Dijkstra using set

vi D(N, 987654321);

// start vertex
set<ii> Q;
D[0] = 0;
Q.insert(ii(0,0));

while(!Q.empty()) {

   // again, fetch the closest to start element
   // from “queue” organized via set
   ii top = *Q.begin();
   Q.erase(Q.begin());
   int v = top.second, d = top.first;

   // here we do not need to check whether the distance
   // is perfect, because new vertices will always
   // add up in proper way in this implementation

   tr(G[v], it) {
        int v2 = it->first, cost = it->second;
        if(D[v2] > D[v] + cost) {
             // this operation can not be done with priority_queue,
             // because it does not support DECREASE_KEY
             if(D[v2] != 987654321) {
                   Q.erase(Q.find(ii(D[v2],v2)));
             }
             D[v2] = D[v] + cost;
             Q.insert(ii(D[v2], v2));
        }
   }
}