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>
typedef long long ll;
using namespace std;

int par[100001];
int high[100001];
//n要素で初期化
void init(int n){
  for(int i=0; i<n; i++){
    par[i]=i;
    high[i]=0;
  }
}
//木の根を求める
int find(int x){
  if(par[x]==x){
    return x;
  }
  else{
    return par[x]=find(par[x]);
  }
}

//xとyの属する集合を併合
void unite(int x,int y){
  x=find(x);
  y=find(y);
  if(x==y)return;
  if(high[x]<high[y]){
    par[x]=y;
  }else{
    par[y]=x;
    if(high[x]==high[y])high[x]++;
  }
}
//xとyが同じ集合に属するか否か
bool same(int x,int y){
  return find(x)==find(y);
}   
int main(){
	int n,m;
  	cin>>n>>m;
  	int a[101],b[101];
  	for(int i=0; i<m; i++){
      cin>>a[i]>>b[i];
      a[i]--; b[i]--;
    }	
  	int ans=0;
  	for(int i=0; i<m; i++){
        init(n);
      	set<int>st;
      	for(int j=0; j<m; j++){
          if(j!=i){unite(a[j],b[j]);}
        }
      	for(int j=0; j<n; j++){
          st.insert(find(j));
        }
      	if(st.size()>1)ans++;
    }
  	cout<<ans<<endl;
}