[CF1294F] Three Paths on a Tree - 树的直径

Description

给定一棵含 \(n\ (3\leq n\leq2\cdot 10^5)\) 个结点的无权树，试找出三个结点 \(u\)、\(v\)、\(w\)，\(\operatorname{s.t.}\) $$\operatorname{card}({u,v\text{ 间的路径}}\cup{v,w\text{ 间的路径}}\cup{w,u\text{ 间的路径}})$$ 最大。输出方案。

Solution

直径以及距离直径的最远点（反证法）

两次 DFS + BFS

#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;

int n;
vector<int> g[N];
int d[N], v[N];

void dfs(int p, int from)
{
    for (int q : g[p])
        if (q != from)
        {
            d[q] = d[p] + 1;
            dfs(q, p);
        }
}

vector<int> path;

bool find_path(int p, int from, int tar)
{
    path.push_back(p);
    if (tar == p)
        return true;
    for (int q : g[p])
        if (q != from)
        {
            if (find_path(q, p, tar))
                return true;
        }
    path.pop_back();
    return false;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    for (int i = 1; i < n; i++)
    {
        int x, y;
        cin >> x >> y;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    dfs(1, 0);
    int p = 0;
    for (int i = 1; i <= n; i++)
        if (d[i] > d[p])
            p = i;
    d[p] = 0;
    dfs(p, 0);
    int q = 0;
    for (int i = 1; i <= n; i++)
        if (d[i] > d[q])
            q = i;

    int len1 = d[q];

    find_path(p, 0, q);

    queue<int> que;
    for (int i = 1; i <= n; i++)
        d[q] = n + 1;
    for (int i : path)
        que.push(i), d[i] = 0, v[i] = 1;

    int last = 0;
    while (que.size())
    {
        int now = que.front();
        que.pop();
        if (now != p && now != q)
            last = now;
        for (int q : g[now])
        {
            if (d[q] > d[now] + 1)
            {
                d[q] = d[now] + 1;
                if (!v[q])
                    v[q] = 1, que.push(q);
            }
        }
    }

    int len2 = d[last];

    cout << len1 + len2 << endl;
    cout << p << " " << q << " " << last << endl;
}