Tree Coordinates
Problem Statement :
We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For example, , where is a metric space. , where . In this challenge, we need a tree-space. You're given a tree, , where is the set of vertices and is the set of edges. Let the function be the distance between two vertices in tree (i.e., is the number of edges on the path between vertices and ). Note that is a metric space. You are given a tree, , with vertices, as well as points in . Find and print the distance between the two furthest points in this metric space! Input Format The first line contains two space-separated positive integers describing the respective values of (the number of vertices in ) and (the number of given points). Each line of the subsequent lines contains two space-separated integers, and , describing edge in . Each line of the subsequent lines contains two space-separated integers describing the respective values of and for point . Constraints Scoring This challenge uses binary scoring, so you must pass all test cases to earn a positive score. Output Format Print a single non-negative integer denoting the maximum distance between two of the given points in metric space .
Solution :
Solution in C :
In C ++ :
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <ctime>
#include <deque>
using namespace std;
#define N 200000
int fa[N][21], dep[N], L[N], R[N], Time, n, m;
vector <int> ve[N], ve1[N];
bool used[N];
int zx, sz;
int sum[N], l[N], c[N];
int ans;
int ma1[N], ma2[N], c1[N], c2[N];
int zhan[N], zhan1[N];
void dfs1(int k, int f) {
fa[k][0] = f;
dep[k] = dep[f] + 1;
for (int i = 1; i <= 20; i++)
fa[k][i] = fa[fa[k][i - 1]][i - 1];
L[k] = ++Time;
for (int i = 0; i < (int) ve[k].size(); i++)
if (ve[k][i] != f)
dfs1(ve[k][i], k);
R[k] = Time;
}
void dfs_size(int k) {
sum[k] = 1;
used[k] = true;
for (int i = 0; i < (int) ve[k].size(); i++)
if (!used[ve[k][i]]) {
dfs_size(ve[k][i]);
sum[k] += sum[ve[k][i]];
}
used[k] = false;
}
void dfs_zx(int k) {
// printf("?? %d %d\n", k, sz);
bool ok = (sz - sum[k] <= sz / 2);
used[k] = true;
for (int i = 0; i < (int) ve[k].size(); i++)
if (!used[ve[k][i]]) {
if (sum[ve[k][i]] > sz / 2)
ok = false;
dfs_zx(ve[k][i]);
}
used[k] = false;
if (ok)
zx = k;
}
void dfs_l(int k) {
used[k] = true;
for (int i = 0; i < (int) ve[k].size(); i++)
if (!used[ve[k][i]]) {
l[ve[k][i]] = l[k] + 1;
dfs_l(ve[k][i]);
}
used[k] = false;
}
void dfs_c(int k) {
used[k] = true;
for (int i = 0; i < (int) ve[k].size(); i++)
if (!used[ve[k][i]]) {
c[ve[k][i]] = c[k];
dfs_c(ve[k][i]);
}
used[k] = false;
}
bool cmp_c(pair<int, int> x, pair<int, int> y) {
return c[x.first] < c[y.first];
}
bool cmp_L(int x, int y) {
return L[x] < L[y];
}
int dist(int x, int y) {
int ans = 0;
if (dep[x] < dep[y])
swap(x, y);
