Cyclic Components（并查集）

链接：http://codeforces.com/contest/977/problem/E

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.

大意：找出有多少个环

思路：算出每个顶点的度，只有度等于2的情况下才能成为环。如果两个顶点的度都为2，使用并查集，如果父节点相同，答案加一，否则联立。

AC代码：

#include <cstdio>   
#include <iostream> 
#include <algorithm> 
#include <cmath> 
#include <cstdlib> 
#include <cstring> 
#include <vector> 
#include <list> 
#include <map> 
#include <stack> 
#include <queue> 
using namespace std; 
#define ll long long
int f[200010];
int sum[200010];
int ans;
struct edge 
{
	int u,v;
};
int Find(int x)
{
	if(f[x] == x)
		return x;
	else
		return f[x] = Find(f[x]);
} 
void unio(int x,int y)
{
	int f1 = Find(x);
	int f2 = Find(y);
	if(f1 != f2)
		f[f1] = f2;
	else
		ans++;
}
int main() 
{
	int n,m;
	while(cin >> n >> m)
	{
		memset(sum,0,sizeof(sum));
		for(int i = 1;i <= n;i++)
			f[i] = i;
		ans = 0;
		struct edge e[m];
		for(int i = 0;i < m;i++)
		{
			cin >> e[i].u >> e[i].v;
			sum[e[i].u]++;
			sum[e[i].v]++;
		}
		for(int i = 0;i < m;i++)
		{
			if(sum[e[i].u] == 2 && sum[e[i].v] == 2)
				unio(e[i].u,e[i].v); 
		}
		cout << ans <<endl;
	}
    //cout << "AC" <<endl; 
    return 0; 
}