Problem:
Score : 500 points

Problem Statement

N players will participate in a tennis tournament. We will call them Player 1, Player 2, \ldots, Player N.

The tournament is round-robin format, and there will be N(N-1)/2 matches in total.
Is it possible to schedule these matches so that all of the following conditions are satisfied? If the answer is yes, also find the minimum number of days required.

Each player plays at most one matches in a day.

Each player i (1 \leq i \leq N) plays one match against Player A_{i, 1}, A_{i, 2}, \ldots, A_{i, N-1} in this order.

Constraints

3 \leq N \leq 1000

1 \leq A_{i, j} \leq N

A_{i, j} \neq i

A_{i, 1}, A_{i, 2}, \ldots, A_{i, N-1} are all different.

Input

Input is given from Standard Input in the following format:

N
A_{1, 1} A_{1, 2} \ldots A_{1, N-1}
A_{2, 1} A_{2, 2} \ldots A_{2, N-1}
:
A_{N, 1} A_{N, 2} \ldots A_{N, N-1}

Output

If it is possible to schedule all the matches so that all of the conditions are satisfied, print the minimum number of days required; if it is impossible, print -1.

Sample Input 1

3
2 3
1 3
1 2

Sample Output 1

3

All the conditions can be satisfied if the matches are scheduled for three days as follows:

Day 1: Player 1 vs Player 2

Day 2: Player 1 vs Player 3

Day 3: Player 2 vs Player 3

This is the minimum number of days required.

Sample Input 2

4
2 3 4
1 3 4
4 1 2
3 1 2

Sample Output 2

4

All the conditions can be satisfied if the matches are scheduled for four days as follows:

Day 1: Player 1 vs Player 2, Player 3 vs Player 4

Day 2: Player 1 vs Player 3

Day 3: Player 1 vs Player 4, Player 2 vs Player 3

Day 4: Player 2 vs Player 4

This is the minimum number of days required.

Sample Input 3

3
2 3
3 1
1 2

Sample Output 3

-1

Any scheduling of the matches violates some condition.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {

	private static List<Integer>[] g;

	private static boolean[] inStack;

	private static boolean[] visited;

	private static long[] memo;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		int[][] a = new int[n][n - 1];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n - 1; j++) {
				a[i][j] = sc.nextInt() - 1;
			}
		}
		Map<Integer, Integer>[] m = new HashMap[n];
		for (int i = 0; i < n; i++) {
			m[i] = new HashMap<>();
		}

		int nodes = 0;
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n - 1; j++) {
				if (a[i][j] < i) {
					a[i][j] = m[a[i][j]].get(i);
				} else {
					m[i].put(a[i][j], nodes);
					a[i][j] = nodes++;
				}
			}
		}

		//for (int i = 0; i < a.length; i++) {
		//	for (int j = 0; j < a[i].length; j++) {
		//		System.out.print(a[i][j]);
		//		System.out.print(" ");
		//	}
		//	System.out.println();
		//}

		g = new ArrayList[n * (n - 1) / 2];
		for (int i = 0; i < g.length; i++) {
			g[i] = new ArrayList<>();
		}

		int[] ds = new int[n * (n - 1) / 2];

		for (int i = 0; i < a.length; i++) {
			for (int j = 1; j < a[i].length; j++) {
				g[a[i][j - 1]].add(a[i][j]);
				ds[a[i][j]]++;
			}
		}

		if (hasCycle()) {
			System.out.println(-1);
			return;
		}

		long answer = 0;
		Arrays.fill(visited, false);
		memo = new long[visited.length];

		for (int i = 0; i < ds.length; i++) {
			if (ds[i] > 0 || visited[i]) continue;
			answer = Math.max(answer, dfs2(i));
		}

		System.out.println(answer);
	}

	private static boolean hasCycle() {
		inStack = new boolean[g.length];
		visited = new boolean[g.length];
		Arrays.fill(inStack, false);
		Arrays.fill(visited, false);
		for (int i = 0; i < g.length; i++) {
			if (visited[i]) continue;
			boolean hasCycle = dfs(i);
			if (hasCycle) return true;
		}

		return false;
	}

	private static boolean dfs(int x) {
		if (inStack[x]) return true;
		if (visited[x]) return false;
		visited[x] = true;
		inStack[x] = true;
		for (int i = 0; i < g[x].size(); i++) {
			boolean a = dfs(g[x].get(i));
			if (a) return true;
		}

		inStack[x] = false;
		return false;

	}

	private static long dfs2(int x) {
		if (visited[x]) return memo[x];
		visited[x] = true;
		memo[x] = 1;
		if (g[x].size() == 0) return memo[x];


		long a = 0;
		for (int i = 0; i < g[x].size(); i++) {
			a = Math.max(a, dfs2(g[x].get(i)));
		}
		memo[x] = a + 1;
		return memo[x];
	}
}