Imagine some made up story here including Afifa (my little sister) and an Intelligent Transport System, which ultimately leads to the main problem.
You are given a connected tree containing N nodes and N - 1 bidirectional edges. That means, you can go from one node to any other node. Every node i has some weight wi. Let’s keep it simple and say that wi = i for the ith node. Let this tree be called A.
Let’s call a node terminal if it has one edge attached to it or no edge at all.
You have to pick a “part” of this tree that’s also a tree. You can pick the entire tree as well. Let this new tree be called B.
fig - 1 denotes an example of tree A. fig - 2 and fig - 3 shows two of the many possible ways to pick tree B. Now, you have to do this in a way so that the cost is maximum.
The cost of choosing tree B = ( minimum weight among the terminals of B ) * ( number of edges in B )
For fig -2, cost is minimum(3, 4, 6) * 3 = 9 and for fig - 3, cost is minimum(1, 6) * 4 = 4. Please note that neither of the costs are maximum.
The first line contains a single integer N (1 ≤ N ≤ 100000) - number of nodes in the tree. Each of the next N - 1 lines contains two integers u and v (1 ≤ u, v ≤ N) - denoting an edge of the tree.
Print the maximum possible cost of new tree B.
7 2 4 1 2 2 5 3 7 4 7 6 7
Sample I/O Explanation:
Here, tree B, marked with bold edges, has two terminals with weight 5 and 6. Number of edges in B equals 4.
So cost is minimum(5, 6) * 4 = 20, which is the maximum possible cost for this case.
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First, take the whole tree A as B. For this, cost would be minimum weight among terminals * n - 1. L... Read more...