Monk and Graph Problem

中文题意：求一个图中的最大连通子图的边的个数。

Monk and his graph problems never end. Here is one more from our own Monk:
Given an undirected graph with $N$ vertices and $M$ edges, what is the maximum number of edges in any connected component of the graph.
In other words, if given graph has $k$ connected components, and $E_1,E_2,....E_k$ be the number of edges in the respective connected component, then find $max(E_1,E_2,....,E_k)$ .

The graph may have multiple edges and self loops. The $N$ vertices are numbered as $1,2,...,N$.

Input Format:
The first line of input consists of two space separated integers $N$ and $M$, denoting number of vertices in the graph and number of edges in the graph respectively.
Following $M$ lines consists of two space separated each $a$ and $b$, denoting there is an edge between vertex $a$ and vertex $b$.

Output Format:
The only line of output consists of answer of the question asked by Monk.

Input Constraints:
$1\le N\le {10}^5$
$0\le M\le {10}^5$
$1\le a,b\le N$

SAMPLE INPUT

 

6 31 22 34 5

SAMPLE OUTPUT

 

2

Explanation

The graph has $3$ connected components :
First component is $1-2-3$ which has $2$ edges.
Second component is $4-5$ which has $1$ edge.
Third component is $6$ which has no edges.

So, answer is $max(2,1,0)=2$

Time Limit:1.0 sec(s) for each input file.

Memory Limit:256 MB

Source Limit:1024 KB

Marking Scheme:Marks are awarded when all the testcases pass.

Allowed Languages:

C, C++, C++14, Clojure, C#, D, Erlang, F#, Go, Groovy, Haskell, Java, Java 8, JavaScript(Rhino), JavaScript(Node.js), Julia, Kotlin, Lisp, Lisp (SBCL), Lua, Objective-C, OCaml, Octave, Pascal, Perl, PHP, Python, Python 3, R(RScript), Racket, Ruby, Rust, Scala, Swift, Visual Basic

#include <bits/stdc++.h>#include <vector>using namespace std;vector <int>adj[100005];bool visited[100005];int mmax = 0;int Count=0;void dfs(int s){visited[s] = 1;for (int i = 0; i<adj[s].size(); ++i){    Count++;if (visited[adj[s][i]]== 0)dfs(adj[s][i]);}}int main(){int nodes, edges, x, y;cin >> nodes >> edges;memset(visited, 0, sizeof(visited));for (int i = 0; i<edges; ++i){cin >> x >> y;adj[x].push_back(y);adj[y].push_back(x);}for (int i=0; i <= nodes; ++i){if (!visited[i])dfs(i);mmax=max(mmax,Count/2);Count=0;}cout << mmax << endl;}