1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
| #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")
struct PairHash {
inline std::size_t operator()(const std::pair<int, int> &v) const { return v.first * 31 + v.second; }
};
#define MAX_N 1010101
#define INF 987654321
#define EPS 1e-9
#define ll long long
#define pll pair<ll, ll>
#define vpll vector<pll>
#define vall3 vector<array<ll,3>>
#define all3 array<ll,3>
#define all5 array<ll,5>
#define vall5 vector<all5>
#define vll vector<ll>
#define vi vector<int>
#define vs vector<string>
#define usll unordered_set<ll>
#define uspll unordered_set<pll, PairHash>
#define umll unordered_map<ll,ll>
#define vvs vector<vs>
#define vvll vector<vll>
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }
using namespace std;
void dfs(ll u, ll par, vpll& res, vvll& adj, ll& leaf) {
if(adj[u].size() == 1) {
if(par != -1) {
res[u] = {leaf, leaf};
leaf++;
} else {
dfs(adj[u][0], u, res, adj, leaf);
res[u] = res[adj[u][0]];
}
} else {
for(auto& v : adj[u]) {
if(v == par) continue;
dfs(v, u, res, adj, leaf);
res[u].first = min(res[u].first, res[v].first);
res[u].second = max(res[u].second, res[v].second);
}
}
}
vpll solve(vpll& A) {
ll n = A.size() + 1, leaf = 1;
vpll res(n, {INT_MAX, INT_MIN});
vvll adj(n);
for(auto& [u, v] : A) {
adj[u-1].push_back(v-1);
adj[v-1].push_back(u-1);
}
dfs(0,-1,res,adj, leaf);
return res;
}
int main() {
ios_base::sync_with_stdio(0);
ios::sync_with_stdio(0);
cin.tie(0);
ll tc = 1;
for (ll i = 1; i <= tc; i++) {
ll n;
cin>>n;
vpll E(n - 1);
for(ll j = 0; j < n - 1; j++) cin>>E[j].first>>E[j].second;
auto res = solve(E);
for(auto& [l, r] : res) cout<<l<<' '<<r<<endl;
}
return 0;
}
|
