Problem:
Score : 300 points

Problem Statement

You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges.

The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i.

An edge whose removal disconnects the graph is called a bridge.

Find the number of the edges that are bridges among the M edges.

Notes

A self-loop is an edge i such that a_i=b_i (1 \leq i \leq M).

Double edges are a pair of edges i,j such that a_i=a_j and b_i=b_j (1 \leq i<j \leq M).

An undirected graph is said to be connected when there exists a path between every pair of vertices.

Constraints

2 \leq N \leq 50

N-1 \leq M \leq min(N(N−1)⁄2,50)

1 \leq a_i<b_i \leq N

The given graph does not contain self-loops and double edges.

The given graph is connected.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

Print the number of the edges that are bridges among the M edges.

Sample Input 1

7 7
1 3
2 7
3 4
4 5
4 6
5 6
6 7

Sample Output 1

4

The figure below shows the given graph:

The edges shown in red are bridges. There are four of them.

Sample Input 2

3 3
1 2
1 3
2 3

Sample Output 2

0

It is possible that there is no bridge.

Sample Input 3

6 5
1 2
2 3
3 4
4 5
5 6

Sample Output 3

5

It is possible that every edge is a bridge.

Solution in Scala (2.13.1) (judge verdict: Accepted)

import scala.collection.mutable.Queue

object Main extends App {
  val sc = new java.util.Scanner(System.in)
  val n, m = sc.nextInt()
  val e = List.fill(m)((sc.nextInt()-1, sc.nextInt()-1))

  def isConnected(s: Int, t: Int, n: Int, edges: Seq[(Int, Int)]): Boolean = {
    val adj = Array.fill(n)(List[Int]())
    for ((u,v) <- edges if (u, v) != (s, t)) {
      adj(u) ::= v
      adj(v) ::= u
    }

    val visited = Array.fill(n)(false)
    visited(s) = true
    val queue = Queue(s)
    while (!visited(t) && queue.nonEmpty) {
      val u = queue.dequeue()
      for (v <- adj(u) if !visited(v)) {
        visited(v) = true
        queue.enqueue(v)
      }
    }
    visited(t)
  }

  val ans = e.count{case (a, b) => !isConnected(a, b, n, e)}
  println(ans)
}