Problem:
Score : 100 points

Problem Statement

There is a directed graph G with N vertices and M edges.
The vertices are numbered 1, 2, \ldots, N, and for each i (1 \leq i \leq M), the i-th directed edge goes from Vertex x_i to y_i.
G does not contain directed cycles.

Find the length of the longest directed path in G.
Here, the length of a directed path is the number of edges in it.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq x_i, y_i \leq N

All pairs (x_i, y_i) are distinct.

G does not contain directed cycles.

Input

Input is given from Standard Input in the following format:

N M
x_1 y_1
x_2 y_2
:
x_M y_M

Output

Print the length of the longest directed path in G.

Sample Input 1

4 5
1 2
1 3
3 2
2 4
3 4

Sample Output 1

3

The red directed path in the following figure is the longest:

Sample Input 2

6 3
2 3
4 5
5 6

Sample Output 2

2

The red directed path in the following figure is the longest:

Sample Input 3

5 8
5 3
2 3
2 4
5 2
5 1
1 4
4 3
1 3

Sample Output 3

3

The red directed path in the following figure is one of the longest:

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.ArrayList;
import java.util.HashSet;
import java.util.Scanner;

public class Main {

	static ArrayList<Integer>[] chld;

	@SuppressWarnings("unchecked")
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt(), m = in.nextInt();

		chld = new ArrayList[n + 1];
		for ( int i = 0; i <= n; i++ ) {
			chld[i] = new ArrayList<Integer>();
		}

		HashSet<Integer> from = new HashSet<Integer>();
		HashSet<Integer> to = new HashSet<Integer>();

		for ( int i = 0; i < m; i++ ) {
			int x = in.nextInt(), y = in.nextInt();
			chld[x].add(y);

			from.add(x);
			to.add(y);
		}

		// 出発点候補
		from.removeAll(to);

//		for ( ArrayList<Integer> list : chld ) {
//			System.out.println(list);
//		}

		//				for ( Integer i : from ) {
		//					System.out.print(i);
		//				}

		int lenmax = 0;
		for ( Integer i : from ) {
			lenmax = Math.max(lenmax, search(i, 0) );
		}
		System.out.println(lenmax);
		in.close();

	}

	private static int search(Integer prnt, int prntlen) {
//		System.out.println("prnt:" + prnt + "len:" + prntlen);

		int nowlen = prntlen + 1;
		if ( chld[prnt].size() == 0 ) {
//			System.out.println();
			return prntlen;
		}
		for ( Integer i : chld[prnt] ) {
			prntlen = Math.max(prntlen, search(i, nowlen));
		}
		return prntlen;
	}
}