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 {
    static boolean canPass[][];
    static boolean visited[];
    static int n;
    
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        int m = sc.nextInt();
        int[] p = new int[m];
        int[] q = new int[m];
        canPass = new boolean[n][n];
        visited = new boolean[n];
        
        for (int i = 0; i < m; i++) {
            p[i] = sc.nextInt() - 1;
            q[i] = sc.nextInt() - 1;
            canPass[p[i]][q[i]] = true;
            canPass[q[i]][p[i]] = true;
        }
        
        int count = 0;
        for (int i = 0; i < m; i++) {
            Arrays.fill(visited, false);
            
            // i番目の辺を通行止めにする
            canPass[p[i]][q[i]] = false;
            canPass[q[i]][p[i]] = false;

            dfs(0);
            
            // System.out.println(Arrays.toString(visited));
            // if (Arrays.asList(visited).contains(false)) {
            //     count++;
            // }
            
            boolean connected = true;
            for (int j = 0; j < n; j++) {
                if (!visited[j]) {
                    connected = false;
                    break;
                }
            }
            if (!connected) {
                count++;
            }
            
            canPass[p[i]][q[i]] = true;
            canPass[q[i]][p[i]] = true;

        }
        System.out.println(count);

    }
    public static void dfs(int p) {
        if (visited[p]) return;
        
        visited[p] = true;
        for (int q = 0; q < n; q++) {
            if (canPass[p][q]) {
                dfs(q);
            }
        }
    }
}