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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    List<List<int[]>> graph = new ArrayList<>();
    int n = Integer.parseInt(sc.next());
    int m = Integer.parseInt(sc.next());
    int a,b;
    for(int i=0;i<n;i++)graph.add(new LinkedList<>());
    
    for(int i=0;i<m;i++){
      a = Integer.parseInt(sc.next())-1;
      b = Integer.parseInt(sc.next())-1;
      graph.get(a).add(new int[]{b,i});
      graph.get(b).add(new int[]{a,i});
    }
    boolean[] isVisited = new boolean[n];
    Queue<Integer> queue = new ArrayDeque<>();
    int res = 0;
    for(int i=0;i<m;i++){
      Arrays.fill(isVisited,false);
      queue.add(0);
      isVisited[0] = true;
      while(queue.size()> 0){
        
        int to = queue.poll();
        
        for(int[] nxt: graph.get(to)){
          if(nxt[1]==i || isVisited[nxt[0]])continue;
          isVisited[nxt[0]] = true;
          queue.add(nxt[0]);
        }
      }
      for(int j=0;j<n;j++){
        if(!isVisited[j]){
          res++;
          break;
        }
      }
    }
    System.out.println(res);
                       
  }
}