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 JavaScript (node.js v5.12) (judge verdict: Accepted)

"use strict"

let n, m, graph, list, visited

//https://abc075.contest.atcoder.jp/tasks/abc075_c
//グラフは隣接行列表現＝二次元配列で表現
//深さ優先探索で 辺を削除後に全部のポイントを訪問できるか

const dfs = v => {
  visited[v] = true
  for (let v2 = 0; v2 < n; v2++) {
    if (visited[v2] || graph[v][v2] == false) continue
    dfs(v2)
  }
}

function main(input) {
  input = input.split(/\n/).map(v => v.split(" ").map(v => ~~v))
  n = input[0][0]
  m = input[0][1]

  list = input.slice(1, input.length)
  graph = Array.apply(null, { length: n }).map(v =>
    Array.apply(null, { length: n }).map(v => false)
  )
  // graphをマーク
  for (let i = 0; i < m; i++) {
    let p1 = list[i][0] - 1
    let p2 = list[i][1] - 1
    graph[p1][p2] = true
    graph[p2][p1] = true
  }

  let ans = 0

  //隣接行列表現を探索
  for (let i = 0; i < n; i++) {
    //訪問済みを初期化
    for (let j = 0; j < n; j++) {
      ///
      visited = Array.apply(null, { length: n }).map(v => false)
      if (graph[i][j] == true) {
        // 辺を見つけたら
        graph[i][j] = graph[j][i] = false
        dfs(i)
        //全点訪問できなかったら == every=(v)=v==true = false
        //橋をカウント
        if (visited.every(v => v == true) != true) ans++
        graph[i][j] = graph[j][i] = true
      }
    }
  }
  console.log(ans / 2) //隣接行列は右上と左下が対象になるので半分にする = ans / 2
}
main(require("fs").readFileSync("/dev/stdin", "utf8"));