CodeForces - 977E，E. Cyclic Components

E. Cyclic Components

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

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 $m$ ($1\le n\le 2\cdot {10}^5$, $0\le m\le 2\cdot {10}^5$) — number of vertices and edges.

The following $m$ lines contains edges: edge $i$ is given as a pair of vertices $v_i$, $u_i$ ($1\le v_i,u_i\le n$, $u_i\ne v_i$). There is no multiple edges in the given graph, i.e. for each pair ($v_i,u_i$) there no other pairs ($v_i,u_i$) and ($u_i,v_i$) in the list of edges.

Output

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

Examples

input

Copy

5 4
1 2
3 4
5 4
3 5

output

Copy

1

input

Copy

17 15
1 8
1 12
5 11
11 9
9 15
15 5
4 13
3 13
4 3
10 16
7 10
16 7
14 3
14 4
17 6

output

Copy

2

Note

In the first example only component $[3,4,5]$ is also a cycle.

The illustration above corresponds to the second example.

由于这种题做得少，虽然知道大体思路，但是具体到实施就不会了，于是看别人的代码     o#（这里、这里）#o    找想法

具体是，用dfs找环（vis标记），对于不成环或者环不标准的标记一下（cls），以后不会再找，能够成环的就加上去。

嗯，让我这么一写，运行时间是人家的代码的4倍，内存也大了1/4,，不过还好是一遍敲出来了。

#include<bits/stdc++.h>
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include<string.h>
#include <queue>
using namespace std;
typedef long long ll;

vector<ll> Ed[200005];

vector<ll>  vis,cls;

int flag;

void dfs(int cur)
{
    if(flag)
        return;
    vis[cur]=1;

    int t=Ed[cur].size();
    if(t!=2)
    {
        flag=-1;
        return;
    }

    for(int i=0;i<t;i++)
    {
        int z=Ed[cur][i];
        if(!vis[z])
        {
            dfs(z);
            if(flag)
                return;
        }
    }
    flag=1;
}

void clear(int cur)
{
    cls[cur]=1;
    int t=Ed[cur].size();

    for(int i=0;i<t;i++)
    {
        if(!cls[Ed[cur][i]])
            clear(Ed[cur][i]);
    }
}


int main()
{
    int n,m;
    cin>>n>>m;
    vis.resize(n+1);
    cls.resize(n+1);
    for(int i=0;i<m;i++)
    {
        ll x,y;
        cin>>x>>y;
        Ed[x].push_back(y);
        Ed[y].push_back(x);
    }
    ll ans=0;
    for(int i=1;i<=n;i++)
    {
        if(!cls[i])
        {
            flag=0;
            dfs(i);
            ans+=max(0,flag);
            clear(i);
        }
    }
    cout<<ans<<endl;

    return 0;
}