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.*;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int m = scan.nextInt();
        Edge[] edges = new Edge[m];
        boolean[][] adj = new boolean[n + 1][n + 1];
        for (int i = 0; i < m; i++) {
            int a = scan.nextInt();
            int b = scan.nextInt();
            edges[i] = new Edge(a, b);
            adj[a][b] = adj[b][a] = true;
        }
        int num = 0;
        for (int i = 0; i < m; i++) {
            Set<Integer> visited = new HashSet<>();
            Queue<Integer> toVisit = new PriorityQueue<>();
            toVisit.add(1);
            visited.add(1);
            int count = 0;
            adj[edges[i].a][edges[i].b] = adj[edges[i].b][edges[i].a] = false;
            while (!toVisit.isEmpty()) {
                int a = toVisit.remove();
                count++;
                for (int j = 0; j < n + 1; j++) {
                    if (adj[a][j] && !visited.contains(j)) {
                        visited.add(j);
                        toVisit.add(j);
                    }
                }
            }
            adj[edges[i].a][edges[i].b] = adj[edges[i].b][edges[i].a] = true;
            if (count != n) {
                num++;
            }
        }
        System.out.println(num);
    }
}

class Edge {
    int a, b;
    boolean active = true;
    public Edge(int c, int d) {
        a = c; b = d;
    }
}