for (int i = 20; i >= 0; i--)
if (dep[fa[x][i]] >= dep[y])
ans += (1 << i), x = fa[x][i];
if (x == y)
return ans;
for (int i = 20; i >= 0; i--)
if (fa[x][i] != fa[y][i])
ans += 1 << (i + 1), x = fa[x][i], y = fa[y][i];
return ans + 2;
}
int lca(int x, int y) {
int ans = 0;
if (dep[x] < dep[y])
swap(x, y);
for (int i = 20; i >= 0; i--)
if (dep[fa[x][i]] >= dep[y])
ans += (1 << i), x = fa[x][i];
if (x == y)
return x;
for (int i = 20; i >= 0; i--)
if (fa[x][i] != fa[y][i])
ans += 1 << (i + 1), x = fa[x][i], y = fa[y][i];
return fa[x][0];
}
void dfs_d(int k) {
// ma1[k] = -1000000000;
// ma2[k] = -1000000000;
// ma1[k] = v[k];
// c1[k] = c[k];
// printf("?? %d\n", k);
for (int i = 0; i < (int) ve1[k].size(); i++) {
int t = ve1[k][i];
dfs_d(t);
ma1[t] += dep[t] - dep[k];
ma2[t] += dep[t] - dep[k];
if (c1[k] != c1[t] || !c1[k] || !c1[t])
ans = max(ans, ma1[k] + ma1[t]);
if (c1[k] != c2[t] || !c1[k] || !c2[t])
ans = max(ans, ma1[k] + ma2[t]);
if (c2[k] != c1[t] || !c2[k] || !c1[t])
ans = max(ans, ma2[k] + ma1[t]);
if (c2[k] != c2[t] || !c2[k] || !c2[t])
ans = max(ans, ma2[k] + ma2[t]);
if (ma1[t] > ma1[k]) {
if (c1[t] == c1[k])
ma1[k] = ma1[t];
else {
ma2[k] = ma1[k];
c2[k] = c1[k];
ma1[k] = ma1[t];
c1[k] = c1[t];
}
}else if (ma1[t] > ma2[k] && c1[t] != c1[k]) {
ma2[k] = ma1[t];
c2[k] = c1[t];
}
if (ma2[t] > ma1[k]) {
if (c2[t] == c1[k])
ma1[k] = ma2[t];
else {
ma2[k] = ma1[k];
c2[k] = c1[k];
ma1[k] = ma2[t];
c1[k] = c2[t];
}
}else if (ma2[t] > ma2[k] && c2[t] != c1[k]) {
ma2[k] = ma2[t];
c2[k] = c2[t];
}
}
}
void doit(vector <pair<int, int> > V) {
int len = 0;
// printf("%d\n", ans);
for (int i = 0; i < (int) V.size(); i++) {
zhan[++len] = V[i].second;
if (c[V[i].first] != c1[V[i].second] || !c[V[i].first] || !c1[V[i].second]) {
ans = max(ans, ma1[V[i].second] + l[V[i].first]);
}
if (c[V[i].first] != c2[V[i].second] || !c[V[i].first] || !c2[V[i].second]) {
ans = max(ans, ma2[V[i].second] + l[V[i].first]);
}
if (l[V[i].first] > ma1[V[i].second]) {
if (c[V[i].first] != c1[V[i].second]) {
ma2[V[i].second] = ma1[V[i].second];
c2[V[i].second] = c1[V[i].second];
ma1[V[i].second] = l[V[i].first];
c1[V[i].second] = c[V[i].first];
}else {
ma1[V[i].second] = l[V[i].first];
}
}else if (l[V[i].first] > ma2[V[i].second] && c[V[i].first] != c1[V[i].second]) {
ma2[V[i].second] = l[V[i].first];
c2[V[i].second] = c[V[i].first];
}
}
// printf("%d\n", ans);
sort(zhan + 1, zhan + len + 1, cmp_L);
for (int i = len; i > 1; i--) {
zhan[++len] = lca(zhan[i - 1], zhan[i]);
}
sort(zhan + 1, zhan + len + 1, cmp_L);
for (int i = 1; i < len; i++) {
ve1[zhan[i]].clear();
}
// for (int i = 1; i <= len; i++)
// printf("%d ", zhan[i]);
// printf("\n");
int len1 = 0;
zhan1[len1 = 1] = zhan[1];
for (int i = 2; i <= len; i++)
if (zhan[i] != zhan[i - 1]) {
