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 Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)

/* Filename: ABC075C.java
 * Author: Mushiyo
 */
import java.util.Scanner;
import java.util.Arrays;

public class Main {

	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		
		while(input.hasNext()){
			int N = input.nextInt();
			int M = input.nextInt();
			int[][] graph = new int[N][N];
			
			for(int i = 0; i < M; ++i){
				int a = input.nextInt() - 1;
				int b = input.nextInt() - 1;
				graph[a][b] = 1;
				graph[b][a] = 1;
			}
			
			int countBridge = 0;
			for(int i = 0; i < graph.length; ++i){
				for(int j = i + 1; j < graph[i].length; ++j){
					if(graph[i][j] == 1){
						graph[i][j] = 0;
						graph[j][i] = 0;
						
						boolean[] isVisited = new boolean[N];
						DFS(i, graph, isVisited);
						
						for(int v = 0; v < isVisited.length; ++v){
							if(isVisited[v] == false){
								++countBridge;
								break;
							}
						}
						
						graph[i][j] = 1;
						graph[j][i] = 1;
					}
				}
			}
			
			System.out.println(countBridge);
		}
	}

	private static void DFS(int current, int[][] graph, boolean[] isVisited){
		isVisited[current] = true;
		
		for(int i = 0; i < graph[current].length; ++i){
			if(graph[current][i] == 1){
				if(isVisited[i] == false){
					DFS(i, graph, isVisited);
				}
			}
		}
	}
}