DP on Trees - Solving For All Roots

Solution - Tree Distances I

It is a common technique to calculate two DP arrays for some DP on trees problems. Usually one DP array is responsible for calculating results within the subtree rooted at $i$. The other DP array calculates results outside of the subtree rooted at $i$.

The focus problem asks us to find for each node the maximum distance to another node. We can divide the problem into two parts.

Define $f[x]$ as the maximum distance from node $x$ to any node in the subtree rooted at $x$, and $g[x]$ as the maximum distance from node $x$ to any node outside of the subtree rooted at $x$. Then the answer for node $x$ is $\max(f[x],g[x])$.

• $f[x]$ can be calculated using a DFS since $f[x]=\max(f[c])+1$, where $c$ is a child of $x$.
• $g[x]$ can also be calculated using a DFS as $g[c]=\max(g[x]+1, f[d]+2)$, where $c$ and $d$ are both children of $x$ with $c \neq d$.

To calculate $g$ in linear time, we can define another array $h$ such that $h[x]$ is the largest distance from node $x$ to any node in the subtree rooted at $x$ excluding the child subtree that contributed to $f[x]$. So if $f[x]$ is transitioned from the branch with $c$, $g[c]=\max(g[x]+1,h[x]+1)$. Otherwise $g[c]=\max(g[x]+1,f[x]+1)$.

Copy
#include <bits/stdc++.h>
using namespace std;

vector<int> graph[200001];
int fir[200001], sec[200001], ans[200001];

void dfs1(int node = 1, int parent = 0) {
	for (int i : graph[node]) if (i != parent) {
		dfs1(i, node);
		if (fir[i] + 1 > fir[node]) {

Copy
import java.util.*;
import java.io.*;

public class Main {
	public static ArrayList <Integer> g[];
	public static Pair maxl1[];
	public static Pair maxl2[];
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int N = Integer.parseInt(br.readLine());

Problems