1 //以一号节点为根节点，求出所有节点到根结点的距离，以及所有点的子节点的个数
2 //然后计算根据已知信息计算所有节点到当前结点的距离
3 //然后扫描n个点，O(n)求解
4 #include<bits/stdc++.h>
5 using namespace std;
6 const int maxn = 50086;
7 struct node {
8 int y, net;
9 }e[maxn << 1];
10 int f[maxn], h[maxn];//h表示以当前节点的根结点的子节点的个数，f表示所有子节点到当前节点的距离和
11 int n;
12 int lin[maxn], len = 0;
13 int id;
14
15 inline int read() {
16 int x = 0, y = 1;
17 char ch = getchar();
18 while(!isdigit(ch)) {
19 if(ch == '-') y = -1;
20 ch = getchar();
21 }
22 while(isdigit(ch)) {
23 x = (x << 1) + (x << 3) + ch - '0';
24 ch = getchar();
25 }
26 return x * y;
27 }
28
29 inline void insert(int xx, int yy) {
30 e[++len].y = yy;
31 e[len].net = lin[xx];
32 lin[xx] = len;
33 }
34
35 int son_num(int x, int fa) {
36 for(int i = lin[x]; i; i = e[i].net) {
37 int to = e[i].y;
38 if(to != fa) h[x] += son_num(to, x) + 1;
39 }
40 return h[x];
41 }
42
43 void everyson_to_one_dis(int x, int fa, int z) {
44 f[1] += z;
45 for(int i = lin[x]; i; i = e[i].net) {
46 int to = e[i].y;
47 if(to != fa) everyson_to_one_dis(to, x, z + 1);
48 }
49 }
50
51 void everyone_to_x_dis(int x, int fa) {
52 f[x] = f[fa] - (h[x] + 1) + (n - h[x] - 1);
53 for(int i = lin[x]; i; i = e[i].net) {
54 int to = e[i].y;
55 if(to != fa) everyone_to_x_dis(to, x);
56 }
57 }
58
59 int main() {
60 memset(lin, 0, sizeof(lin));
61 n = read();
62 for(int i = 1; i < n; ++i) {
63 int x, y;
64 x = read(), y = read();
65 insert(x, y);
66 insert(y, x);
67 }
68 son_num(1, 0);
69 for(int i = lin[1]; i; i = e[i].net)
70 everyson_to_one_dis(e[i].y, 1, 1);
71 for(int i = lin[1]; i; i = e[i].net)
72 everyone_to_x_dis(e[i].y, 1);
73 id = 1;
74 for(int i = 2; i <= n; ++i)
75 if(f[id] > f[i]) id = i;
76 cout << id << ' ' << f[id] << '\n';
77 return 0;
78 }
