CF 977E Cyclic Components

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.

$\text{[math]}$ 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 $m$ ( $1 \leq n \leq 2 \cdot 10^{5}$ , $0 \leq m \leq 2 \cdot 10^{5}$ ) — number of vertices and edges.

The following $m$ lines contains edges: edge $i$ is given as a pair of vertices $v_{i}$ , $u_{i}$ ( $1 \leq v_{i}, u_{i} \leq n$ , $u_{i} \neq v_{i}$ ). 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n个点，m条无向边，找有几个环。（定义：t个点，t条边，首尾依次相接，不含有其他边，围成个圈）

【分析】

网上的思路：利用并查集查找环的个数

我的思路:dfs判环，稍加修饰——

加个记忆化操作：显然每个点要么是一个环上的一点，要么不是；

1、当这个点的度数不为2，一定不是

2、当这个点的邻点不是，一定不是

【代码】

#include<vector>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int N=2e5+5;
int n,m,du[N],f[N];bool vis[N];
vector<int> e[N];
inline void Init(){
	scanf("%d%d",&n,&m);
	for(int i=1,x,y;i<=m;i++){
		scanf("%d%d",&x,&y);
		e[x].push_back(y);
		e[y].push_back(x);
		du[x]++;du[y]++;
	} 
}
int dfs(int x,int fa){
	int &now=f[x];
	if(~now) return now;
	now=1;
	if(du[x]!=2) return now=0;
	if(vis[x]) return now=1;
	vis[x]=1;
	for(int i=0;i<e[x].size();i++){
		int v=e[x][i];
		if(x!=fa) now&=dfs(v,x);
		if(!now) return now;
	}
	return now;
}
inline void Solve(){
	memset(f,-1,sizeof f);
	int ans=0;
	for(int i=1;i<=n;i++){
		if(!vis[i]){
			if(dfs(i,0)){
				ans++;
			}
		}
	}
	printf("%d\n",ans);
}
int main(){
	Init();
	Solve();
	return 0;
}

__EOF__

本文作者：shenben
本文链接：https://www.cnblogs.com/shenben/p/10497641.html
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