while (len1 > 0 && R[zhan1[len1]] < L[zhan[i]])
len1 --;
zhan1[++len1] = zhan[i];
ve1[zhan1[len1 - 1]].push_back(zhan1[len1]);
}
dfs_d(zhan[1]);
// for (int i = 0; i < (int) V.size(); i++)
// for (int j = 0; j < i; j++)
// if (c[V[i].first] != c[V[j].first] || !c[V[i].first] || !c[V[j].first]) {
// // printf("xx %d %d\n", l[V[i].first] + l[V[j].first], dist(V[i].second, V[j].second));
// ans = max(ans, l[V[i].first] + l[V[j].first] + dist(V[i].second, V[j].second));
// }
for (int i = 1; i <= len; i++) {
// v[V[i].second] = -1000000000;
ma1[zhan[i]] = ma2[zhan[i]] = -1000000000;
}
}
void doit1(vector <pair<int, int> > V) {
for (int i = 0; i < (int) V.size(); i++)
for (int j = 0; j < i; j++)
// if (c[V[i].first] != c[V[j].first] || !c[V[i].first] || !c[V[j].first]) {
// printf("xx %d %d\n", l[V[i].first] + l[V[j].first], dist(V[i].second, V[j].second));
ans = max(ans, dist(V[i].first, V[j].first) + dist(V[i].second, V[j].second));
// }
}
void dfs(int k, vector <pair<int, int> > V) {
// printf("?? %d\n", k);
if (V.size() <= 1)
return ;
if (V.size() <= 100) {
doit1(V);
return ;
}
dfs_size(k);
sz = sum[k];
dfs_zx(k);
k = zx;
l[k] = 0;
dfs_l(k);
c[k] = 0;
used[k] = true;
int tot = 0;
for (int i = 0; i < (int) ve[k].size(); i++)
if (!used[ve[k][i]]) {
tot += 1;
c[ve[k][i]] = tot;
dfs_c(ve[k][i]);
}
used[k] = false;
doit(V);
sort(V.begin(), V.end(), cmp_c);
int q = 0;
used[k] = true;
vector <int> vv;
for (int i = 0; i <= tot; i++) {
while (q < (int) V.size() && c[V[q].first] <= i)
q += 1;
vv.push_back(q);
}
tot = 0;
for (int i = 0; i < (int) ve[k].size(); i++)
if (!used[ve[k][i]]) {
tot += 1;
vector <pair<int, int> > V1;
// printf("?? %d %d %u\n", vv[tot - 1], vv[tot], V.size());
for (int j = vv[tot - 1]; j < vv[tot]; j++)
V1.push_back(V[j]);
dfs(ve[k][i], V1);
}
used[k] = false;
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i < n; i++) {
int x, y;
scanf("%d%d", &x, &y);
// x = i;
// y = i + 1;
ve[x].push_back(y);
ve[y].push_back(x);
}
vector <pair<int, int> > V;
for (int i = 1; i <= m; i++) {
int x, y;
scanf("%d%d", &x, &y);
V.push_back(make_pair(x, y));
}
dfs1(1, 0);
for (int i = 1; i <= n; i++)
ma1[i] = ma2[i] = -1000000000;
dfs(1, V);
printf("%d\n", ans);
}
In Java :
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;
public class Solution {
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
static int[][] buildSparseTable(int[] arr) {
int pow = 1;
while ((1 << pow) < arr.length) pow++;
int[][] result = new int[arr.length][pow];
for (int i = 0; i < arr.length; i++) result[i][0] = arr[i];
for (int j = 1; j <= pow; j++) {
for (int i = 0; i + (1 << j) <= arr.length; i++) {
result[i][j] = Math.min(result[i][j-1],
result[i + (1 << (j-1))][j-1]);
}
}
return result;
}
/*
* Complete the treeCoordinates function below.
