size(); for(auto node: dep[max_dep]) { std::sort(hkr[node]->begin(), hkr[node]->end()); if(hkr[node]->size() != expected_size) return false; for(int i = 1; i < hkr[node]->size(); ++i) { if(hkr[node]->at(i) - hkr[node]->at(i - 1) != dep[max_dep]....
设第ii次操作后叶子节点个数是cnticnti,叶子节点的深度期望是fifi,那么第ii次新加的节点期望深度和就是(fi−1+1)×m(fi−1+1)×m,并且fi=fi−1×cnti−1+(fi−1+1)×mcntifi=fi−1×cnti−1+(fi−1+1)×mcnti,显然cnti=i⋅(m−1)+1cnti=i⋅(m−1)+1。 复杂度Θ(k...
push_back({u, i}); deg[u]++, deg[v]++; } vector<int> dfn(n, -1), low(n, -1), ban(n); int tot = 0; function<void(int, int)> tarjan = [&](int cur, int pre) { dfn[cur] = low[cur] = tot++; for (auto &[nex, j] : g[cur]) { if (nex != pre) { if ...
size(); for(auto node: dep[max_dep]) { std::sort(hkr[node]->begin(), hkr[node]->end()); if(hkr[node]->size() != expected_size) return false; for(int i = 1; i < hkr[node]->size(); ++i) { if(hkr[node]->at(i) - hkr[node]->at(i - 1) != dep[max_dep]....