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 <cstdio>
#include <string>
#include <algorithm>
#include <utility>
#include <cmath>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define rep(i, n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
const ll mod = 1000000007;

class UnionFind {
public:
    vector<int> parent;  //親ノード
    vector<int> Siz;  //集合の大きさ

    // コンストラクタ
    UnionFind(int n): parent(n), Siz(n, 1) {
        rep(i, n) {
            parent[i] = i;
        }
    }

    // n要素で初期化
    void init(int n) {
        parent.resize(n);
        Siz.assign(n, 1);
        rep(i, n) {
            parent[i] = i;
        }
    }

    // xの属する集合の根を求める
    int root(int x) {
        if(parent[x] == x) {
            return x;
        }
        else {
            return parent[x] = root(parent[x]);
        }
    }

    // xとyの属する集合を併合
    void unite(int x, int y) {

        x = root(x);
        y = root(y);
        if(x == y) {
            return;
        }

        if(Siz[x] < Siz[y]) {
            swap(x, y);
        }

        Siz[x] += Siz[y];
        parent[y] = x;
    }

    // xとyが同じ集合に属するかどうか
    bool same(int x, int y) {
        return root(x) == root(y);
    }

    // xの属する集合の要素数を求める
    int size(int x) {
        return Siz[root(x)];
    }
};

int main() {
    int n, m;
    cin >> n >> m;

    UnionFind Uni(n);

    vector<int> a(m), b(m);
    rep(i, m) {
        cin >> a[i] >> b[i];
        a[i]--;  b[i]--;
    }

    int ans = 0;
    rep(i, m) {

        Uni.init(n);

        rep(j, m) {
            if(i == j) {
                continue;
            }
            Uni.unite(a[j], b[j]);
        }

        if(Uni.size(0) < n) {
            ans++;
        }
    }

    cout << ans << endl;
    return 0;
}