Cyclic Components CodeForces - 977E（bfs+思维）

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.

题意：给出n个点，m条边（无向），问这个图中的单环有多少个（单环，即只有一个环，列二中有一个环不算单环）

思路：如果是单环，那么任意节点必定有且只有两条边与之相连。那么，dfs来做就好了。

#include "iostream"
#include "vector"
using namespace std;
const int Max=2e5+10;
vector<int> G[Max];
bool flag,vis[Max];
void dfs(int x){
    vis[x]=1;
    if (G[x].size()!=2) flag=0;//判断节点所连的边数是否为2
    for (int i=0;i<G[x].size();i++)
        if (!vis[G[x][i]]) dfs(G[x][i]);
}
int main(){
    ios::sync_with_stdio(false);
    int n,m,u,v,ans=0;
    cin>>n>>m;
    for(int i=0;i<Max;i++) G[i].clear();
    for(int i=0;i<m;i++){
        cin>>u>>v;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    for(int i=1;i<=n;i++){
        flag=1;
        if (!vis[i]){
            dfs(i);
            if(flag) ans++;
        }
    }
    cout<<ans<<endl;
    return 0;
}