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 Python (3.4.3) (judge verdict: Runtime Error)

// 深さ優先探索　計算量：O(N+M)

#include <bits/stdc++.h>
typedef long long ll;
#define MOD 1000000007

using namespace std;

const int limit = 50;

int n,m;
int a[limit],b[limit];

bool graph[limit][limit];
bool visited[limit];

void dfs(int v) {
    visited[v] = true;
    for (int v2 = 0; v2 < n; ++v2) {
        if (graph[v][v2] == false) continue; // vからv2へ路があるなら次の処理へ
        if (visited[v2] == true) continue; // v2を訪れたことがないならv2へ進みdfs
        dfs(v2);
    }
}

int main(){
    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;
    }
    int ans = 0;
    for (int i = 0; i < m; i++)
    {
        graph[a[i]][b[i]] = graph[b[i]][a[i]] = false; // 1つ目の辺を取り除く
        for (int j = 0; j < n; j++)
        {
            visited[j] = false; // visitedを初期化
        }
        dfs(0);
        bool bridge = false;
        for (int j = 0; j < n; j++)
        {
            if(visited[j] == false) bridge = true; // 訪れていない点が1つでも存在する => ”橋”が存在する
        }
        if(bridge) ans++;
        graph[a[i]][b[i]] = graph[b[i]][a[i]] = true; //取り除いた辺を元に戻す
        
        
    }
    cout << ans << endl;
    return 0;
}