*/
static int treeCoordinates(int n, int[][] edges, int[][] points) {
ArrayList<Integer>[] nodes = new ArrayList[n + 1];
for (int i = 1; i <= n; i++) nodes[i] = new ArrayList<Integer>();
for (int[] edge : edges) {
nodes[edge[0]].add(edge[1]);
nodes[edge[1]].add(edge[0]);
}
// Find diameter (two BFS)
int root = 0;
int tail = 0;
{
class Entry {
int node;
int dist;
public Entry(int node, int dist) {
this.node = node;
this.dist = dist;
}
}
LinkedList<Entry> Q = new LinkedList<Entry>();
boolean[] visited = new boolean[n + 1];
visited[1] = true;
Q.offer(new Entry(1, 0));
int maxDist = 0;
int farNode = 1;
while (Q.size() > 0) {
Entry cur = Q.poll();
if (cur.dist > maxDist) {
maxDist = cur.dist;
farNode = cur.node;
}
for (int neighbor : nodes[cur.node]) {
if (visited[neighbor]) continue;
visited[neighbor] = true;
Q.offer(new Entry(neighbor, cur.dist + 1));
}
}
root = farNode;
Q = new LinkedList<Entry>();
visited = new boolean[n + 1];
visited[root] = true;
Q.offer(new Entry(root, 0));
maxDist = 0;
farNode = root;
while (Q.size() > 0) {
Entry cur = Q.poll();
if (cur.dist > maxDist) {
maxDist = cur.dist;
farNode = cur.node;
}
for (int neighbor : nodes[cur.node]) {
if (visited[neighbor]) continue;
visited[neighbor] = true;
Q.offer(new Entry(neighbor, cur.dist + 1));
}
}
tail = farNode;
}
//System.out.println("root = " + root + ", tail = " + tail);
// Euler tour
int[] eulerTour = new int[2*n - 1];
final int[] depth = new int[n + 1];
int[] eulerLevels = new int[2*n - 1];
int[] eulerIndex = new int[n + 1];
boolean[] visited = new boolean[n + 1];
int[] S = new int[n];
int spos = 0;
S[0] = root;
int pos = 0;
int[] neighborCount = new int[n + 1];
while (spos > -1) {
int cur = S[spos--];
if (!visited[cur]) {
depth[cur] = spos + 1;
eulerIndex[cur] = pos;
visited[cur] = true;
}
eulerLevels[pos] = spos + 1;
eulerTour[pos] = cur;
pos++;
while (neighborCount[cur] < nodes[cur].size()) {
if (visited[nodes[cur].get(neighborCount[cur])]) {
neighborCount[cur]++;
continue;
}
int next = nodes[cur].get(neighborCount[cur]);
//parent[next] = cur;
S[++spos] = cur;
S[++spos] = next;
neighborCount[cur]++;
break;
}
}
//System.out.println(Arrays.toString(eulerTour));
//System.out.println(Arrays.toString(eulerLevels));
//System.out.println(Arrays.toString(eulerIndex));
// Sparse table
int[][] lookup = buildSparseTable(eulerLevels);
//for (int i = 0; i < lookup.length; i++) {
// for (int j = 0; j < lookup[0].length; j++) {
// System.out.print(lookup[i][j] + " ");
// }
// System.out.println();
//}
class Entry implements Comparable<Entry> {
int x;
int y;
int val;
public Entry(int x, int y, int val) {
this.x = x;
this.y = y;
this.val = val;
}
public int compareTo(Entry other) {
return val - other.val;
}
}
List<Entry> list1 = new ArrayList<Entry>(points.length);
List<Entry> list2 = new ArrayList<Entry>(points.length);
List<Entry> list3 = new ArrayList<Entry>(points.length);
List<Entry> list4 = new ArrayList<Entry>(points.length);
for (int i = 0; i < points.length; i++) {
int x = points[i][0];
int y = points[i][1];
int xLcaLevel;
{
int start = Math.min(eulerIndex[x], eulerIndex[tail]);
int end = Math.max(eulerIndex[x], eulerIndex[tail]);
int pow = 0;
while (1 << (pow + 1) <= (end - start)) pow++;
xLcaLevel = Math.min(lookup[start][pow],
lookup[end + 1 - (1<<pow)][pow]);
}
int yLcaLevel;
{
int start = Math.min(eulerIndex[y], eulerIndex[tail]);
int end = Math.max(eulerIndex[y], eulerIndex[tail]);
int pow = 0;
while (1 << (pow + 1) <= (end - start)) pow++;
yLcaLevel = Math.min(lookup[start][pow],
lookup[end + 1 - (1<<pow)][pow]);
}
int val1 = depth[x] + depth[y];
