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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main() {
    int n,m;
    cin >> n >> m;
    vector<pair<int,int>> P;
    for(int i = 0; i < m; i++) {
        int a,b;
        cin >> a >> b;
        P.push_back(make_pair(a,b));
    }
    int ans = 0;
    for(int i = 0; i < m; i++) {
        bool ok = true;
        int x = P[i].first;
        int y = P[i].second;
        for(int j = 0; j < m; j++) {
            if(j == i) {
                continue;
            }
            if(P[j].first == x) {
                int point = j;
                int cur = P[j].second;
                for(int k = 0; k < m; k++) {
                    if(k == point) {
                        continue;
                    }
                    if(P[k].first == cur && P[k].second == y) {
                        ok = false;
                        break;
                    }
                    else if(P[k].second == cur && P[k].first == y) {
                        ok = false;
                        break;
                    }
                }
            }
            else if(P[j].second == x) {
                int point = j;
                int cur = P[j].first;
                for(int k = 0; k < m; k++) {
                    if(k == point) {
                        continue;
                    }
                    if(P[k].first == cur && P[k].second == y) {
                        ok = false;
                        break;
                    }
                    else if(P[k].second == cur && P[k].first == y) {
                        ok = false;
                        break;
                    }
                }
            }
        }
        if(ok) {
            ans++;
        }
    }
    cout << ans << endl;
    return 0;
}