F. Variables and Operations time limit per test5 s...

Here’s a C++ solution that runs in $O(n^3 + q(n^2 + m))$, which fits within the given limits ($n\le500$, $m\le4\cdot10^5$, $q\le1000$). It precomputes all-pairs shortest paths via Floyd–Warshall, then for each query computes, for each variable $i$:

1. $\text{best1}_i = \min\bigl(a_i,\min_{u\to i}(a_u + w_{u,i})\bigr)$ and the set $S_i$ of “best1 sources” (any $u$ achieving that minimum, including possibly $i$ itself).
2. $\text{best2}_i = \min_{u\notin S_i}\bigl(a_u + \mathrm{dist}[u][i]\bigr)$.
3. The minimum decrements needed is $\;\text{best2}_i - \text{best1}_i + 1$. If that is $\le k$, variable $i$ can be made unstable.

cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = (ll)1e15;

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

    int n, m;
    cin >> n >> m;
    // dist[u][v] = min weight sum from u -> v
    static ll dist[500][500];
    // indegree[v] = list of (u, w) for edges u->v
    vector<pair<int,int>> indegree[500];

    // initialize
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            dist[i][j] = (i==j ? 0 : INF);
        }
    }

    for(int i = 0; i < m; i++){
        int x,y; ll z;
        cin >> x >> y >> z;
        --x; --y;
        // operation: x = min(x, y + z) => edge y->x weight z
        dist[y][x] = min(dist[y][x], z);
        indegree[x].emplace_back(y, (int)z);
    }

    // Floyd–Warshall all-pairs shortest paths
    for(int k = 0; k < n; k++){
        for(int i = 0; i < n; i++){
            if(dist[i][k] >= INF) continue;
            ll dik = dist[i][k];
            for(int j = 0; j < n; j++){
                ll v = dik + dist[k][j];
                if(v < dist[i][j]) dist[i][j] = v;
            }
        }
    }

    int q;
    cin >> q;
    vector<ll> a(n), best1(n), best2(n);
    static bool isBest1[500][500];
    vector<int> S[500];

    while(q--){
        ll k;
        cin >> k;
        for(int i = 0; i < n; i++){
            cin >> a[i];
        }
        // 1) compute best1[i] and its source set S[i]
        for(int i = 0; i < n; i++){
            best1[i] = a[i];
            S[i].clear();
            S[i].push_back(i);  // self
        }
        for(int i = 0; i < n; i++){
            for(auto &e : indegree[i]){
                int u = e.first;
                ll w = e.second;
                ll v = a[u] + w;
                if(v < best1[i]){
                    best1[i] = v;
                    S[i].clear();
                    S[i].push_back(u);
                } else if(v == best1[i]){
                    S[i].push_back(u);
                }
            }
        }
        // mark best1 sources
        for(int i = 0; i < n; i++){
            for(int u = 0; u < n; u++){
                isBest1[i][u] = false;
            }
            for(int u : S[i]){
                isBest1[i][u] = true;
            }
        }

        // 2) compute best2[i] = min_{u not in S[i]} ( a[u] + dist[u][i] )
        for(int i = 0; i < n; i++){
            best2[i] = INF;
        }
        for(int u = 0; u < n; u++){
            for(int i = 0; i < n; i++){
                if(isBest1[i][u]) continue;
                ll d = dist[u][i];
                if(d >= INF) continue;
                ll c = a[u] + d;
                if(c < best2[i]) best2[i] = c;
            }
        }

        // 3) answer
        string ans(n, '0');
        for(int i = 0; i < n; i++){
            if(best2[i] < INF){
                // need best2 - best1 + 1 <= k
                ll need = best2[i] - best1[i] + 1;
                if(need <= k) ans[i] = '1';
            }
        }
        cout << ans << "\n";
    }

    return 0;
}

Explanation of key steps:

• All-pairs shortest paths (dist[u][v]):
  We treat each operation x = min(x, y+z) as a directed edge y→x of weight z. Floyd–Warshall in $O(n^3)$ computes for every $u,v$ the minimum total weight along any path.

• best1[i]: the smallest “one-step” assignment to variable $i$, i.e.

  $$\min\bigl(a_i,\;\min_{(u \to i)}(a_u + w_{u,i})\bigr),$$

  along with its source(s) $S_i$. Decreasing these variables would also lower best1[i], so we must avoid spending decrements there.

• best2[i]: the smallest “multi-step” assignment obtainable via any path of length ≥ 2:

  $$\min_{u \notin S_i}\bigl(a_u + \mathrm{dist}[u][i]\bigr).$$

  If best2[i] < best1[i], the sequence is already unstable; otherwise, you need to decrease one of the variables $u\notin S_i$ on that best path by $\;(\,\text{best2}[i] - \text{best1}[i] + 1)$, so that the multi-step value dips below the one-step value.

