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.Text;
using System.Collections.Generic; 
using System.Linq; 
using E = System.Linq.Enumerable; 
class Program {
    static List<long>[] SNS;
    static long N; static void
    Main(string[] args) {
        //入力を受け取る
        var NM = Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToArray();
            N = NM[0];
        var M = NM[1];
        //橋の関係を記録するリスト
        SNS = new List<long>[N + 1];
        for (long i = 0; i <= N; i++) {
            SNS[i] = new List<long>();
        }
        var AB = new long[M][];
        for(long i = 0; i < M; i++) {
            AB[i] =
            Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToArray();
            var A = AB[i][0];
            var B = AB[i][1];
            SNS[A].Add(B);
            SNS[B].Add(A);
            }
            //幅優先探索でSNS上の友達の友達（深さが2まで求める)
            var ans=0;
            for (long i = 1; i <= M; i++) {
              SNS[AB[i-1][0]].Remove(AB[i-1][1]);
SNS[AB[i-1][1]].Remove(AB[i-1][0]);
              //iの友人関係を調べ
              if(WFS(1)<N){
                ans++;
              }
              
SNS[AB[i-1][0]].Add(AB[i-1][1]);
SNS[AB[i-1][1]].Add(AB[i-1][0]);              
               }
               Console.WriteLine(ans);
     }
               
     //numは起点にする値         
     static long WFS(long num) {
         var depth = new long[N+1];
         var queue = new MyQueue();
         var count = 1L;
         queue.Enque(num);
         depth[num] = 1;
         while(!(queue.isEmpty())){ 
             var top = queue.Deque(); 
             foreach (long item in SNS[top]){
                 //探索済みか
                 if (depth[item] == 0){
                     depth[item] = depth[top] + 1;
                     queue.Enque(item);
                     count++;
                 }
             }
             
         }
         return count;
      } 
                   }
                   class MyQueue { public long[] datas = new long[100000]; public int enpoint = -1; public int depoint = -1; public void Enque(long data) { this.enpoint++; datas[this.enpoint] = data; } public long Deque() { this.depoint++; return datas[this.depoint]; } public bool isEmpty() { return (this.depoint == this.enpoint); } }