(Analysis by Nick Wu)

In this problem, we have several cows who are lining up to enter Farmer John's farm. Each cow enters the line at some time and requires some amount of questioning before being allowed to enter the farm. We want to know when every cow has entered the farm.

Intuitively, if we know the order that each cow arrives in, we can process the cows in that order. We can track when their questioning starts - it's the later of their arrival time and when the questioning of the previous cow ended.

Given the above, it remains to figure out the order in which the cows appear. For this, while we haven't yet processed all the cows, we can find the cow that has arrived the earliest and process that cow next.

import java.io.*;
import java.util.*;
public class cowqueue {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new FileReader("cowqueue.in"));
		PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("cowqueue.out")));
		// read in the number of cows
		int n = Integer.parseInt(br.readLine());
		// allocate arrays to store entry times and duration of questioning times
		int[] enter = new int[n];
		int[] duration = new int[n];
		// allocate an array to store if a cow has already been processed
		boolean[] processed = new boolean[n];
		// read in the entry and questioning times
		for(int i = 0; i < n; i++) {
			StringTokenizer st = new StringTokenizer(br.readLine());
			enter[i] = Integer.parseInt(st.nextToken());
			duration[i] = Integer.parseInt(st.nextToken());
		int nextAvailableTime = 0;
		// loop over the cows
		for(int a = 0; a < n; a++) {
			int nextToEnter = -1;
			for(int i = 0; i < n; i++) {
				// if a cow has not been processed and it has the earlier entry time of all cows seen so far, it is next to enter
				if(!processed[i] && (nextToEnter == -1 || enter[i] < enter[nextToEnter])) {
					nextToEnter = i;
			// process that cow, tracking exactly when questioning ends
			processed[nextToEnter] = true;
			if(enter[nextToEnter] > nextAvailableTime) {
				nextAvailableTime = enter[nextToEnter] + duration[nextToEnter];
			else {
				nextAvailableTime = nextAvailableTime + duration[nextToEnter];