CF977E Cyclic Components(并查集判环)
题面：
You are given an undirected graph consisting of nn vertices and mm 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 aa is connected with a vertex bb, a vertex bb is also connected with a vertex aa). An edge can't connect vertex with itself, there is at most one edge between a pair of vertices.
Two vertices uu and vv belong to the same connected component if and only if there is at least one path along edges connecting uu and vv.
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 66 connected components, 22 of them are cycles: [7,10,16][7,10,16] and [5,11,9,15][5,11,9,15].
Input：
The first line contains two integer numbers nn and mm (1≤n≤2⋅1051≤n≤2⋅105, 0≤m≤2⋅1050≤m≤2⋅105) — number of vertices and edges.
The following mm lines contains edges: edge ii is given as a pair of vertices vivi, uiui (1≤vi,ui≤n1≤vi,ui≤n, ui≠viui≠vi). There is no multiple edges in the given graph, i.e. for each pair (vi,uivi,ui) there no other pairs (vi,uivi,ui) and (ui,viui,vi) in the list of edges.
Output：
Print one integer — the number of connected components which are also cycles.
Example:
Input
5 4 1 2 3 4 5 4 3 5
Output
1
Input
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
2
Note：
In the first example only component [3,4,5][3,4,5] is also a cycle.
The illustration above corresponds to the second example.
分析：
本题目现已加入太长不看系列，大意就是给你一些联通边，问你其中有几个环。我第一想到的就是并查集，但是这题被放在DFS专题里，DFS也是可以做的，但我还是选择了更熟练（不用动脑子）的并查集判环。
并查集判环大致就是在两个点进行unite操作时先判断一下父节点是否相同，不相同的话就正常把两个点放进一个集合，如果相同，说明这两个点已经处于一个集合中了，再把这两个点联通，也就出现了环。
AC代码：
#include
#include
#include
using namespace std;
const int maxn = 2E5 +5;
int n, m, ans;
int par[maxn], sum[maxn];
struct edge{
int u, v;
};
void init(int n){
for(int i = 0; i < n; i++) par[i] = i;
}
int find(int x){
return par[x] == x ? x : par[x] = find(par[x]);
}
void unite(int x, int y){
x = find(x);
y = find(y);
if(x != y) par[x] = y;
else ans++;
}
int main(){
while(cin>>n>>m){
memset(sum, 0, sizeof(sum));
init(n);
ans = 0;
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)
unite(e[i].u, e[i].v);
}
cout<