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: Runtime Error)

#include <bits/stdc++.h>
using namespace std; 
void dfs(int v,bool **G,bool *visited,int N){
  visited[v]=true;
  for(int v2=0;v2<N;v2++){
    if(G[v][v2]==false)continue;
    if(visited[v2]==true)continue;
    dfs(v2,G,visited,N);
  }
}
int main(){
  int N,M,ans=0;
  cin>>N>>M;
  bool **G=new bool*[N];
  for(int i=0;i<N;i++){
    G[i]= new bool[N];
  }
  bool visited[N];
  bool bridge;
  int a[N],b[N];
  for(int i=0;i<N;i++){
    for(int j=0;j<N;j++){
      G[i][j]=false;
    }
  }
  for(int i=0;i<M;i++){
    cin>>a[i]>>b[i];
    a[i]--;
    b[i]--;
    G[a[i]][b[i]]=G[b[i]][a[i]]=true;
  }
  for(int i=0;i<M;i++){
    G[a[i]][b[i]]=G[b[i]][a[i]]=false;
    for(int j=0;j<N;++j) visited[j]=false;
    dfs(0,G,visited,N);
    bool bridge=false;
    for(int j=0;j<N;++j) if(visited[j]==false)bridge=true;
    if(bridge)ans++;
    G[a[i]][b[i]]=G[b[i]][a[i]]=true;
  }
  cout<<ans<<endl;
  for(int i=0;i<N;i++){
    delete[] G[i];
  }
  delete[] G;
  return 0;
}