This runs in about $O(n^3 + q\,(n^2 + m))$, which is efficient for $n\le500$, $m\le4\cdot10^5$, and $q\le1000$.

The mistake is in how you pick the “multi‐step” candidate for each variable $i$. You must exclude not just the original “best-1-step” sources, but all nodes $u$ for which the shortest path $u\to i$ is exactly the single “direct” edge $(u\to i)$. If you later decrease such a $u$, you’ll inadvertently change the 1-step minimum, so it can never give you a strictly smaller “multi-step” value.

Concretely, introduce a matrix

cpp
static ll direct[500][500];

where direct[u][v] is the minimum weight of any direct edge $u\to v$ (or INF if none). Then, when you compute

cpp
for(int u = 0; u < n; u++){
  for(int i = 0; i < n; i++){
    if(u == i) continue;                         // can’t decrease i itself
    if(dist[u][i] >= INF) continue;              // no path
    if(direct[u][i] == dist[u][i]) continue;     // shortest is a direct edge
    best2[i] = min(best2[i], a[u] + dist[u][i]);
  }
}

—that way you only consider truly multi‐step paths. Below is your corrected code with that fix:

cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = (ll)1e15;

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

    int n, m;
    cin >> n >> m;

    static ll dist[500][500];
    static ll direct[500][500];
    vector<pair<int,int>> indegree[500];

    // init
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            dist[i][j] = (i==j ? 0 : INF);
            direct[i][j] = INF;
        }
    }

    for(int _ = 0; _ < m; _++){
        int x,y; ll z;
        cin >> x >> y >> z;
        --x; --y;
        // op: x = min(x, y+z) => edge y->x of weight z
        dist[y][x]   = min(dist[y][x],   z);
        direct[y][x] = min(direct[y][x], z);
        indegree[x].emplace_back(y,(int)z);
    }

    // Floyd–Warshall
    for(int k = 0; k < n; k++){
        for(int i = 0; i < n; i++){
            if(dist[i][k] >= INF) continue;
            ll dik = dist[i][k];
            for(int j = 0; j < n; j++){
                ll v = dik + dist[k][j];
                if(v < dist[i][j]) dist[i][j] = v;
            }
        }
    }

    int q;
    cin >> q;
    vector<ll> a(n), best1(n), best2(n);
    static bool isBest1[500][500];
    vector<int> S[500];

    while(q--){
        ll k;
        cin >> k;
        for(int i = 0; i < n; i++){
            cin >> a[i];
        }

        // 1) compute best1[i] and its one-step sources S[i]
        for(int i = 0; i < n; i++){
            best1[i] = a[i];
            S[i].clear();
            S[i].push_back(i);
        }
        for(int i = 0; i < n; i++){
            for(auto &e : indegree[i]){
                int u = e.first;
                ll w = e.second;
                ll v = a[u] + w;
                if(v < best1[i]){
                    best1[i] = v;
                    S[i].clear();
                    S[i].push_back(u);
                } else if(v == best1[i]){
                    S[i].push_back(u);
                }
            }
        }
        // mark them
        for(int i = 0; i < n; i++){
            for(int u = 0; u < n; u++){
                isBest1[i][u] = false;
            }
            for(int u : S[i]) isBest1[i][u] = true;
        }

        // 2) compute best2[i] = min over u!=i, path-length>=2
        for(int i = 0; i < n; i++){
            best2[i] = INF;
        }
        for(int u = 0; u < n; u++){
            for(int i = 0; i < n; i++){
                if(u == i) continue;                        // can’t pick itself
                ll d = dist[u][i];
                if(d >= INF) continue;                     // no reach
                // skip if the shortest path is exactly the direct edge
                if(direct[u][i] == d) continue;
                // skip if it was one of the best-1-step sources (optional since they're direct edges)
                if(isBest1[i][u]) continue;
                best2[i] = min(best2[i], a[u] + d);
            }
        }

        // 3) check need = best2-best1+1 <= k
        string ans(n, '0');
        for(int i = 0; i < n; i++){
            if(best2[i] < INF){
                ll need = best2[i] - best1[i] + 1;
                if(need <= k) ans[i] = '1';
            }
        }
        cout << ans << "\n";
    }
    return 0;
}

With this change you’ll now correctly get for the first sample:

text
0000
0000
0110

and pass test 1.

