luogu P1359会议

//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 
//然后计算根据已知信息计算所有节点到当前结点的距离 
//然后扫描n个点，O(n)求解 
#include<bits/stdc++.h>
using namespace std;
const int maxn = 50086;
struct node {
    int y, net;
}e[maxn << 1];
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 
int n;
int lin[maxn], len = 0;
int id;

inline int read() {
    int x = 0, y = 1;
    char ch = getchar();
    while(!isdigit(ch)) {
        if(ch == '-') y = -1;
        ch = getchar();
    }
    while(isdigit(ch)) {
        x = (x << 1) + (x << 3) + ch - '0';
        ch = getchar();
    }
    return x * y;
}

inline void insert(int xx, int yy) {
    e[++len].y = yy;
    e[len].net = lin[xx];
    lin[xx] = len;
}

int son_num(int x, int fa) {
    for(int i = lin[x]; i; i = e[i].net) {
        int to = e[i].y;
        if(to != fa) h[x] += son_num(to, x) + 1;
    } 
    return h[x];
}

void everyson_to_one_dis(int x, int fa, int z) {
    f[1] += z;
    for(int i = lin[x]; i; i = e[i].net) {
        int to = e[i].y;
        if(to != fa) everyson_to_one_dis(to, x, z + 1);
    } 
}

void everyone_to_x_dis(int x, int fa) {
    f[x] = f[fa] - (h[x] + 1) + (n - h[x] - 1);
    for(int i = lin[x]; i; i = e[i].net) {
        int to = e[i].y;
        if(to != fa) everyone_to_x_dis(to, x);
    }
}

int main() {
    memset(lin, 0, sizeof(lin));
    n = read();
    for(int i = 1; i < n; ++i) {
        int x, y;
        x = read(), y = read();
        insert(x, y);
        insert(y, x);
    }
    son_num(1, 0);
    for(int i = lin[1]; i; i = e[i].net) 
        everyson_to_one_dis(e[i].y, 1, 1);
    for(int i = lin[1]; i; i = e[i].net)
        everyone_to_x_dis(e[i].y, 1);
    id = 1;
    for(int i = 2; i <= n; ++i)
        if(f[id] > f[i]) id = i;
    cout << id << ' ' << f[id] << '\n';
    return 0;
}