list1.add(new Entry(x, y, val1));
int val2 = -depth[x] - depth[y] + 2*xLcaLevel + 2*yLcaLevel;
list2.add(new Entry(x, y, val2));
int val3 = depth[x] + depth[y] - 2*xLcaLevel;
list3.add(new Entry(x, y, val3));
int val4 = -depth[x] - depth[y] + 2*yLcaLevel;
list4.add(new Entry(x, y, val4));
}
Collections.sort(list1, Collections.reverseOrder());
Collections.sort(list2);
Collections.sort(list3, Collections.reverseOrder());
Collections.sort(list4);
//int maxDist = Math.max(list1.get(0).val - list2.get(0).val, list3.get(0).val - list4.get(0).val);
int maxDist = 0;
for (int i = 0; i < points.length; i++) {
// ith increasing diagonal
boolean shouldContinue = false;
for (int j = 0; j <= i; j++) {
Entry e1 = list1.get(i-j);
Entry e2 = list2.get(j);
int potential12 = e1.val - e2.val;
if (potential12 > maxDist) {
shouldContinue = true;
int x1 = e1.x;
int y1 = e1.y;
int x2 = e2.x;
int y2 = e2.y;
int xLcaLevel;
{
int start = Math.min(eulerIndex[x1], eulerIndex[x2]);
int end = Math.max(eulerIndex[x1], eulerIndex[x2]);
int pow = 0;
while (1 << (pow + 1) <= (end - start)) pow++;
xLcaLevel = Math.min(lookup[start][pow],
lookup[end + 1 - (1<<pow)][pow]);
}
int yLcaLevel;
{
int start = Math.min(eulerIndex[y1], eulerIndex[y2]);
int end = Math.max(eulerIndex[y1], eulerIndex[y2]);
int pow = 0;
while (1 << (pow + 1) <= (end - start)) pow++;
yLcaLevel = Math.min(lookup[start][pow],
lookup[end + 1 - (1<<pow)][pow]);
}
int actual12 = depth[x1] + depth[x2] - 2*xLcaLevel
+ depth[y1] + depth[y2] - 2*yLcaLevel;
maxDist = Math.max(maxDist, actual12);
}
Entry e3 = list3.get(i-j);
Entry e4 = list4.get(j);
int potential34 = e3.val - e4.val;
if (potential34 > maxDist) {
shouldContinue = true;
int x3 = e3.x;
int y3 = e3.y;
int x4 = e4.x;
int y4 = e4.y;
int xLcaLevel;
{
int start = Math.min(eulerIndex[x3], eulerIndex[x4]);
int end = Math.max(eulerIndex[x3], eulerIndex[x4]);
int pow = 0;
while (1 << (pow + 1) <= (end - start)) pow++;
xLcaLevel = Math.min(lookup[start][pow],
lookup[end + 1 - (1<<pow)][pow]);
}
int yLcaLevel;
{
int start = Math.min(eulerIndex[y3], eulerIndex[y4]);
int end = Math.max(eulerIndex[y3], eulerIndex[y4]);
int pow = 0;
while (1 << (pow + 1) <= (end - start)) pow++;
yLcaLevel = Math.min(lookup[start][pow],
lookup[end + 1 - (1<<pow)][pow]);
}
int actual34 = depth[x3] + depth[x4] - 2*xLcaLevel
+ depth[y3] + depth[y4] - 2*yLcaLevel;
maxDist = Math.max(maxDist, actual34);
}
}
if (!shouldContinue) break;
}
return maxDist;
}
public static void main(String[] args) throws IOException {
BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));
Reader reader = new Reader();
int n = reader.nextInt();
int m = reader.nextInt();
int[][] edges = new int[n-1][2];
for (int edgesRowItr = 0; edgesRowItr < n-1; edgesRowItr++) {
edges[edgesRowItr][0] = reader.nextInt();
edges[edgesRowItr][1] = reader.nextInt();
}
int[][] points = new int[m][2];
for (int pointsRowItr = 0; pointsRowItr < m; pointsRowItr++) {
points[pointsRowItr][0] = reader.nextInt();
points[pointsRowItr][1] = reader.nextInt();
}
int result = treeCoordinates(n, edges, points);
bufferedWriter.write(String.valueOf(result));
bufferedWriter.newLine();
bufferedWriter.close();
}
}
In Python3 :
#!/bin/python3
import sys
n,m = input().strip().split(' ')
n,m = [int(n),int(m)]
edges = []
for edges_i in range(n-1):
edges_t = [int(edges_temp) for edges_temp in input().strip().split(' ')]
edges.append(edges_t)
points = []
for points_i in range(m):
points_t = [int(points_temp) for points_temp in input().strip().split(' ')]
points.append(points_t)
# your code goes here
class Node:
def __init__(self,i):
self.id = i
self.neighbors = set()
