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: Runtime Error)

using System;
using System.Collections.Generic;
using System.Linq;

namespace lineAralgebra
{
	class Program
	{
		static void Main(string[] args)
		{
			var input = Console.ReadLine().Split().Select(int.Parse).ToArray();
			var N = input[0];
			var M = input[1];

			var A = new int[N];
			var B = new int[N];

			for (int i = 0; i < M; i++)
			{
				input = Console.ReadLine().Split().Select(int.Parse).ToArray();
				A[i] = input[0] - 1;
				B[i] = input[1] - 1;
			}

			var ans = 0;

			for (var i = 0; i < M; i++)
			{
				var uf = new uf(N);
				for (var j = 0; j < M; j++)
				{
					if (j != i)
					{
						uf.Union(A[j], B[j]);

						if (uf.Count() == 1) break; 	
					}
				}
				if (uf.Count() != 1) ans++;
			}

			Console.WriteLine(ans);
		}

	}

	public class uf
	{
		private int[] ids;
		private int[] size;
		private int count;

		public uf(int n)
		{
			this.ids = new int[n];
			this.size = new int[n];

			count = n;
			for (int i = 0; i < ids.Length; i++)
			{
				ids[i] = i;
				size[i] = 1;
			}
		}

		public bool Connected(int p, int q)
		{
			int pRoot = Find(p);
			int qRoot = Find(q);

			return pRoot == qRoot;
		}

		public int Count()
		{
			return count;
		}

		// public int Find(int p)
		// {

		// 	while (ids[p] != p)
		// 	{
		// 		ids[p] = ids[ids[p]];
		// 		p = ids[p];
		// 	}

		// 	return p;
		// }

		public int Find(int p) {
			if (ids[p] == p) return p;
			else {
				return ids[p] = Find(ids[p]);
			}
		}

		public void Union(int q, int p)
		{
			int rootQ = Find(q);
			int rootP = Find(p);

			if (rootP == rootQ) return;

			if (size[rootP] < size[rootQ])
			{
				ids[rootP] = rootQ;
				size[rootQ] += size[rootP];
			}
			else
			{
				ids[rootQ] = rootP;
				size[rootP] += size[rootQ];
			}
			count--;
		}

		// public int Try(int q, int p) {
		// 	int rootQ = Find(q);
		// 	int rootP = Find(p);

		// 	if (rootP == rootQ) return -1;

		// 	var tempSize = new int[size.Length];
		// 	var tempIds = new int[ids.Length];

		// 	Array.Copy(size, tempSize, size.Length);
		// 	Array.Copy(ids, tempIds, ids.Length);

		// 	retur	
		// }

	}
}