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: Wrong Answer)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using System.Text;
using System.Globalization;
using System.Numerics;

class Myon
{
    //chokudaiさんのコードをコピペ
    static Scanner cin;
    public Myon() { }
    public static int Main()
    {
        //Console.SetOut(new Printer(Console.OpenStandardOutput()));
        cin = new Scanner();
        new Myon().calc();

        return 0;
    }


    public void calc()
    {
        int n = cin.nextInt();

        //System.Diagnostics.Stopwatch sw = new System.Diagnostics.Stopwatch();
        //sw.Start();

        int m = cin.nextInt();
        List<int> a = new List<int>(m);
        List<int> b = new List<int>(m);
        for (int i = 0; i < m; i++)
        {
            a.Add(cin.nextInt());
            b.Add(cin.nextInt());
        }

        int ans = 0;
        List<int> bridge = new List<int>();
        for(int i = 0; i < m; i++)
        {
            tempConnected = new bool[m+1];
            AddConnected(0, a, b, i);
            bool allconnected = true; for(int j = 1;j<m+1;j++) if (!tempConnected[j]) { allconnected = false;break; }
            if (!allconnected)
            {
                //bridge.Add(i + 1);
                ans++;
            }
        }
        showL(ans);

        //sw.Stop();
        //showL($"処理時間: {sw.ElapsedMilliseconds} [msec]");


    }

    
    bool[] tempConnected;
    public void AddConnected(int index, List<int>a,List<int>b, int ignoreingBridge = -1)
    {
        //i番目のノードにつながるノードをconnectedにする
        for(int i = 0; i < a.Count; i++)
        {
            if (i == ignoreingBridge) continue;
            if (a[i] == index && !tempConnected[b[i]]) { tempConnected[b[i]] = true; AddConnected(b[i], a, b,ignoreingBridge); }
            else if (b[i] == index && !tempConnected[a[i]]) { tempConnected[a[i]] = true; AddConnected(a[i], a, b,ignoreingBridge); }
        }
   
    }
    public void backetSort(ref int[] a)
    {
        //max a << a.len に対して有効なソート
        //よってa = int[] とする
        long n = a.Length;
        var count = new int[n];
        var loc = new long[n];//loc[i] は i が終わる 位置 max index where c[index] = loc[i]
        for (long i = 0; i < n; i++) count[a[i]]++;
        loc[0] = count[0] - 1;
        for (long i = 0; i < n-1; i++) loc[i+1] = loc[i] + count[i+1]; 
        long index = 0;

        for(int i = 0; i < n; i++)
        {
            while (index <= loc[i]) a[index++] = i; //incex <= loc[i] == index < loc[i] + 1
        }
    }

    public static void showL<T>(T s) { Console.WriteLine(s); }
}

class Printer : StreamWriter
{
    public override IFormatProvider FormatProvider { get { return CultureInfo.InvariantCulture; } }
    public Printer(Stream stream) : base(stream, new UTF8Encoding(false, true)) { }
    public Printer(Stream stream, Encoding encoding) : base(stream, encoding) { }
}

class Scanner
{
    string[] s;
    int i;

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

    public Scanner()
    {
        s = new string[0];
        i = 0;
    }

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

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

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

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

}

class SMath
{
    
    /// <summary>
    /// 素因数分解して、n番目に見つかった素因数をoutput[n-1]に返します。
    /// 12 なら 2,2,3 です。
    /// </summary>
    /// <param name="n"></param>
    /// <returns></returns>
    public static List<long> Factorization(long n)
    {
        if (n <= 0) return null;
        List<long> output = new List<long>();
        int i = 2;
        long max = (long)Math.Sqrt(n); ;
        while (n > 1)
        {
            if (i > max) { output.Add(n); break; }
            while (n % i == 0)
            {
                output.Add(i);
                n /= i;
            }
            i++;
        }
        return output;
    }

    /// <summary>
    /// 素因数分解して、n番目に見つかった素因数をoutput[n-1]に返します。
    /// 84 = (2^2 * 3 * 7) なら 2, 1, 0, 1 　remainder = 1です。(打ち切られない）
    /// 13 = 0,0,...0 remainder = 2019
    /// ただし、試行を打ち切った時に限り残りをremainderに返す。(remainderは素数または1となる)
    /// </summary>
    /// <param name="n"></param>
    /// <param name="remainder"></param>
    /// <returns></returns>
    public static List<long> Factorization(long n, out long remainder)
    {
        //output[i] が i-1番目の素数の数を示す
        //ただし、試行を打ち切った時に限り残りをremainderに返す。(remainderは素数または1となる)
        remainder = 1;
        if (n <= 0) return null;
        List<long> output = new List<long>();
        long tempn = 2;
        long max = (long)Math.Sqrt(n);
        bool[] isNOTPrime = new bool[max+1];
        while (n > 1)
        {

            if (tempn > max) { remainder = n; break; }
            if (!isNOTPrime[tempn])
            {
                output.Add(0);
                while (n % tempn == 0)
                {
                    output[output.Count - 1]++;
                    n /= tempn;
                }
                for (long j = tempn * 2; j < isNOTPrime.Length; j += tempn) isNOTPrime[j] = true;
            }
            tempn++;
        }
        return output;
    }

    /// <summary>
    /// i番目の素数をoutput[i-1]に返します。
    /// ただしnより大きい素数については判定しません。
    /// </summary>
    /// <param name="n"></param>
    /// <returns></returns>
    public static List<long> GetPrimeNumbers(long n)
    {
        if (n <= 0) return null;
        List<long> output = new List<long>();
        long tempn = 2;
        bool[] isNOTPrime = new bool[n + 1];
        while (n > 1)
        {
            if (tempn > n) { break; }
            if (!isNOTPrime[tempn])
            {
                output.Add(tempn);
                for (long j = tempn * 2; j < isNOTPrime.Length; j += tempn) isNOTPrime[j] = true;
            }
            tempn++;
        }
        return output;
    }


    private static long[][] combMemo;
    private static int maxn_combination = -1;
    public static long Combination(int n, int k)
    {
        if (n < 0 || k > n) return 0;
        if (n == k || k == 0) return 1;
        if (k > n/2) return Combination(n, n - k);
        if (maxn_combination < n)
        {
            maxn_combination = n;
            combMemo = new long[n+1][];
        }
        if (combMemo[n] == null) combMemo[n] = new long[n/2 + 1];
        if (combMemo[n][k] != 0) return combMemo[n][k];
        else
        {
            combMemo[n][k] = Combination(n - 1, k) + Combination(n - 1, k - 1);
            return combMemo[n][k];
        }
        
    }
    public static long Gcd(long a, long b)
    {
        if (a <= 0) return -1;
        //if (a > b) return gcd(b, a);
        if (b == 0) return a;
        return Gcd(b, a % b);
    }
    public static long Lcm(long a,long b)
    {
        return a / Gcd(a, b) * b;
    }
}