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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <map>
#include <stack>
#include <queue>
#include <set>

using namespace std;
#define ARRAY_LENGTH(array) (sizeof(array) / sizeof(array[0]))

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N,M;
	cin >> N >> M;
	int G[N][N],tmp[N][N];
	for(int i = 0;i < N;i++){
		for(int j = 0;j < N;j++){
			G[i][j] = 0;
			tmp[i][j] = 0;
		}
	}
	for(int i = 0;i < M;i++){
		int a,b;
		cin >> a >> b;
		G[a-1][b-1] = 1;
		G[b-1][a-1] = 1;
		tmp[a-1][b-1] = 1;
		tmp[b-1][a-1] = 1;
	}

	int count = 0;
	int reached[N];
	for(int i = 0;i < N;i++) reached[i] = 0;
	for(int i = 0;i < N-1;i++){
		for(int j = i+1;j < N;j++){
			if(G[i][j] == 1){
				tmp[i][j] = 0;
				tmp[j][i] = 0;
				stack<int> s;
				s.push(i);
				int x = i;
				int y = 0;
				reached[x] = 1;
				while(!s.empty()){
					if(tmp[x][y] == 1){
						if(reached[y] == 0){
							s.push(y);
							reached[y] = 1;
							tmp[x][y] = 0;
							tmp[y][x] = 0;
						}
					}
					if(y >= N){
						x = s.top();
						s.pop();
						y = 0;
					}else{
						y++;
					}
				}
				for(int k = 0;k < N;k++){
					if(reached[k] == 0){
						count++;
						break;
					}
				}
				for(int k = 0;k < N;k++){
					for(int l = 0;l < N;l++){
						tmp[k][l] = G[k][l];
					}
					reached[k] = 0;
				}
			}
		}

	}
	cout << count << endl;

}