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;

vector<int> v[100];
vector<pair<int,int> > ed;
int l[100];

int fi(int nw)
{
	if(l[nw]==nw)return nw;
	return l[nw]=fi(l[nw]);
}

void uni(int a,int b)
{
	l[a]=b;
}

main()
{
	int n,m;
	scanf("%d %d",&n,&m);
	for(int i=0;i<m;i++)
	{
		int a,b;
		scanf("%d %d",&a,&b);
		v[a].push_back(b);
		v[b].push_back(a);
		ed.push_back(make_pair(a,b));
	}
	int ans=0;
	for(int i=0;i<m;i++)
	{
		for(int j=0;j<n;j++)
		{
			l[j]=j;
		}
		for(int j=0;j<m;j++)
		{
			if(i==j)
				continue;
			if(fi(ed[j].first)!=fi(ed[j].second))
			{
				uni(fi(ed[j].first),fi(ed[j].second));
			}
		}
		if(fi(ed[i].first)!=fi(ed[i].second))
			ans++;
	}
	printf("%d",ans);
}