1021 deep root (25 points)

1. Use a random dot x Conduct dfs,dfs Record the distance in x Farthest point set
2. Take a random point from the set , Conduct dfs, Get the farthest point set again
3. The intersection of two sets + Is the farthest node

Liunuo ：
The main idea of the topic ： give n Nodes （1~n） Between n side , Ask if it can form a tree , If it cannot be constructed, the number of connected components it has is output , If it can form a tree , Output the height of the deepest tree , Root node of tree . If there are more than one , Output from small to large .
analysis ： First, the depth first search determines how many connected components it has . If there are more than one , Then output Error: x components, If only one , Just two deep first searches , Start with a node dfs Keep the nodes with the highest height , Then start at any of these nodes dfs Get the nodes with the highest height , The union of these two sets of nodes is the result

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <set>
using namespace std;
const int n = 1e5 + 10;
int N;
vector<int> v[n];
bool dp[n];
// set<int> ans;
set<int> ans;
vector<int> temp;
int maxt;
void dfs(int k, int l)
{
    
    dp[k] = 1;
    if (maxt < l)
    {
    
        maxt = l;
        temp.clear();
        temp.push_back(k);
    }
    if (maxt == l)
        temp.push_back(k);
    for (auto &e : v[k])
        if (!dp[e])
            dfs(e, l + 1);
}
int main()
{
    
    cin >> N;
    int x, y;
    for (int i = 1; i < N; i++)
    {
    
        cin >> x >> y;
        v[x].push_back(y);
        v[y].push_back(x);
    }
    int cnt = 0;
    for (int i = 1; i <= N; i++)
    {
    
        if (!dp[i])
        {
    
            cnt++;
            dfs(i, 1);
            if (i == 1)
                for (auto &e : temp)
                    ans.insert(e);
        }
    }

    if (cnt != 1)
        printf("Error: %d components", cnt);
    else
    {
    
        temp.clear();
        maxt = 0;
        memset(dp, 0, sizeof dp);
        dfs(*ans.begin(), 1);
        for (auto &e : temp) 
            ans.insert(e);
        for (auto &e : ans)
            cout << e << endl;
    }

    return 0;
}