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

Solution outline

We model every operation

xi ← min( axi , ayi + zi )

as a directed edge

yi  →  xi  (cost = zi)

Because zi ≥ 0, applying an operation can only decrease the value of its target; values never go up again.

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1. What is the largest value a variable can finish with?

If we run all edges that enter i first, each of them uses the initial value of its start‐vertex, so after those first moves

hi = min( ai , min over edges y→i ( ay + wy,i ) )

No ordering can leave i larger than that, therefore hi is the maximum final value of i.

(store hi as upper[i] in the code).

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2. How small can it become?

To make i as small as possible we pick a path of length ≥ 2
s → … → t → i

text
      cost = dist[s][i]          (sum of the edge costs)
      produced value = as + dist[s][i]

Run every edge on the path from s to t first – that drives t down – and run the last edge t → i at the very end; then i finishes with that produced value. Taking the shortest such path gives

lo = min over s≠i, path length≥2 ( as + dist[s][i] )

(we obtain every dist[·][·] once in advance with Floyd-Warshall).

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3. When is a variable i unstable?

i is unstable iff two different orders give two different results, i.e. lo < hi.

Because you may lower some starting values before executing anything, let us try to push lo below hi by decreasing a single source vertex s.

We never touch

• i itself – that would also lower hi;
• a vertex that has an edge directly to i and already gives hi – that would lower hi by exactly the same amount.

For any other s

need(s,i) = max( 0 , as + dist[s][i] − hi + 1 )

units have to be subtracted from as.
The smallest of those (need[i]) is the total number of unit decrements that turns i into an i-unstable variable.

i is answer 1 in a query with budget k iff need[i] ≤ k, otherwise 0.

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4. Pre-computation (once)

• w[y][x] – smallest weight of a direct edge y→x (∞ if none)
• dist – all–pairs shortest paths on that graph
  (Floyd-Warshall, n ≤ 500 → 125 million relaxations, fine)
  Both tables are O(n²) memory.

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5. Per query work ( q ≤ 1000, n = 500 )

• read ai, k
• hi ← ai
hi ← min( hi , ay + w[y][i] ) for all y
• for every i
  • scan all s
    • skip s = i
    • skip dist[s][i] = ∞
    • skip s if it owns a direct edge s→i and as + w[s][i] = hi
    • need = max(0, as + dist[s][i] − hi + 1)
    • keep the minimum
  • output 1 if that minimum ≤ k, else 0

Two dense n² scans → about 0.5 million operations per query, easily fast enough.

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6. Correctness proof (sketch)

hi is attainable: run every in-edge of i first, then anything else.
No later step can increase i, hence it never exceeds hi.

lo is attainable: take the shortest path s → … → t → i (length ≥ 2).
Execute all edges up to t first, then the rest of the graph, finally edge t→i. At that moment t already has its minimal value, so i receives as + dist[s][i].

Need formula: decreasing a source s that is not the sole direct originator of hi changes lo by exactly the same amount while leaving hi unchanged; the formula is the minimal positive amount that makes lo < hi.

Thus i is unstable iff lo < hi, and the algorithm finds the least total decrements that achieve that, proving both soundness and completeness.

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Reference implementation (GNU C++17)

cpp
#include <bits/stdc++.h>
using namespace std;
using int64 = long long;
const int64 INF = (int64)4e18;

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

    int n, m;
    if (!(cin >> n >> m)) return 0;
    vector<vector<int64>> w(n, vector<int64>(n, INF));        // direct edge
    vector<vector<int64>> dist(n, vector<int64>(n, INF));     // shortest path
    for (int i = 0; i < n; ++i) dist[i][i] = 0;

    for (int i = 0; i < m; ++i) {
        int x, y; int z;
        cin >> x >> y >> z;
        --x; --y;                               // 0-based
        if (z < w[y][x]) {                      // y → x
            w[y][x]   = z;
            dist[y][x] = z;
        }
    }

    /* Floyd-Warshall */
    for (int k = 0; k < n; ++k)
        for (int i = 0; i < n; ++i) {
            if (dist[i][k] == INF) continue;
            for (int j = 0; j < n; ++j) {
                if (dist[k][j] == INF) continue;
                auto nd = dist[i][k] + dist[k][j];
                if (nd < dist[i][j]) dist[i][j] = nd;
            }
        }

    int q;  cin >> q;
    vector<int64> a(n), upper(n);
    string answer;
    while (q--) {
        int64 k;  cin >> k;
        for (int i = 0; i < n; ++i) cin >> a[i];

        /* 1.  compute hi = upper[i] */
        for (int i = 0; i < n; ++i) upper[i] = a[i];
        for (int y = 0; y < n; ++y) {
            const int64 ay = a[y];
            for (int x = 0; x < n; ++x) if (w[y][x] != INF) {
                int64 cand = ay + w[y][x];
                if (cand < upper[x]) upper[x] = cand;
            }
        }

        answer.clear(); answer.reserve(n);
        for (int i = 0; i < n; ++i) {
            int64 need = INF;

            for (int s = 0; s < n; ++s) {
                if (s == i) continue;
                int64 d = dist[s][i];
                if (d == INF) continue;

                /* if s→i exists and uses hi, skip (path length would be 1) */
                if (w[s][i] != INF && a[s] + w[s][i] == upper[i]) continue;

                int64 delta = a[s] + d - upper[i] + 1;
                if (delta < 0) delta = 0;
                if (delta < need) need = delta;
            }

            answer.push_back( (need <= k) ? '1' : '0' );
        }
        cout << answer << '\n';
    }
    return 0;
}

The program follows exactly the reasoning above:

• pre-computation
  – w for single-edge costs, dist for all-pairs shortest paths.

• per query
  – compute every hi in O(n²),
  – find the minimal required decrements for each variable in O(n²),
  – output the 0/1 string.

Time-complexity

text
pre-processing:   O(n³)        ≤ 125 ·10⁶   operations
each query:       O(n²)        ≈ 5 ·10⁵     operations

Memory-usage

text
dist, w  :   n² · 8  bytes  ≈ 2 MB
aux arrays:  negligible

Well within the limits (n ≤ 500, m ≤ 4·10⁵, q ≤ 1000).