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: Wrong Answer)

'use strict'

class Main {
  solve() {
    const size = input.nextIntArr();
    console.log(size);
    const nodes = input.nextIntRange(size[1]);
    const edges = [];
    for (const n of nodes) {
      edges[n[0]] = edges[n[0]] ? edges[n[0]] : [];
      edges[n[1]] = edges[n[1]] ? edges[n[1]] : [];
      edges[n[0]].push(n[1]);
      edges[n[1]].push(n[0]);
    }

    let cnt = 0;
    for (let i = 0; i < size[1]; ++i) {
      const n = nodes[i];
      const stack = [n[0]];
      const visited = [];
      while (stack.length > 0) {
        const e = stack.pop();
        visited[e] = true;
        for (let _e of edges[e]) {
          if (e == n[0] && _e == n[1]) { continue; }
          if (visited[_e]) {continue;}
          stack.push(_e);
        }
      }
      if (!visited[n[1]]) { ++cnt; }
    }
    console.log(cnt);
  }
}

class Input {
  constructor() {
    this.cursor = 0;
    this.inputAll();
  }

  inputAll() {
    this.data = (require("fs").readFileSync("/dev/stdin", "utf8")).split('\n');
    return this.data;
  }

  /* 1行 or 複数行文字列 */
  nextLine(n) {
    if (n) {
      const ret = this.data.slice(this.cursor, this.cursor + n);
      this.cursor += n;
      return ret;
    }
    return this.data[this.cursor++];
  }
  /* 一行文字配列 */
  nextStrArr() { return this.nextLine().split(' '); } 
  /* 一行整数配列 */
  nextIntArr() { return this.nextStrArr().map((e) => parseInt(e)); }
  /* 一行浮動小数点配列 */
  nextFloatArr() { return this.nextStrArr().map((e) => parseFloat(e)); }
  /* 複数行整数配列 */
  nextIntLine(n) { return this.nextLine(n).map((e) => parseInt(e)) }
  /* 複数行浮動小数点配列 */
  nextFloatLine(n) { return this.nextLine(n).map((e) => parseFloat(e)) }
  /* 複数行文字列二次元配列 */
  nextStrRange(n) { return this.nextLine(n).map((line) => line.split(' ')); }
  /* 複数行整数二次元配列 */
  nextIntRange(n) { return this.nextLine(n).map((line) => line.split(' ').map((e) => parseInt(e))); }
  /* 複数行浮動小数点数二次元配列 */
  nextFloatRange(n) {
    return this.nextLine(n).map((line) => line.split(' ').map((e) => parseFloat(e)));
  }
} 
const input = new Input();
(new Main()).solve();