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++ (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
using namespace std;

int n,m;
int a[51],b[51];
int graph[51][51];
int visited[51];

void dfs(int v){
        visited[v] = true;
        for(int v2=0;v2<n;v2++){
                if(visited[v2]==true)    continue;
                if(graph[v][v2]==false) continue;
                dfs(v2);
        }
}

int main(){
        int ans = 0;
        cin >> n >> m;
        for(int i=0;i<m;i++){
                 cin >> a[i] >> b[i];
                 a[i]--,b[i]--;
                 graph[a[i]][b[i]] = graph[b[i]][a[i]] = true;
        }
        for(int i=0;i<m;i++){
                graph[a[i]][b[i]] = graph[b[i]][a[i]] = false;
                for(int j=0;j<n;j++)    visited[j] = false;
                dfs(0);
                bool bridge=false;
                for(int j=0;j<n;j++)    if(visited[j]==false) bridge = true;
                if(bridge)      ans++;
                 graph[a[i]][b[i]] = graph[b[i]][a[i]] = true;
        }
        cout << ans << endl;
        return 0;
}