原题传送门
求环的个数
直接dfs处理出每个连通块
然后如果一个连通块的所有点的度数都为2,那么是一个环
Code:
#include <bits/stdc++.h>
#define maxn 200010
using namespace std;
struct Edge{
int to, next;
}edge[maxn << 1];
int num, head[maxn], n, m, d[maxn], vis[maxn], ans, cnt;
vector <int> loop[maxn];
inline int read(){
int s = 0, w = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
return s * w;
}
void addedge(int x, int y){ edge[++num] = (Edge){y, head[x]}; head[x] = num; }
void dfs(int u){
loop[cnt].push_back(u); vis[u] = 1;
for (int i = head[u]; i; i = edge[i].next){
int v = edge[i].to;
if (!vis[v]) dfs(v);
}
}
int main(){
n = read(), m = read();
for (int i = 1; i <= m; ++i){
int x = read(), y = read();
addedge(x, y), addedge(y, x);
++d[x], ++d[y];
}
for (int i = 1; i <= n; ++i)
if (!vis[i]) ++cnt, dfs(i);
for (int i = 1; i <= cnt; ++i){
int flag = 1;
for (int j = 0; j < loop[i].size(); ++j)
if (d[loop[i][j]] != 2){ flag = 0; break; }
ans += flag;
}
printf("%d\n", ans);
return 0;
}