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.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        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;
        }

        int count = 0;
        for (int i = 0 ; i < m ; i++) {
            UnionFindTree uf = new UnionFindTree(n);
            for (int j = 0 ; j < m ; j++) {
                if (i != j) {
                    uf.union(a[j], b[j]);
                }
            }
            if (!uf.same(a[i], b[i])) count++;
        }

        System.out.println(count);

    }

}

class UnionFindTree {
    int[] parent;
    int[] rank;

    public UnionFindTree(int size) {
        this.parent = new int[size];
        this.rank = new int[size];

        for (int i = 0 ; i < size ; i++) {
            makeSet(i);
        }
    }
    /*
     * UnionFindを初期化
     * はじめはすべての頂点が根
     * はじめはすべての高さは1
     */
    public void makeSet(int i) {
        parent[i] = i;
        rank[i] = 0;
    }

    /*
     * 引数の属する木のルートのIDを返す
     */
    public int find(int i) {
        if (i != parent[i]) {
            parent[i] = find(parent[i]);
        }
        return parent[i];
    }

    /*
     * 引数を同じ集合に追加する。
     * その際に木構造もいい感じにいじくる。
     */
    public void union(int x, int y) {
        int xRoot = find(x);
        int yRoot = find(y);

        if (rank[xRoot] > rank[yRoot]) {
            parent[yRoot] = xRoot;
        } else if (rank[xRoot] < rank[yRoot]) {
            parent[xRoot] = yRoot;
        } else if (xRoot != yRoot) {
            parent[yRoot] = xRoot;
            rank[xRoot]++;
        }
    }

    /*
     * 同じ木に属しているか
     */
    public boolean same(int x, int y) {
        return find(x) == find(y);
    }
}