tuple<ve, ValueT> solve(ve& edges, int nodes) {
sort(edges.begin(), edges.end(), edge_compare);
ve tree;
DisjointSet ds(nodes);
ValueT sum = 0;
for (Edge e : edges) {
if (!ds.sameSet(e.u, e.v)) {
ds.join(e.u, e.v);
tree.push_back(e);
sum += e.w;
} else if (tree.size() == nodes - 1) {
break;
}
}
return tuple<ve, ValueT>{tree, sum};
}
// Chinese remainder theorem: find z such that
// z % x[i] = a[i] for all i. Note that the solution is
// unique modulo M = lcm_i (x[i]). Return (z,M). On failure, M = -1.
pii chinese_remainder_theorem(const ve<int> &x, const ve<int> &a) {
pii ret = make_pair(a[0], x[0]);
for (int i = 1; i < x.size(); i++) {
ret = chinese_remainder_theorem(ret.second, ret.first, x[i], a[i]);
if (ret.second == -1) break;
}
return ret;
}
void printEdges(ve & edges){
for (int i = 0; i < edges.size(); i++){
edge e = edges[i];
printf("Edge to %i of length %f\n", e.first, e.second);
}
}
