(Analysis by Nick Wu)

In this problem, we have a series of observations about cows crossing a road and we want to count the number of times that we have confirmed that a cow has crossed the road.

Imagine that we only have one cow, and we have the list of observations for that cow. The only time when we can confirm that a cow has crossed the road is when some observation has the cow on one side of the road, and the observation immediately following that one has the cow on the other side of the road.

In the actual problem, we have 10 different cows. Since different cows are independent of each other, we can track their last known locations independently and then accumulate the total number of times we have confirmed that a cow has had a crossing.

import java.io.*;
import java.util.*;
public class crossroad {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new FileReader("crossroad.in"));
		PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("crossroad.out")));
		// read in the number of observations
		int n = Integer.parseInt(br.readLine());
		// create an array to track the last known appearance of each cow
		// lastSeen[cow] == 0 means we haven't seen the cow yet
		// lastSeen[cow] == 1 means we have last seen the cow on the left side of the road
		// lastSeen[cow] == 2 means we have last seen the cow on the right side of the road
		int[] lastSeen = new int[11];
		// track the number of crossings that are confirmed
		int crossings = 0;
		for(int i = 0; i < n; i++) {
			// read in the data for one observation
			StringTokenizer st = new StringTokenizer(br.readLine());
			int index = Integer.parseInt(st.nextToken());
			int current = Integer.parseInt(st.nextToken())+1;
			// if we have seen a cow already and the last side we saw it on is different
			// from the current side, then it is a confirmed crossing
			if(lastSeen[index] > 0 && lastSeen[index] != current) {
			// update the last side we have seen the given cow on
			lastSeen[index] = current;
		// print the answer