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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int N,M;
int a[50];
int l[50];
int r[50];
int ans=0;

void init();
int root(int x);
void unite(int x,int y);
int query(int x,int y);
int judge();

int main(){
	scanf("%d%d",&N,&M);
	
	int i;
	for(i=0;i<M;i++){
		scanf("%d%d",&l[i],&r[i]);
		l[i]--;
		r[i]--;
	}
	
	for(i=0;i<M;i++){
		init();
		
		int j;
		for(j=0;j<M;j++){
			if(j==i)continue;
			unite(l[j],r[j]);
		}
		
		if(judge()==0)ans++;
	}
	
	printf("%d\n",ans);
	return 0;
}

void init(){
	for(int i=0;i<N;i++)a[i]=i;
}

int root(int x){
	if(a[x]==x)return x;
	else return a[x]=root(a[x]);
}

void unite(int x,int y){
	x=root(x);
	y=root(y);
	if(x==y)return;
	a[x]=y;
}

int query(int x,int y){
	if(root(x)==root(y))return 1;
	return 0;
}

int judge(){
	for(int i=0;i<N;i++){
		for(int j=i+1;j<N;j++){
			if(query(i,j)==0)return 0;
		}
	}
	return 1;
}