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 Haskell (GHC 7.10.3) (judge verdict: Accepted)

import qualified Data.ByteString.Char8 as BC8
import qualified Data.Maybe as M
-- import qualified Data.IntMap.Lazy as M
import Control.Monad
import Data.List
import Debug.Trace

main :: IO ()
main = do
    [n, m] <- map (fst . M.fromJust . BC8.readInt) . BC8.words <$> BC8.getLine
    edges <- replicateM m ((\[a, b]->(a,b)) . map (fst . M.fromJust . BC8.readInt) . BC8.words <$> BC8.getLine)
    -- print edges
    print . length $ filter (\i -> not (isConnectedGraph (edges \\ [edges!!i]) n)) [0..m-1]

isConnectedGraph :: [(Int, Int)] -- グラフ
                 -> Int          -- 頂点の数
                 -> Bool
isConnectedGraph graph n = (==0) . length $ [1..n] \\ isConnectedGraph' (graph++map (\(f,t)->(t,f)) graph) [] (fst $ head graph)

isConnectedGraph' :: [(Int, Int)] -> [Int] -> Int -> [Int]
isConnectedGraph' [] already pos = pos:already
isConnectedGraph' graph already pos = let
    ns = filter (\(f,t)->pos==f && not (elem t already)) graph
    in if null ns
        then nub $ pos:already
        else nub . foldr (++) [] $  map (\(f,t) -> isConnectedGraph' graph (nub $ pos:already ++ map snd ns) t) ns