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>
using namespace std;
#define ll long long
#define INF 100000000000000
int main(){
	int n,m,ans=0;cin>>n>>m;
	
	vector<vector<int>> G(n);//隣接行列
	
	vector<pair<int,int>> P;
	for(int i=0;i<m;i++){
		int a,b;cin>>a>>b;
		a--;b--;
		G[a].push_back(b);
		G[b].push_back(a);
		P.push_back(make_pair(a,b));//辺を記憶
	}

	for(int i=0;i<P.size();i++){//各辺について
		vector<int> visit(n,0);//各辺を訪れたかどうかの判定(visit:1  !visit:0  切り離し:-1)
		
		//dfs
		stack<int> st;
		st.push(P[i].first);
		visit[P[i].first]=visit[P[i].second]=-1;//辺を一つ切り離す(-1が切り離されたところ)
		
		while(st.size()!=0){
	
			int now=st.top();
			st.pop();
			visit[now]=1;//visit

			visit[now]=1;//visitしたならば1
			for(auto next:G[now]){
				if(now==P[i].first && next==P[i].second) continue;//切り離し
				if(visit[next]==1) continue;//already visit

				st.push(next);
			}
		}

		bool bridge=false;
		for(int j=0;j<n;j++){
			if(visit[j]==0 || visit[j]==-1){bridge=true;}//切り離して訪れていないところがあれば橋
		}
		if(bridge){ans++;}
	}


	cout<<ans<<endl;
	return 0;
}