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 (2.11.7) (judge verdict: Accepted)

import scala.util.control.Breaks

object Main{
  def main(args:Array[String]) = {
    val sc = new java.util.Scanner(System.in)
    val N = sc.nextInt
    val M = sc.nextInt

    val as = Array.ofDim[Int](M)
    val bs = Array.ofDim[Int](M)

    for(i <- 0 to M-1){
      val a = sc.nextInt-1
      val b = sc.nextInt-1

      as(i) = a
      bs(i) = b
    }
    //println(as.deep)
    //println(bs.deep)

    var ans = 0
    for(ii <- 0 to M-1){
      val a = as(ii)
      val b = bs(ii)

      val d:Array[Array[Int]] = Array.fill(N,N)(1000)
      //println("...." + ii)
      for(j <- 0 to N-1){
        //println(j)
        d(j)(j) = 0
        //println(j)
      }
      for(j <- 0 to M-1) d(as(j))(bs(j)) = 1
      for(j <- 0 to M-1) d(bs(j))(as(j)) = 1
      d(a)(b) = 1000
      d(b)(a) = 1000

      //println("-------------------------------")
      //d.foreach(inp => println(inp.deep))


      for(k <- 0 to N-1){
        for(i <- 0 to N-1){
          for(j <- 0 to N-1){
            if(d(i)(j) > d(i)(k) + d(k)(j)) d(i)(j) = d(i)(k) + d(k)(j)
          }
        }
      }

      var f = false
      for(i <- 0 to N-1) for(j <- i+1 to N-1) if(d(i)(j) >= 1000) f=true
      if(f) ans = ans + 1
      //println(f)
      //d.foreach(inp => println(inp.deep))
    }
    println(ans)
  }
}