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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)


import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.util.NoSuchElementException;

public class Main implements Runnable { // クラス名はMain1

	static int mod = 1000000000 + 7; //10^9+7
	int[] dx = { 1, 0, -1, 0 }, dy = { 0, 1, 0, -1 }; //4 directions

	public static void main(String[] args) {
		Thread.setDefaultUncaughtExceptionHandler((t, e) -> System.exit(1));
		new Thread(null, new Main(), "", 1024 * 1024 * 1024).start(); // 16MBスタックを確保して実行
	}

	public void run() {
		PrintWriter out = new PrintWriter(System.out);
		InputReader sc = new InputReader(System.in);

		try {
			/**
			 * Bit 全探索
				//Nは要素数 (bit全探索)
				//計算量はO(2^N) Nが0-20~25 (30は多分無理)くらいまでしか使えない
				Out: for (int bit = 0; bit < (1 << N); bit++) {
					System.out.println(Integer.toBinaryString(bit));
					In: for (int j = 0; j < N; j++) {
						System.out.println(1 & (bit >> j));
						//1が立ってれば対象。例えば101だったら1桁目と3桁目は選ばれている
						//「二進数iをjだけ右にビットシフトしたときの、iの1桁目と1とでビット論理積をとって、演算結果が1であるなら処理を実行する」
						if ((1 & bit >> j) == 1) {
						}
					}
				}
			 *
			 * 表示系
			 *   System.out.println();
			 *   System.out.println("i:" + i + " ,j:" + j);
			 *   System.out.println(Arrays.toString(A));
			 *
			 * Sort (Arrays)
			 *  Integer[] A = new Integer[N];
			 *    Desc : Arrays.sort(A , Collections.reverseOrder());
			 *    Asc  : Arrays.sort(A);
			 *
			 * 定義系
			 *   int[] s = new int[M];
			 *   ArrayDeque<Integer> deq= new ArrayDeque<>();//先頭と末尾への要素の追加・先頭と末尾の要素の取り出しと削除が定数時間
			 *   ArrayList<Integer> arr = new ArrayList<Integer>();//定数時間
			 *   HashMap<Integer,Integer> hm = new HashMap<>(); //定数時間
			 *
			 *   //要素の追加と、最小の要素の取り出しと削除が対数時間
			 *   PriorityQueue<Integer> pq = new PriorityQueue<>();//昇順
			 *   PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder()); //逆順。降順
			 *
			 *   TreeSet<Integer> ts = new TreeSet<>();//対数時間(重複はできない)
			 *
			 * 文字系
			 *   String S = sc.next().trim();
			 *   for (char i = 'a'; i <= 'z'; i++) {}
			 *   S.substring(1,2);//あいうえお → い
			 *   S.substring(0,1); // 1文字目をとりたいとき
			 *   文字列のSort
			 *    String sorted = S.chars().sorted().collect(StringBuilder::new,StringBuilder::appendCodePoint, StringBuilder::append) .toString();
			 */
			int N = sc.nextInt();
			int M = sc.nextInt();
			int[] a = new int[M];
			int[] b = new int[M];

			for (int i = 0; i < M; i++) {
				a[i] = sc.nextInt();
				b[i] = sc.nextInt();
			}

			long ans = 0;
			for (int i = 0; i < M; i++) {
				UnionFindTree uft = new UnionFindTree(N);
				for (int j = 0; j < M; j++) {
					if (i != j) {
						uft.unite(a[j] - 1, b[j] - 1);
					}
				}
				//out.println(i + " " + uft.getGroup());
				if (uft.getGroup() > 1) {
					ans++;
				}
			}

			out.println(ans);
			out.close();

		} catch (ArithmeticException ae) {
			ae.printStackTrace();
		} catch (Exception e) {
			e.printStackTrace();
		}

	}

	public static long comb(long a, long b) {
		if (a < b)
			return 0;
		long res = 1;
		long inv = pow(fact(Math.min(b, a - b)), mod - 2);
		for (long i = a; i > a - Math.min(b, a - b); i--) {
			res *= i;
			res %= mod;
		}
		res *= inv;
		res %= mod;
		return res;
	}

	public static long pow(long x, long n) {
		long res = 1;
		while (n > 0) {
			if ((n & 1) == 1) {
				res *= x;
				res %= mod;
			}
			x *= x;
			x %= mod;
			n >>= 1;
		}
		return res;
	}

	public static long fact(long n) {
		long res = 1;
		while (n > 0) {
			res *= n;
			res %= mod;
			n--;
		}
		return res;
	}

