cond() has complexity $$$O(n^2 \log n)$$$.

A faster solution would be just calling cond() once, with $$$m = n$$$, and return $$$i$$$ if the set size is $$$n$$$, but it's still too slow.

Because this

auto cond = [](const vector<pair<int, int>> &edges, int m, int n) {
        set<int> s{1};
        for (int i = 0; i < m; ++i)
        {
            for (auto [u, v] : edges)
            {
                if (s.count(u))
                {
                    s.insert(v);
                }
                if (s.count(v))
                {
                    s.insert(u);
                }
            }
        }
        return s.size() == n;
    };

works in $$$O(m \cdot n\log n)$$$ or, in general, $$$\Omega(ans \cdot n\log n)$$$ (that is, the cond() complexity is $$$O(n^2\log n)$$$, because $$$ans = O(n)$$$)