E. Cyclic Components（并查集？）

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：给你n个顶点，m条边，求单圈环的个数。（只有一个圈的环）

题解：通过观察发现单圈环里的顶点的度都为2，所以并查集找连通图，map存顶点度数，遍历查找单圈环。

#include<iostream>  
#include<string.h>  
#include<algorithm>  
#include<cmath>  
#include<map>  
#include<string>  
#include<stdio.h>  
#include<vector>  
#include<stack>
#include<set>
using namespace std;
#define INIT ios::sync_with_stdio(false)
#define LL long long int

struct node {
	int id;
	int count;
};
int dp[2 * 100005];
vector<int>mm[200005];
void init() {
	for (int i = 0;i < 2 * 100005;i++) {
		dp[i] = i;
	}
}

int _find(int x) {
	if (dp[x] != x) {
		return dp[x] = _find(dp[x]);
	}
	return dp[x];
}

void _union(int  x, int y) {
	int xx = _find(x);
	int yy = _find(y);
	if (xx < yy) {
		dp[xx]= yy;
	}
	else {
		dp[yy] = xx;
	}
}

int main() {
	int n, m;
	map<int, int>mp;//存每个节点的度
	while (cin >> n >> m)
	{
		mp.clear();
		init();
		int a, b;
		int ans = 0;
		for (int i = 0;i < m;i++) {
			cin >> a >> b;
			mp[a]++;
			mp[b]++;
			if (_find(a) != _find(b)) {
				_union(a, b);
			}
		}
		//通过父节点把连通图存进一个vector数组
		for (int i = 1;i <= n;i++) {
			mm[_find(i)].push_back(i);
		}
		for (int i = 1;i <= n;i++) {
			//连通图至少要有3个点
			if (mm[i].size() > 2) {
				int flag = 1;
				for (int j = 0;j < mm[i].size()&&flag;j++) {
					//只要有一个度不为2就跳出
					if (mp[mm[i][j]] != 2) {
						flag = 0;
					}
				}
				if (flag)ans++;
			}
		}
		cout << ans << endl;
	}
	return 0;
}