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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using static System.Math;
using static System.Console;
using System.Text;

class Program
{
    static void Main(string[] args)
    {
        solve();
    }
    static Scanner cin;
    static int N, M;
    static Dictionary<int, List<int>> G = new Dictionary<int, List<int>>();
    static int[] A, B;
    static bool[] isVisited;
    static void input()
    {
        cin = new Scanner();
        N = cin.nextInt();
        M = cin.nextInt();
        A = new int[M];
        B = new int[M];
        for(int i = 0; i < M; i++)
        {
            int a = cin.nextInt() - 1;
            int b = cin.nextInt() - 1;
            A[i] = a;
            B[i] = b;
            if (!G.ContainsKey(a))
                G.Add(a, new List<int>());

            G[a].Add(b);

            if (!G.ContainsKey(b))
                G.Add(b, new List<int>());

            G[b].Add(a);
        }
        isVisited = new bool[N];
    }
    static void solve()
    {
        input();
        int ans = 0;
        for(int i = 0; i < M; i++)
        {
            isVisited = new bool[N];
            DFS(0, i);
            if (isVisited.Contains(false))
                ans++;
        }
        WriteLine(ans);
    }
    class Scanner
    {
        string[] s;
        int i;

        char[] cs = new char[] { ' ' };

        public Scanner()
        {
            s = new string[0];
            i = 0;
        }
        
        public string next()
        {
            if (i < s.Length) return s[i++];
            string st = Console.ReadLine();
            while (st == "") st = Console.ReadLine();
            s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
            if (s.Length == 0) return next();
            i = 0;
            return s[i++];
        }

        public int nextInt()
        {
            return int.Parse(next());
        }

        public int[] ArrayInt(int N, int add = 0)
        {
            int[] Array = new int[N];
            for(int i = 0; i < N; i++)
            {
                Array[i] = nextInt() + add;
            }
            return Array;
        }

        public long nextLong()
        {
            return long.Parse(next());
        }

        public long[] ArrayLong(int N, long add = 0)
        {
            long[] Array = new long[N];
            for(int i = 0; i < N; i++)
            {
                Array[i] = nextLong() + add;
            }
            return Array;
        }

        public double nextDouble()
        {
            return double.Parse(next());
        }

        public double[] ArrayDounble(int N, int add = 0)
        {
            double[] Array = new double[N];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextDouble() + add;
            }
            return Array;
        }
    }
    static void DFS(int v, int disUseEdge)
    {
        isVisited[v] = true;
        foreach(var u in G[v])
        {
            if (!isVisited[u] && ((u != A[disUseEdge] || v != B[disUseEdge]) && (v != A[disUseEdge] || u != B[disUseEdge]) ))
                DFS(u, disUseEdge);
        }
    }
}