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 static java.lang.System.*;
public class Main {
    public static void main(String[]$){
        Scanner sc = new Scanner(in);
        int n=sc.nextInt();
        int[] node=new int[n+1];
        //頂点1,2,3,...,n
        //探索済み→node[i]=-1,未探索→node[i]=0
        //0はジャマなのではじめから探索済みにしておく
        node[0]=-1;
        //各頂点iに対し、そことつながっている頂点j,k,...をbridge[i]=j,k....で表す
        ArrayList<Integer>[] bridge=new ArrayList[n+1];
        for (int i = 1; i <=n ; i++) {
            bridge[i]=new ArrayList<Integer>();
        }
        int m=sc.nextInt();
        int[] a=new int[m+1];
        int[] b=new int[m+1];
        for (int i = 1; i <=m ; i++) {
            a[i] = sc.nextInt();
            b[i] = sc.nextInt();
        }
        for(int i=1;i<=m;i++){
            bridge[a[i]].add(b[i]);
            bridge[b[i]].add(a[i]);
        }
        int ans=0;
        for (int i = 1; i <=m; i++) {
            //i番目の辺を削除
            bridge[a[i]].remove((Integer)b[i]);
            bridge[b[i]].remove((Integer)a[i]);
            //条件を満たすか確認、満たすなら++ans
            if(bfs(node,bridge))++ans;
            //i番目の辺を復元
            bridge[a[i]].add(b[i]);
            bridge[b[i]].add(a[i]);
        }
        out.println(ans);
    }
    //グラフをBFS
    static boolean bfs(int[] node,ArrayList[] x){
        Queue<Integer> q=new ArrayDeque<>();
        //未探索→探索済みの回数をカウント(新たに頂点を探索したら++sum)
        //つまり、sumは「探索することができた頂点の個数」である
        int n=node.length-1;
        //初期化
        for(int i=1;i<=n;i++){
            node[i]=0;
        }
        //もし全体が連結ならば最終的にsum==nとなるはず
        int sum=0;
        //頂点1から探索開始
        q.add(1);
        while(q.size()>0) {
            int tempnode = q.poll();
            //未探索かどうか判定
            if (node[tempnode] == 0) {
                //探索済みにする
                node[tempnode] = -1;
                ++sum;
                //隣接する頂点をQueueに入れる（BFSですね）
                for (Object k : x[tempnode]) {
                    q.add((Integer) k);
                }
            }
        }
        //グラフが非連結になっているかどうか判定（非連結になっていれば題意を満たします）
        return sum!=n;
    }
}