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: Accepted)

#include<bits/stdc++.h>
#define rep(i,n) for(int i = 0;i < (n);i++)
typedef long long ll;
using namespace std;
const int limit = 55;
bool graph[limit][limit];
bool visited[limit];

void dfs(int now ,int n){
   visited[now] = true;
   rep(i,n){
      if(visited[i] == true)continue;
      if(graph[now][i] == true){
         dfs(i,n);
      }
   }
}

int main(){
   int n ,m;
   cin >> n >> m;
   vector<int> a(m),b(m);
   rep(i,m){
      cin >> a[i] >> b[i];
      graph[a[i]-1][b[i]-1] = graph[b[i]-1][a[i]-1] = true;
   }
   int ans = 0;

   rep(i,m){
      graph[a[i]-1][b[i]-1] = graph[b[i]-1][a[i]-1] = false;
      rep(j,n)visited[j] = false;

      dfs(0,n);
      bool bridge = false;
      rep(j,n)if(visited[j] == false)bridge = true;
      if(bridge)ans++;
      graph[a[i]-1][b[i]-1] = graph[b[i]-1][a[i]-1] = true;
   }
   cout << ans << endl;
   return 0;
}