HackerRank "Kruskal (MST): Really Special Subtree"

Kruskal Algorithm is based on Union-Find - quite intuitive.

#include <vector>
#include <iostream>
#include <queue>
#include <unordered_map>
#include <unordered_set>
using namespace std;

struct Edge
{
    Edge() :s(0), t(0), d(0) {}
    Edge(unsigned rs, unsigned rt, unsigned rd) : s(rs), t(rt), d(rd) {}

    unsigned s;
    unsigned t;
    unsigned d;

    bool operator()(const Edge &e1, const Edge &e2)
    {
        if (e1.d != e2.d)    return e1.d > e2.d;
        return (e1.s + e1.d + e1.t) > (e2.s + e2.d + e2.t);
    }
};

//    Union-Set
unordered_map<unsigned, unsigned> pm; // smaller as parent
unsigned get_union_id(unsigned id)
{
    if (pm.find(id) == pm.end())
    {
        pm[id] = id;
        return id;
    }

    unsigned ret = id;
    while(ret != pm[ret]) ret = pm[ret];
    return ret;
}

void union_2id(unsigned id0, unsigned id1)
{
    unsigned p0 = get_union_id(id0);
    unsigned p1 = get_union_id(id1);

    if(p0 < p1)         pm[p1] = id0;
    else if(p0 > p1)     pm[p0] = id1;
}
//////////////////////////////////////

int main()
{
    long n, m; cin >> n >> m;

    priority_queue<Edge, vector<Edge>, Edge> q, qs;

    //    from -> to -> length
    unordered_map<unsigned, unordered_map<unsigned, unsigned>> g;
    for (int i = 0; i < m; i++)
    {
        unsigned a, b, d; cin >> a >> b >> d;
        q.push(Edge(min(a,b), max(a,b), d));
        g[a][b] = g[b][a] = d;
    }

    unsigned long long ret = 0;

    //    Step 1
    unsigned s; cin >> s;
    for (auto &kv : g[s])
    {
        unsigned a = s;
        unsigned b = kv.first;
        qs.push(Edge(min(a, b), max(a, b), kv.second));
    }
    //    pick first
    const Edge &first = qs.top();
    ret += first.d;
    union_2id(first.s, first.t);

    while (!q.empty())
    {
        Edge picked = q.top(); q.pop();

        if (get_union_id(picked.s) == get_union_id(picked.t)) continue;
        union_2id(picked.s, picked.t);

        ret += picked.d;
    }
    cout << ret << endl;
    return 0;
}