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 x[50][50] = { 0 };
int ab[50][2] = { 0 };
int cnt[50] = { 0 };

void check(int n,int N) {
	cnt[n] = 1;

	for (int i = 0; i < N; i++) {
		if ((x[n][i] == 1) && (cnt[i] == 0)&&(i!=n)) {
			check(i, N);
		}
	}
}

int main(void) {
	int N, M;
	int ans = 0;

	scanf("%d %d", &N, &M);
	for (int i = 0; i < M; i++) {
		scanf("%d %d", &ab[i][0], &ab[i][1]);
	}
	
	for (int i = 0; i < M; i++) {
		for (int j = 0; j < N; j++) {
			for (int k = 0; k < N; k++) x[j][k] = 0;
		}

		for (int j = 0; j < M; j++) {
			int a = ab[j][0] - 1;
			int b = ab[j][1] - 1;
			if (j != i) {
				x[a][b] = 1;
				x[b][a] = 1;
			}
		}

		for (int j = 0; j < N; j++) {
			for (int k = 0; k < N; k++) cnt[k] = 0;
			check(j, N);
			int cn = 0;
			for (int k = 0; k < N; k++) {
				if (cnt[k] == 1) cn++;
			}
			if (cn < N) {
				ans++;
				break;
			}
		}
	}

	printf("%d\n", ans);

	return 0;
}