给定n个结点m条边的简单无向图，判断该图是否存在鱼形状的子图：有一个环，其中有一个结点有另外两条边，连向不在环内的两个结点。若有，输出子图的连边

题目

思路：

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define fi first
#define se second
#define lson p << 1
#define rson p << 1 | 1
const int maxn = 1e6 + 5, inf = 1e18 * 3, maxm = 4e4 + 5, mod = 1e9 + 7;
int a[maxn], b[maxn];
int n, m;
string s;
bool vis[maxn];
vector<int> G[maxn];
int from, to;
vector<int> tmp, vec;
int deg[maxn];void dfs(int u, int p){if(u == to && p == from) return;if(vis[u]) return;vis[u] = 1;tmp.pb(u);if(u == to){vec = tmp;return;}for(auto v : G[u]){dfs(v, u);}tmp.pop_back();
}
void solve(){int res = 0;int k, x, q;cin >> n >> m;for(int i = 1; i <= n; i++){G[i].clear();deg[i] = 0;}for(int i = 1; i <= m; i++){int u, v;cin >> u >> v;G[u].pb(v);G[v].pb(u);deg[u]++;deg[v]++;}for(int u = 1; u <= n; u++){if(deg[u] < 4) continue;for(auto v : G[u]){for(int i = 1; i <= n; i++){vis[i] = 0;}from = u;to = v;vec.clear();tmp.clear();dfs(u, u);if(vec.empty()) continue;vector<int> extra;tmp.clear();for(auto x : G[u]){if(x == vec.back() || x == *(vec.begin() + 1)) continue;if(find(vec.begin(), vec.end(), x) == vec.end()){//不在环内extra.pb(x);if(extra.size() == 2) break;}else{tmp.pb(x);//在环内,且与u相连}}vector<int> vt;if(extra.size() < 2){for(auto x : vec){vt.pb(x);if(find(tmp.begin(), tmp.end(), x) != tmp.end()){//在tmp内break;}}extra.pb(vec.back());for(int i = vec.size() - 1; i >= 0; i--){int x = vec[i];if(find(tmp.begin(), tmp.end(), x) != tmp.end()){extra.pb(x);break;}}// extra.pb(vec.end() - 2) 不能这样写，因为这个结点不一定与u相连// extra.pb(tmp.back()); 不能这样写，因为tmp的顺序跟环的结点顺序不一致extra.resize(2);vec = vt;}cout << "Yes\n";cout << vec.size() + 2 << '\n';int m = vec.size();for(int i = 0; i < vec.size(); i++){cout << vec[i] << ' ' << vec[(i + 1) % m] << '\n';}cout << u << ' ' << extra[0] << '\n';cout << u << ' ' << extra[1] << '\n';return;}}cout << "No\n";
}signed main(){ios::sync_with_stdio(0);cin.tie(0);int T = 1;cin >> T;while (T--){solve();}return 0;
}