	// 高速なScanner
	static class InputReader {
		private InputStream in;
		private byte[] buffer = new byte[1024];
		private int curbuf;
		private int lenbuf;

		public InputReader(InputStream in) {
			this.in = in;
			this.curbuf = this.lenbuf = 0;
		}

		public boolean hasNextByte() {
			if (curbuf >= lenbuf) {
				curbuf = 0;
				try {
					lenbuf = in.read(buffer);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (lenbuf <= 0)
					return false;
			}
			return true;
		}

		private int readByte() {
			if (hasNextByte())
				return buffer[curbuf++];
			else
				return -1;
		}

		private boolean isSpaceChar(int c) {
			return !(c >= 33 && c <= 126);
		}

		private void skip() {
			while (hasNextByte() && isSpaceChar(buffer[curbuf]))
				curbuf++;
		}

		public boolean hasNext() {
			skip();
			return hasNextByte();
		}

		public String next() {
			if (!hasNext())
				throw new NoSuchElementException();
			StringBuilder sb = new StringBuilder();
			int b = readByte();
			while (!isSpaceChar(b)) {
				sb.appendCodePoint(b);
				b = readByte();
			}
			return sb.toString();
		}

		public int nextInt() {
			if (!hasNext())
				throw new NoSuchElementException();
			int c = readByte();
			while (isSpaceChar(c))
				c = readByte();
			boolean minus = false;
			if (c == '-') {
				minus = true;
				c = readByte();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res = res * 10 + c - '0';
				c = readByte();
			} while (!isSpaceChar(c));
			return (minus) ? -res : res;
		}

		public long nextLong() {
			if (!hasNext())
				throw new NoSuchElementException();
			int c = readByte();
			while (isSpaceChar(c))
				c = readByte();
			boolean minus = false;
			if (c == '-') {
				minus = true;
				c = readByte();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res = res * 10 + c - '0';
				c = readByte();
			} while (!isSpaceChar(c));
			return (minus) ? -res : res;
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public int[] nextIntArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		public long[] nextLongArray(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}

		public char[][] nextCharMap(int n, int m) {
			char[][] map = new char[n][m];
			for (int i = 0; i < n; i++)
				map[i] = next().toCharArray();
			return map;
		}
	}
}

class UnionFindTree {

	int[] parent; //インデックスにとノードを対応させ、そのルートノードのインデックスを格納
	int[] rank; //parentと同様に、木の高さを格納
	int[] size; //木の要素の数を格納。最新はparentのindexに
	int group;

	public UnionFindTree(int n) {
		this.parent = new int[n];
		this.rank = new int[n];
		this.size = new int[n];
		group = n;

		for (int i = 0; i < n; i++) {
			makeSet(i);
		}
	}

	public void makeSet(int i) {
		parent[i] = i;
		rank[i] = 0; //集合の高さ
		size[i] = 1;
	}

	public void unite(int x, int y) {
		int xRoot = find(x);
		int yRoot = find(y);
		if (xRoot == yRoot)
			return;
		group--;
		//xが属する木の方が大きい場合
		if (rank[xRoot] > rank[yRoot]) {
			parent[yRoot] = xRoot; //yの親をxに更新
			size[xRoot] += size[yRoot];

		} else if (rank[xRoot] < rank[yRoot]) {
			parent[xRoot] = yRoot;
			size[yRoot] += size[xRoot];
		} else {
			parent[yRoot] = xRoot;
			rank[xRoot]++; //同じ高さの木がルートの子として着くから大きさ++;
			size[xRoot] += size[yRoot];
		}
	}

	//引数の属する木のルートのidを返す
	public int find(int i) {
		if (i != parent[i]) {
			parent[i] = find(parent[i]);
		}
		return parent[i];
	}

	//同じ木に属しているか
	public boolean same(int x, int y) {
		return find(x) == find(y);
	}

	// via Oland https://atcoder.jp/contests/abc120/submissions/4452920
	public int size(int i) {
		return size[find(i)];
	}

	public int getGroup() {
		return group;
	}
}

class Exm implements Comparable<Exm> {

	public int a;
	public int b;

	@Override
	public int compareTo(Exm o) {

		//return this.a - o.a;//昇順
		// return o.a - this.a;//降順

		//複数項目で並び替え。aの降順、aが同じならbの降順
		if (this.a == o.a) {
			return o.b - this.b;
		} else {
			return o.a - this.a;
		}
	}
}