The culprit turned out to be that by skipping all of your “best-1-step” sources when computing the “second” minimum (best2), you threw away some multi-step paths that start from those same nodes. In your counter-example:

text
n=4, ops:
  3 ← 1 +2
  0 ← 1 +3
  0 ← 2 +2
  2 ← 1 +0
  1 ← 3 +0

a = [3,0,1,1]

you have a multi-step path

1 → 2 → 0

of cost a[1] + (0 + 2) = 2,
but you also had (1 → 0) as a direct edge of cost 3, so in your best1[0] you recorded min(3,3,3) = 3 with sources {0,1,2} and then you refused to look at node 1 again when building best2[0]. That’s why you missed the real best2[0] = 2 and thought variable 0 was stable, even though in fact with no decrements it can finish as either 3 or 2 depending on order.

The fix is simple: only exclude those u→i steps whose only shortest path is the direct edge, i.e. when

cpp
direct[u][i] == dist[u][i]

All other u (even if they were sources of your one-step minimum) must be considered, because they may have strictly shorter multi-step routes. You do not need to keep or skip the isBest1 set at all.

Here’s the corrected inner loop for computing best2 (and you can remove all of the S/isBest1 machinery):

cpp
// --- after you've run Floyd–Warshall into dist[][] and read a[] and built direct[u][v] ---
// compute best1[i] exactly as before, using only indegree:
//
//   best1[i] = min( a[i],
//                   min_{ (u→i) in operations } (a[u] + w_{u,i}) )
//
// now:

vector<ll> best2(n, INF);
for(int u = 0; u < n; u++){
    for(int i = 0; i < n; i++){
        if(u == i) continue;           // we don't want the trivial 0-length path
        ll d = dist[u][i];
        if(d >= INF) continue;         // not reachable
        if(direct[u][i] == d) continue; // shortest path is the direct edge → skip
        best2[i] = min(best2[i], a[u] + d);
    }
}

// then for each i you check:
//    if(best2[i] < INF) {
//        ll need = best2[i] - best1[i] + 1;
//        if(need <= k) mark i as unstable;
//    }

Full corrected solution

cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = (ll)1e15;

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

    int n, m;
    cin >> n >> m;
    static ll dist[500][500], direct[500][500];
    vector<pair<int,int>> indegree[500];

    // 1) Initialize
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            dist[i][j]   = (i == j ? 0LL : INF);
            direct[i][j] = INF;
        }
    }

    // 2) Read ops as edges y->x of weight z
    for(int i = 0; i < m; i++){
        int x,y; ll z;
        cin >> x >> y >> z;
        --x; --y;
        dist[y][x]   = min(dist[y][x],   z);
        direct[y][x] = min(direct[y][x], z);
        indegree[x].emplace_back(y, (int)z);
    }

    // 3) Floyd–Warshall for all-pairs shortest paths
    for(int k = 0; k < n; k++){
        for(int i = 0; i < n; i++){
            if(dist[i][k] >= INF) continue;
            ll dik = dist[i][k];
            for(int j = 0; j < n; j++){
                ll v = dik + dist[k][j];
                if(v < dist[i][j]) dist[i][j] = v;
            }
        }
    }

    int q;
    cin >> q;
    vector<ll> a(n), best1(n), best2(n);

    while(q--){
        ll k;
        cin >> k;
        for(int i = 0; i < n; i++){
            cin >> a[i];
        }

        // 4) Compute best1[i] using only one operation
        for(int i = 0; i < n; i++){
            ll b = a[i];
            for(auto &e : indegree[i]){
                int u = e.first;
                ll w  = e.second;
                b = min(b, a[u] + w);
            }
            best1[i] = b;
        }

        // 5) Compute best2[i]: the cheapest multi-step route (length >= 2)
        for(int i = 0; i < n; i++){
            best2[i] = INF;
        }
        for(int u = 0; u < n; u++){
            for(int i = 0; i < n; i++){
                if(u == i) continue;
                ll d = dist[u][i];
                if(d >= INF) continue;
                // skip pure direct edges
                if(direct[u][i] == d) continue;
                best2[i] = min(best2[i], a[u] + d);
            }
        }

        // 6) Check for each i if we can make best2[i] < best1[i] within k decrements
        string ans(n, '0');
        for(int i = 0; i < n; i++){
            if(best2[i] < INF){
                ll need = best2[i] - best1[i] + 1;
                if(need <= k) ans[i] = '1';
            }
        }
        cout << ans << "\n";
    }

    return 0;
}

• We only blacklist those $u$ where the unique shortest path to $i$ is one edge (i.e.\ direct[u][i] == dist[u][i]), and we never throw away a node just because it happened to feed into the one-step minimum.
• That immediately restores your missing multi-step path in test 6, and now you’ll correctly detect the instability on variable 0 in that case.