I was trying this problem HOLI but I am not able to come up with a good solution. I was thinking of pairing up the leaf nodes of the longest path in the tree. After pairing up, I remove the two paired nodes and continue with the next longest path. But this would give a O(n^2) approach resulting TLE. Can somebody tell me an efficient approach for the solving problem ? I would be happy if somebody helps me in this. Thank you in advance.

• +6

 » 7 years ago, # | ← Rev. 2 →   +7 The idea is the following:Let's take a look at one edge of the tree. If this edge is removed, there are two sub-trees instead of the initial tree. Let's assume that one of these sub-trees contains x vertices. Then there're exactly y = n — x vertices in another sub-tree. There're at most 2 * min(x, y) paths going through this edge. So all we need to know is how many vertices there are in each of two sub-trees if this edge is removed. (For edge i, let it be x[i] and y[i]) It's clear that the answer is the sum for all edges: 2 * min(x[i], y[i]) * weight[i] (weight[i] — weight of the i-th edge).There's only one thing left to do to solve the problem: calculate x[i] and y[i] for all edges quickly. It can be done with simple sub-tree dp(we need to know the number of vertices in a sub-tree). This solution works in O(N) and it's pretty simple to implement. Here's my code.
•  » » 7 years ago, # ^ |   +5 Thank you for the reply. Can you please explain how did you get that there're at most 2 * min(x, y) paths going through an edge ? I am not getting this part.
•  » » » 7 years ago, # ^ |   +6 Let's assume that for i-th edge x[i] > y[i]. If more than y[i] paths go through this edge, there'll be y[i] houses for people to stay in and more than y[i] people. So they will have to share a house.
•  » » » » 7 years ago, # ^ |   +5 Yes, I got it. The problem really simplifies a lot after this deduction which is not trivial. Got it accepted with this approach. Thanks a lot for the help.
•  » » » 2 years ago, # ^ |   0 pigeonhole principle
•  » » 2 years ago, # ^ |   0 hey wanted to know how u came up with the idea
•  » » 6 months ago, # ^ |   0 I am unable to understand two things- 1.purpose of cnt array 2. what are we achieving using min(cnt[to],n-cnt[to])*2*w[v][i]
 » 12 months ago, # |   0 It may be a silly doubt but are we sure doing c = 2 * (x, n — x) * weight will give max? Will we always find a configuration as I m thinking what if due to max contribution c of some edge, max contribution of some other edge may reduce.
•  » » 11 months ago, # ^ |   0 This approach is based on pigeon hole principle.
•  » » » 11 months ago, # ^ |   0 What he is asking is how to prove there always exists a configuration (or pairing) in which every edge gets traversed exactly $2 \cdot \min(x, n - x)$ times.Does someone have a nice proof for this?
 » 5 weeks ago, # | ← Rev. 5 →   -10 http://zobayer.blogspot.com/2014/01/spoj-holi.htmlGot from above site contains awesome explanation w.r.t to pigeon hole principle -------------- Given a weighted tree, consider there are N people in N nodes. You have to rearrange these N people such that everyone is in a new node, and no node contains more than one people and maximizes the cost. Here cost is the distance travelled for each person.  First observation is, in order to maximize cost, all edges will be used to travel around. So, if we can calculate how many times each edge will be used, we can calculate the cost. Now for any edge Ei, we can partition the whole tree into two subtrees, if one side has n nodes, the other side will have N — n nodes. Also, note that, min(n, N-n) people will be crossing the edge from each side. Because if more people cross the edge, then by pigeon-hole principle in one side, we will get more people than available node which is not allowed in the problem statement. So, Ei will be used a total of 2 * min(n, N-n) times.  Thus the final result is: cost = ∑ 2*min(ni, N-ni)*wieght(Ei) | for each edge Ei  Here, ni is the count for nodes in one side of the edge Ei and N-ni on the other side. So, we can use simple DFS algorithm to solve the problem. During the traversal, just count number of nodes in the subtree for an edge u⇒v and calculate result for each edge in the process. 
•  » » 5 weeks ago, # ^ | ← Rev. 2 →   -10 But I am getting WA on SPOJ HOLI — Holiday Accommodation #include using namespace std; typedef long long int ll; // constraints // T (1 ≤ T ≤ 10) // N (2 ≤ N ≤ 105) // X, Y, Z (1 ≤ X, Y ≤ N, 1 ≤ Z ≤ 106), class Graph{ int V; list> *adj; public: Graph(int V) { this->V=V; adj=new list>[V+1]; } void addEdge(ll u,ll v,ll wt) { adj[u].push_back({v,wt}); adj[v].push_back({u,wt}); } void print_count(ll*count) { for(ll i=1;i<=V;i++) { cout< "; for(auto nbr_pair:adj[i]) { cout<<"("<>T; while(t<=T) { int u,v,wt,V; cin>>V; // create a graph Graph g(V); pairp; for(int i=0;i>u>>v>>wt; g.addEdge(u,v,wt); } // Maximize it distance travelled so for that // every edge has to be travelled at max using greedy backed by pigeon hole // so we do dfs and count cost int cost=g.dfs(); cout<<"Case #"<