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| #include <bits/stdc++.h>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
using namespace std;
struct PairHash {
inline std::size_t operator()(const std::pair<int, int> &v) const { return v.first * 31 + v.second; }
};
#define Code ios_base::sync_with_stdio(false);
#define By ios::sync_with_stdio(0);
#define Sumfi cout.tie(NULL);
using ll = long long;
using ld = long double;
const ld PI = 3.14159265358979323846;
const ll INF = 1e18;
const ld EPS = 1e-9;
const ll MAX_N = 101010;
const ll mod = 1e9+7;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef array<ll,3> all3;
typedef array<ll,5> all5;
typedef vector<all3> vall3;
typedef vector<all5> vall5;
typedef vector<ld> vld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef unordered_set<ll> usll;
typedef unordered_set<pll, PairHash> uspll;
typedef unordered_map<ll, ll> umll;
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,m,n) for(ll i=n;i>=m;i--)
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
#define pyes cout<<"Yes\n";
#define pno cout<<"No\n";
#define pneg1 cout<<"-1\n";
template <typename T>
void print(T &&t) { cout << t << "\n"; }
template<typename T>
void printv(vector<T>v){ll n=v.size();rep(i,0,n)cout<<v[i]<<" ";cout<<"\n";}
ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }
ll __lcm(ll x, ll y) { return x * y / __gcd(x,y); }
using namespace std;
vll adj[MAX_N];
void dfs(ll u, ll par, ll root, vpll& res) {
if(u != root and adj[u].size() == 1) res.push_back({root, u});
else {
for(auto& v : adj[u]) {
if(v == par) continue;
dfs(v,u,root,res);
}
}
}
vpll solve(vpll& E, ll& n) {
vll count(MAX_N);
for(auto [u, v] : E) {
adj[u].push_back(v);
adj[v].push_back(u);
count[u]++;
count[v]++;
}
ll root = -1;
rep(i,1,n+1) {
if(root != -1 and count[i] > 2) return {};
if(count[i] > 2) root = i;
}
if(root == -1) {
rep(i,1,n+1) {
if(count[i] == 1) root = i;
}
}
vpll res;
dfs(root, -1, root, res);
return res;
}
int main() {
Code By Sumfi
cout.precision(12);
ll tc = 1;
for (ll i = 1; i <= tc; i++) {
ll n;
cin>>n;
vpll A(n-1);
rep(i,0,n-1) cin>>A[i].first>>A[i].second;
auto res = solve(A, n);
if(res.empty()) pno
else {
pyes
print(res.size());
for(auto [u, v] : res)
cout<<u<<' '<<v<<endl;
}
}
return 0;
}
|
