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.ArrayDeque;
import java.util.Arrays;
import java.util.Deque;
import java.util.Scanner;

public class Main {
    static long dp[] = new long[1000000];

    public static void main(String[] args) {
        Scanner std = new Scanner(System.in);
        int n = std.nextInt();
        int m = std.nextInt();
        int[] from = new int[m];
        int[] to = new int[m];

        for (int i = 0; i < m; i++) {
            from[i] = std.nextInt() - 1;
            to[i] = std.nextInt() - 1;
        }

//        PriorityQueue<Integer> deque = new PriorityQueue<>(n);
        Deque<Integer> deque = new ArrayDeque<>();
        long ans = 0;
        boolean[] visited = new boolean[n];
        for (int k = 0; k < n; k++) {
            if (visited[k]) {
                continue;
            }
            deque.add(k);
            while (!deque.isEmpty()) {
                int v = deque.poll();
                visited[v] = true;
                for (int i = 0; i < m; i++) {
                    int f = from[i];
                    if (f == v) {
                        int t = to[i];
                        long cost = dp[v] + 1;

                        dp[t] = Math.max(dp[t], cost);
                        if (!deque.contains(t)) {
                            deque.add(t);
                        }
                    }
                }
            }
            ans = Math.max(Arrays.stream(dp).max().getAsLong(), ans);
        }

        System.out.println(ans);
    }
}