(Analysis by Nick Wu)

In this problem, we have $N$ cows scattered on the plane and want to compute the minimum distance $D$ such that all cows are able to communicate with each other (perhaps through some intermediate cows), given that two cows can only communicate directly if the distance between them is less than or equal to $D$.

We can model this problem as a graph problem where each cow is a vertex connected to every other cow with an edge with the weight being the distance between the two cows. We therefore want to compute the minimum edge weight such that the graph is connected.

We can solve this problem using a union-find data structure. A simple union-find data structure supports two operations:

$Find(x)$ - return a unique identifier for vertex $x$. $Find(x) = Find(y)$ if and only if $x$ and $y$ are in the same connected component.

$Merge(x, y)$ - connect $x$ and $y$ with an edge.

Let's assume that we have a sufficiently efficient implementation of a union-find data structure. Given the original $N$ vertices are all pairwise disconnected, we can loop over the edges in nondecreasing order by weight and add an edge between two vertices if they are not already in the same component. After we add $N-1$ edges, the graph is necessarily connected, and the last edge gives us our answer.

It remains to implement the union-find data structure efficiently. We will describe an implementation that supports both operations in essentially $O(1)$.

We start by describing an approach that supports both operations in linear time, which will not be fast enough for our purposes.

We start by annotating each node with a "parent" field - each node starts out with itself being the parent. When we call $Find(x)$, either $x$ is its own parent, in which case we return $x$, or we return $Find(parent(x))$. When we call $Merge(x, y)$, we set the parent of $Find(x)$ to $Find(y)$.

Intuitively, what we are doing here is building a rooted tree and when we want to merge two trees, we have the root of one tree point to the root of the other tree.

The reason that this runs in linear time is that the depth of the tree may be linear. There are a couple ways to improve the performance here - if we always are careful to set the parent of the root of the smaller tree to the root of the larger tree, then the depth of the tree is bounded to $O(\log N)$. Another approach we can take is path compression - when we compute $Find(parent(x))$, we can memoize that value.

My Java solution uses the latter approach.

import java.io.*;
import java.util.*;
public class moocast {
		BufferedReader br = new BufferedReader(new FileReader("moocast.in"));
		PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("moocast.out")));
		int n = Integer.parseInt(br.readLine());
		int[] x = new int[n];
		int[] y = new int[n];
			StringTokenizer st = new StringTokenizer(br.readLine());
			x[i] = Integer.parseInt(st.nextToken());
			y[i] = Integer.parseInt(st.nextToken());
		parent = new int[n];
		ArrayList<Edge> edges = new ArrayList<Edge>();
			parent[i] = i;
				int distance = (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]);
				edges.add(new Edge(i, j, distance));
		int lastWeight = 0;
		int numComponents = n;
		for(Edge curr: edges) {
			if(find(curr.i) != find(curr.j)) {
				merge(curr.i, curr.j);
				lastWeight = curr.w;
				if(--numComponents == 1) {
	static int[] parent;
	public static int find(int curr) {
		return parent[curr] == curr ? curr : (parent[curr] = find(parent[curr]));
	public static void merge(int x, int y) {
		parent[find(x)] = find(y);
	static class Edge implements Comparable<Edge> {
		public int i, j, w;
		public Edge(int a, int b, int c){
		public int compareTo(Edge e) {
			return w - e.w;