Algorithm
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.
The first line contains two integer numbers n and m (1≤n≤2⋅105, 0≤m≤2⋅105) — number of vertices and edges.
The following m lines contains edges: edge i is given as a pair of vertices vi, ui (1≤vi,ui≤n, ui≠vi). There is no multiple edges in the given graph, i.e. for each pair (vi,ui) there no other pairs (vi,ui) and (ui,vi) in the list of edges.
Print one integer — the number of connected components which are also cycles.
5 4
1 2
3 4
5 4
3 5
1
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
2
In the first example only component [3,4,5] is also a cycle.
The illustration above corresponds to the second example.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
using namespace std;
int const N = 2e5 + 1;
int n, m, res;
bool vis[N];
vector<vector<int> > g;
vector<int> all;
void DFS(int u) {
vis[u] = true;
all.push_back(u);
for(int i = 0;i < g[u].size(); ++i)
if(!vis[g[u][i]])
DFS(g[u][i]);
}
int main() {
cin >> n >> m;
g.resize(n);
for(int i = 0, a, b; i < m; ++i) {
cin >> a >> b;
--a, --b;
g[a].push_back(b);
swap(a, b);
g[a].push_back(b);
}
for(int i = 0; i < n; ++i) {
if(!vis[i]) {
all.clear();
DFS(i);
bool ok = true;
for(int j = 0; j < all.size(); ++j)
if(g[all[j]].size() != 2) {
ok = false;
break;
}
res += (ok && all.size() > 2);
}
}
cout << res << endl;
return 0;
}
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Input
1 2
3 4
5 4
3 5
Output
Demonstration
Codeforces Solution-E. Cyclic Components-Solution in C, C++, Java, Python