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: Runtime Error)

import java.util.Scanner;
import java.util.Arrays;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        boolean[][] g = new boolean[n][n];
        int[] a = new int[m];
        int[] b = new int[m];
        for(int i = 0;i < m;i++){
            a[i] = sc.nextInt()-1;
            b[i] = sc.nextInt()-1;
            g[a[i]][b[i]] = true;
            g[b[i]][a[i]] = true;
        }

        int answer = 0;
        for(int i = 0;i < m;i++){
            g[a[i]][b[i]] = false;
            g[b[i]][a[i]] = false;
            boolean[] v = new boolean[n];
            dfs(i,n,g,v);
            boolean bridge = false;
            //System.out.println(Arrays.toString(v));
            for(int j = 0;j < n;j++){
                if(!v[j]){
                    bridge = true;
                    break;
                }
            }
            if(bridge){
                answer++;
            }
            g[a[i]][b[i]] = true;
            g[b[i]][a[i]] = true;
        }
        System.out.println(answer);
    }
    public static void dfs(int i,int n,boolean[][] g,boolean[] v){
        v[i] = true;
        for(int j = 0;j < n;j++){
            if(!g[i][j]){
                continue;
            }
            if(v[j]){
                continue;
            }
            dfs(j,n,g,v);
        }
        return;
    }
}