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 <iostream>
#include <vector>
#include <string>
#include <algorithm>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)

using namespace std;
typedef long long int lli;
const int MAX=1e5;

int N,M;
int a[1250],b[1250];
bool G[50][50];

//頂点vから始めて到達できるものにフラグ立て
bool vis[50];
void dfs(int v){
	vis[v]=true;
	rep(u,N)if(G[v][u]&&!vis[u]) dfs(u);
  	return;
}

int main(){
	cin>>N>>M;
	rep(i,M){
		cin>>a[i]>>b[i];
        a[i]--;
      	b[i]--;
		G[a[i]][b[i]]=G[b[i]][a[i]]=true;
	}
	int bridge=0;
	rep(i,M){
		G[a[i]][b[i]]=false;
        rep(j,N) vis[j]=false;
		dfs(a[i]);
		if(!vis[b[i]]) bridge++;
		G[a[i]][b[i]]=true;
	}
	cout<<bridge;
}