nodes = [Node(i) for i in range(n)]
for from_id,to_id in edges:
nodes[from_id-1].neighbors.add(nodes[to_id-1])
nodes[to_id-1].neighbors.add(nodes[from_id-1])
distances = [[10**5] * n for _ in range(n)]
for i in range(n):
distances[i][i] = 0
def set_distance(i, j, new_dist):
distances[i][j] = new_dist
distances[j][i] = new_dist
# find pair-wise distances with bfs <-- O(n^2)
for center_node in nodes:
cur_dist = 1
marked = set([center_node])
cur_level = set([center_node])
while cur_level:
next_level = set()
for node in cur_level:
for neighbor in node.neighbors:
if not neighbor in marked:
marked.add(neighbor)
next_level.add(neighbor)
set_distance(center_node.id, neighbor.id, cur_dist)
cur_level = next_level
cur_dist += 1
# now go through the points
max_dist = 0
for p1 in range(m):
x1, y1 = points[p1]
x1 -= 1
y1 -= 1
for p2 in range(p1 + 1, m):
x2, y2 = points[p2]
x2 -= 1
y2 -= 1
new_dist = distances[x1][x2] + distances[y1][y2]
if new_dist > max_dist:
max_dist = new_dist
print(max_dist)
View More Similar Problems
Pair Sums
Given an array, we define its value to be the value obtained by following these instructions: Write down all pairs of numbers from this array. Compute the product of each pair. Find the sum of all the products. For example, for a given array, for a given array [7,2 ,-1 ,2 ] Note that ( 7 , 2 ) is listed twice, one for each occurrence of 2. Given an array of integers, find the largest v
View Solution →Lazy White Falcon
White Falcon just solved the data structure problem below using heavy-light decomposition. Can you help her find a new solution that doesn't require implementing any fancy techniques? There are 2 types of query operations that can be performed on a tree: 1 u x: Assign x as the value of node u. 2 u v: Print the sum of the node values in the unique path from node u to node v. Given a tree wi
View Solution →Ticket to Ride
Simon received the board game Ticket to Ride as a birthday present. After playing it with his friends, he decides to come up with a strategy for the game. There are n cities on the map and n - 1 road plans. Each road plan consists of the following: Two cities which can be directly connected by a road. The length of the proposed road. The entire road plan is designed in such a way that if o
View Solution →Heavy Light White Falcon
Our lazy white falcon finally decided to learn heavy-light decomposition. Her teacher gave an assignment for her to practice this new technique. Please help her by solving this problem. You are given a tree with N nodes and each node's value is initially 0. The problem asks you to operate the following two types of queries: "1 u x" assign x to the value of the node . "2 u v" print the maxim
View Solution →Number Game on a Tree
Andy and Lily love playing games with numbers and trees. Today they have a tree consisting of n nodes and n -1 edges. Each edge i has an integer weight, wi. Before the game starts, Andy chooses an unordered pair of distinct nodes, ( u , v ), and uses all the edge weights present on the unique path from node u to node v to construct a list of numbers. For example, in the diagram below, Andy
View Solution →Heavy Light 2 White Falcon
White Falcon was amazed by what she can do with heavy-light decomposition on trees. As a resut, she wants to improve her expertise on heavy-light decomposition. Her teacher gave her an another assignment which requires path updates. As always, White Falcon needs your help with the assignment. You are given a tree with N nodes and each node's value Vi is initially 0. Let's denote the path fr
View Solution →