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.*;
import java.io.*;

class Edge{
	int from, to;
	public Edge(int from, int to) {
		super();
		this.from = from;
		this.to = to;
	}
}

public class Main {
	static int mod = (int) 1e9 + 7;
	static int DX[] = { -1, 0, 1, 0 }, DY[] = { 0, -1, 0, 1 };
	static int n, m, to[][];
	public static void main(String[] args) {
		FastScanner fs = new FastScanner(System.in);
		n = fs.nextInt();
		m = fs.nextInt();
		to = new int[n][n];
		List<Edge> list = new ArrayList<>();
		for(int i=0;i<m;i++) {
			int a = fs.nextInt()-1, b = fs.nextInt()-1;
			to[a][b]=1;
			to[b][a]=1;
			list.add(new Edge(a,b));
		}
		int ans = 0;
		for(int i=0;i<m;i++) {
			int a = list.get(i).from;
			int b = list.get(i).to;
			to[a][b]=0;
			to[b][a]=0;
			if(!dfs(0,new int[n]))ans++;
			to[a][b]=1;
			to[b][a]=1;
		}
		System.out.println(ans);
	}
	static boolean dfs(int v,int visited[]) {
		visited[v]=1;
		boolean last = true;
		boolean res = false;
		for(int i=0;i<n;i++) {
			if(to[v][i]==1&&visited[i]==0) {
				last = false;
				res |= dfs(i,visited);
			}
		}
		if(last) {
			for(int i=0;i<n;i++) {
				if (visited[i]==0)return false;
			}
			return true;
		}
		return res;
	}
}


//高速なScanner
class FastScanner {
	private BufferedReader reader = null;
	private StringTokenizer tokenizer = null;

	public FastScanner(InputStream in) {
		reader = new BufferedReader(new InputStreamReader(in));
		tokenizer = null;
	}

	public String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public String nextLine() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken("\n");
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}