E. Cyclic Components

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我是靠谱客的博主寂寞马里奥，最近开发中收集的这篇文章主要介绍E. Cyclic Components，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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You are given an undirected graph consisting of n vertices and m

edges. Your task is to find the number of connected components which are cycles.

Here are some definitions of graph theory.

An undirected graph consists of two sets: set of nodes (called vertices) and set of edges. Each edge connects a pair of vertices. All edges are bidirectional (i.e. if a vertex a

is connected with a vertex b, a vertex b is also connected with a vertex a

). An edge can't connect vertex with itself, there is at most one edge between a pair of vertices.

Two vertices u

and v belong to the same connected component if and only if there is at least one path along edges connecting u and v

A connected component is a cycle if and only if its vertices can be reordered in such a way that:

the first vertex is connected with the second vertex by an edge,
the second vertex is connected with the third vertex by an edge,
...
the last vertex is connected with the first vertex by an edge,
all the described edges of a cycle are distinct.

A cycle doesn't contain any other edges except described above. By definition any cycle contains three or more vertices.

$\text{[math]}$ There are 6 connected components, 2 of them are cycles: [7,10,16] and [5,11,9,15]

. Input

The first line contains two integer numbers n

and m ( 1≤n≤2⋅105, 0≤m≤2⋅105

) — number of vertices and edges.

The following m

lines contains edges: edge i is given as a pair of vertices vi, ui ( 1≤vi,ui≤n, ui≠vi). There is no multiple edges in the given graph, i.e. for each pair ( vi,ui) there no other pairs ( vi,ui) and ( ui,vi

) in the list of edges.

Output

Print one integer — the number of connected components which are also cycles.

Examples

Input

Output

Input

Output

Note

In the first example only component [3,4,5]

is also a cycle.

The illustration above corresponds to the second example.

题意：给你n个点，m条边，问其中环的个数

分析：1、环的定义是每个点的度为二，也就是不能有多余的分支

2、dfs找到同一个集合的点，然后判断所有点的度是否为2，来判断是否成环

3、判断点的度可以用邻接表中每个点对应的大小

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#include <vector>
vector<int> g[200010];
vector<int> di;
int vis[200010]={0};
void dfs(int x)
//找到同一个集合的点
{
vis[x]=1;
//避免重复查找
di.push_back(x);
//存储同一个集合的点
for(int i=0;i<g[x].size();i++)
//遍历对应的邻接表
{
if(vis[g[x][i]]==0)
dfs(g[x][i]);
}
}
int main()
{
int n,m,x,y,i,j,cnt=0;
scanf("%d%d",&n,&m);
for(i=0;i<m;i++)
{
scanf("%d%d",&x,&y);
g[x].push_back(y);
//存储邻接表
g[y].push_back(x);
}
for(i=1;i<=n;i++)
{
if(!vis[i])
{
di.clear();
//清空存储仓
dfs(i);
for(j=0;j<di.size();j++)
//判断是否成环，所有点的度均为2
{
if(g[di[j]].size()!=2)
break;
}
if(j==di.size())
cnt++;
}
}
printf("%dn",cnt);
return 0;
}