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.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int M = sc.nextInt();
    int[][] graph = new int[N+1][N+1];
    int[][] edge = new int[M][2];
    for (int i = 0; i < M; i++) {
      int a = sc.nextInt();
      int b = sc.nextInt();
      graph[a][b] = 1;
      graph[b][a] = 1;
      edge[i][0] = a;
      edge[i][1] = b;
    }
    
    int bridge = 0;
    for (int i = 0; i < M; i++) {

      int aa = edge[i][0];
      int bb = edge[i][1];

      graph[aa][bb] = 0;
      graph[bb][aa] = 0;

      int[] visited = new int[N+1];
      dfs(graph, 1, visited);
      boolean visiteAll = true;
      for (int j = 1; j <= N; j++) {
        if (visited[j] != 1) {
          visiteAll = false;
          break;
        }
      }
      if (!visiteAll) {
        bridge++;
      }
      
      graph[aa][bb] = 1;
      graph[bb][aa] = 1;
    }

    System.out.println(bridge);
  }
  
  private static void dfs(int[][] graph, int v, int[] visited) {
    visited[v] = 1;
    for (int i = 1; i < graph[v].length; i++) {
      if (graph[v][i] == 1 && visited[i] != 1) {
        dfs(graph, i, visited);
      }
    }
  }
}