S - Cyclic Components （并查集的理解）

Posted 2020-12-04 acgoto

Description

You are given an undirected graph consisting of nn vertices and mm 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 aa is connected with a vertex bb, a vertex bb is also connected with a vertex aa). An edge can‘t connect vertex with itself, there is at most one edge between a pair of vertices.Two vertices uu and vv belong to the same connected component if and only if there is at least one path along edges connecting uu and vv.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 66 connected components, 22 of them are cycles: [7,10,16]and [5,11,9,15].

Input

The first line contains two integer numbers nn and mm (1≤n≤2?10^5, 0≤m≤2?10^5) — number of vertices and edges.

The following mm lines contains edges: edge ii 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

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）中所有点的度是否为2，并且集合中元素的个数至少为3（这样才能构成单环）即可。

AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=200005;
 4 int n,m,a,b,c,cnt,fa[maxn],Deg[maxn];vector<int> vec[maxn];
 5 void init(){//初始化
 6     for(int i=1;i<=n;++i)fa[i]=i;
 7 }
 8 int findt(int x){
 9     int per=x,tmp;
10     while(fa[per]!=per)per=fa[per];
11     while(x!=per){tmp=fa[x];fa[x]=per;x=tmp;}//路径压缩
12     return x;
13 }
14 void unite(int x,int y){
15     x=findt(x),y=findt(y);
16     if(x!=y)fa[x]=y;
17 }
18 int main(){
19     cin>>n>>m;
20     init();cnt=0;
21     memset(Deg,0,sizeof(Deg));
22     for(int i=1;i<=n;++i)vec[i].clear();//清空
23     while(m--){
24         cin>>a>>b;
25         unite(a,b);
26         Deg[a]++;Deg[b]++;//每个节点的度加1
27     }
28     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
29         vec[findt(i)].push_back(i);
30     for(int i=1;i<=n;++i){
31         if(vec[i].size()>2){//构成单环的点的个数至少为3个
32             bool flag=false;
33             for(size_t j=0;j<vec[i].size();++j)
34                 if(Deg[vec[i][j]]!=2){flag=true;break;}//如果度有不是2的，直接退出
35             if(!flag)cnt++;//如果是单环，计数器就加1
36         }
37     }
38     cout<<cnt<<endl;
39     return 0;
40 }