Cyclic Components （并查集）

Description

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≤n≤$2⋅10^5$, 0≤m≤$2⋅10^5$) — 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.

Input

5 4
1 2
3 4
5 4
3 5

Output

1

Input

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

2

Note

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

The illustration above corresponds to the second example.

解题思路：并查集的运用。判断单环的条件为判断每个集合（连通分量，同一个祖先节点）中所有点的度数是否都为2，并且该集合中元素的个数至少为3个，满足这两个条件才可构成单环。

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=200005;
int n,m,a,b,c,cnt,fa[maxn],Deg[maxn];vector<int> vec[maxn];
void init(){//初始化
    for(int i=1;i<=n;++i)fa[i]=i;
}
int findt(int x){
    int per=x,tmp;
    while(fa[per]!=per)per=fa[per];
    while(x!=per){tmp=fa[x];fa[x]=per;x=tmp;}//路径压缩
    return x;
}
void unite(int x,int y){
    x=findt(x),y=findt(y);
    if(x!=y)fa[x]=y;
}
int main(){
    cin>>n>>m;
    init();cnt=0;
    memset(Deg,0,sizeof(Deg));
    for(int i=1;i<=n;++i)vec[i].clear();//清空
    while(m--){
        cin>>a>>b;
        unite(a,b);
        Deg[a]++;Deg[b]++;//每个顶点的度数加1
    }
    for(int i=1;i<=n;++i)//把同一个祖先所有的节点放在一个邻接表中
        vec[findt(i)].push_back(i);
    for(int i=1;i<=n;++i){
        if(vec[i].size()>2){//构成单环的点的个数至少为3个
            bool flag=false;
            for(size_t j=0;j<vec[i].size();++j)
                if(Deg[vec[i][j]]!=2){flag=true;break;}//如果度数不为2的，直接退出
            if(!flag)cnt++;//如果是单环，计数器就加1
        }
    }
    cout<<cnt<<endl;
    return 0;
}