v] = 1;
            for (int u = 0; u < n; ++u)
                if (w[u][v] < INF && a[u] + w[u][v] == best)
                    isMin[u][v] = 1;
        }

        /* for every variable decide (in)stability            */
        answer.clear();
        for (int v = 0; v < n; ++v) {
            long long need = INF;

            for (int s = 0; s < n; ++s) {
                if (s == v || isMin[s][v]) continue;        // cannot use
                if (dist[s][v] == INF)      continue;       // no path

                long long cand = a[s] + dist[s][v];
                long long delta = cand - hi[v];
                long long cost = (delta < 0) ? 0 : delta + 1;
                if (cost < need) need = cost;
                if (need == 0) break;       // early exit
            }
            answer.push_back( (need <= K) ? '1' : '0' );
        }
        cout << answer << '\n';
    }
    return 0;
}

The code follows exactly the reasoning sketched above:

• Edge compression to w.
• Floyd-Warshall to obtain the all-pairs shortest-path matrix dist.
• For each query
  • compute hi(i) and mark every source that attains it (isMin),
  • find the cheapest source–path that can beat that hi(i) without touching the same minimiser,
  • compare the required budget with k.

If the required budget does not exceed k we output 1 (the variable can indeed be made i-unstable), otherwise 0.