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 main(){
		int n,m,i,ans=0;
		int a[1000][3];
		scanf("%d%d",&n,&m);
		for(i=0;i<m;i++){
				scanf("%d%d",&a[i][0],&a[i][1]);
				a[i][2]=0;
		}
		for(i=0;i<m;i++){
				a[i][2]=1;
				int p[10000]={0},j,flag[51]={0};
				p[0]=a[i][0];flag[a[i][0]-1]=1;
				int st=0,fin=1;
				while(st<=fin){
						for(j=0;j<m;j++){
								if((p[st]==a[j][0]&&a[j][2]!=1&&flag[a[j][1]-1]!=1)){
							//			printf("%d %d\n",a[j][0],a[j][1]);
										p[fin]=a[j][1];
										flag[a[j][1]-1]=1;
										fin++;
								}
								if((p[st]==a[j][1]&&a[j][2]!=1&&flag[a[j][0]-1]!=1)){
							//			printf("%d %d\n",a[j][1],a[j][0]);
										p[fin]=a[j][0];
										flag[a[j][0]-1]=1;
										fin++;
								}
						}
						st++;
				}
				a[i][2]=0;
				int f=0;
				for(j=0;j<n;j++){
	//					printf("%d\n",flag[j]);
						if(flag[j]==0)f=1;
						if(flag[j]==1){
								flag[j]=0;
						}
				}
				if(f==1)ans++;
		}
		printf("%d\n",ans);
		return 0;
}