Cyclic Components (dfs)

最新推荐文章于 2022-12-03 20:24:44 发布

Mannix_Y

最新推荐文章于 2022-12-03 20:24:44 发布

阅读量219

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/Mannix_Y/article/details/80247509

本文介绍了一个图论问题的解决方法，即在一个给定的无向图中找出所有构成环状结构的连通组件的数量。文章提供了完整的代码实现，并通过DFS算法进行遍历验证。

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
inputCopy
4
2
4
4
5
outputCopy
1
inputCopy
15
8
12
11
9
15
5
13
13
3
16
10
7
3
4
6
outputCopy
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：有n个点，m条连线，求循环组件的个数

题解：使之循环，满足的点有且仅链接两条边，再用dfs判定是否循环

#include<iostream>
#include<cstdio>
#include<map>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstring>
#include<climits>
#include<set>
#include<cmath>
#include<stack>
#define ll long long
#define MT(a,b) memset(a,0,sizeof(a))
using namespace std;
const int maxn = 2e5+5;
ll a,b;
int vis[maxn];
ll x=0;
int sum=0;
int n,m,l=0;
map<ll ,vector<ll>>mp;
bool flag=false;

void dfs(ll p)
{
    vis[p]=1;
    if(mp[p].size()!=2)return;
    if(l>=2&&(mp[p][0]==x||mp[p][1]==x))sum++;
    for(auto i=mp[p].begin();i!=mp[p].end();i++)
    {
        if(!vis[*i])
        {
            vis[*i]=1;
            l++;
            dfs(*i);
            l--;
        }
    }
}

int main()
{
    cin>>n>>m;
    for(ll i=0;i<m;i++)
    {
        cin>>a>>b;
        mp[a].insert(mp[a].end(),b);
        mp[b].insert(mp[b].end(),a);
    }
    MT(vis,0);
    for(auto i=mp.begin();i!=mp.end();i++)
    {
        flag=false;
        l=0;
        if(vis[i->first])continue;
        x=i->first;
        dfs(i->first);
    }
    cout<<sum<<endl;
    return 0;
}