# Super Maximum Cost Queries

### Problem Statement :

```Victoria has a tree, T , consisting of N nodes numbered from 1 to N. Each edge from node Ui to Vi  in tree T has an integer weight, Wi.

Let's define the cost, C, of a path from some node X to some other node Y as the maximum weight ( W ) for any edge in the unique path from node X to Y node .

Victoria wants your help processing Q queries on tree T, where each query contains 2 integers, L and R, such that L <= R . For each query, she wants to print the number of different paths in T  that have a cost, C , in the inclusive range [ L , R ] .

It should be noted that path from some node X to some other node Y is considered same as path from node Y to X  i.e  { X, Y }is same as { Y, X } .

Input Format

The first line contains 2 space-separated integers, N (the number of nodes) and Q (the number of queries), respectively.
Each of the N -1  subsequent lines contain 3 space-separated integers, U , V , and W, respectively, describing a bidirectional road between nodes U and V which has weight W.
The Q subsequent lines each contain 2 space-separated integers denoting L and R.

Constraints

1  <=  N,  Q  < = 10 ^5
1  <=   U, V  <=  N
1  <=  W   <=  10 ^ 9
1  <=  L  <=  R  <=  10^9

Output Format

For each of the Q queries, print the number of paths in T having cost C  in the inclusive range  [ L, R ] on a new line.```

### Solution :

```                            ```Solution in C :

In C++ :

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<cassert>
#define PB push_back
#define MP make_pair
#define sz(v) (in((v).size()))
#define forn(i,n) for(in i=0;i<(n);++i)
#define forv(i,v) forn(i,sz(v))
#define fors(i,s) for(auto i=(s).begin();i!=(s).end();++i)
#define all(v) (v).begin(),(v).end()
using namespace std;
typedef long long in;
typedef vector<in> VI;
typedef vector<VI> VVI;
struct unifnd{
VI ht,pr,ss;
in fnd(in a){
in ta=a;
while(a!=pr[a])a=pr[a];
in tt=ta;
while(ta!=a){
tt=pr[ta];
pr[ta]=a;
ta=tt;
}
return a;
}
in uni(in a, in b){
a=fnd(a);
b=fnd(b);
if(a==b)return 0;
if(ht[b]<ht[a])swap(a,b);
pr[a]=b;
in r=ss[a]*ss[b];
ss[b]+=ss[a];
ht[b]+=(ht[a]==ht[b]);
return r;
}
void ini(in n){
ht.resize(n);
pr.resize(n);
ss.resize(n);
forn(i,n){
ht[i]=0;
ss[i]=1;
pr[i]=i;
}
}
};
VI ans;
unifnd cf;
struct ev{
in typ,u,v,w;
ev(in a=0, in b=0, in c=0, in d=0){
typ=a;
u=b;
v=c;
w=d;
}
bool operator<(const ev cp)const{
if(w!=cp.w)
return w<cp.w;
if(typ!=cp.typ)
return typ<cp.typ;
return 0;
}
};
in sm=0;
void prev(ev tp){
if(tp.typ==1){
ans[tp.u]+=tp.v*sm;
return;
}
sm+=cf.uni(tp.u,tp.v);
}
vector<ev> evs;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
in n,q;
cin>>n>>q;
cf.ini(n);
ans.resize(q,0);
in ta,tb,tc;
forn(i,n-1){
cin>>ta>>tb>>tc;
--ta;
--tb;
evs.PB(ev(0,ta,tb,tc));
}
forn(i,q){
cin>>ta>>tb;
evs.PB(ev(1,i,-1,ta-1));
evs.PB(ev(1,i,1,tb));
}
sort(all(evs));
forv(i,evs)
prev(evs[i]);
forv(i,ans)
cout<<ans[i]<<"\n";
return 0;
}

In Java :

import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

import static java.lang.System.out;

class WeightCount {
private int weight;
private long count;

public WeightCount(int weight, long count) {
this.weight = weight;
this.count = count;
}

public int getWeight() {
return weight;
}
public long getCount() {
return count;
}

public void setWeight(int weight) {
this.weight = weight;
}
public void setCount(long count) {
this.count = count;
}

public static int lower(WeightCount[] array, int size, int key) {
if (array == null || size < 0)
return -1;

if (size == 0)
return 0;

int l = 0;
int r = size - 1;

int mid, weight;
while ((r - l) > 1) {
mid = l + ((r - l) >> 1);
weight = array[mid].getWeight();
if (weight > key)
r = mid - 1;
else if (weight < key)
l = mid;
else
r = mid;
}

if (array[l].getWeight() > key)
return l - 1;
if (key == array[l].getWeight() ||
array[r].getWeight() > key)
return l;
return r;
}
}

class Edge implements Comparable<Edge> {
private int u;
private int v;
private int w;

public Edge(int u, int v, int w) {
this.u = u;
this.v = v;
this.w = w;
}

public int getU() {
return u;
}
public int getV() {
return v;
}
public int getW() {
return w;
}

public void setU(int u) {
this.u = u;
}
public void setV(int v) {
this.v = v;
}
public void setW(int w) {
this.w = w;
}

public int compareTo(Edge e) {
if (e != null) {
int tmp = e.getW();
if (w < tmp)
return -1;
if (w > tmp)
return 1;
}

return 0;
}
}

class DisjointSet {
private static final int DEFAULT_SIZE = 31;

private int[] idx;
private int[] size;
private int n;
private int components;

public DisjointSet(int n) {
if (n < 1)
n = DEFAULT_SIZE;

idx = new int[n + 1];
size = new int[n + 1];

this.n = n;
components = n;

for (int i = n; i > 0; i--) {
idx[i] = i;
size[i] = 1;
}
}

public DisjointSet() {
this(DEFAULT_SIZE);
}

private int root(int i) {
if (i < 1 || i > n)
return 0;

int p = i;
while (idx[p] != p)
p = idx[p];

int tmp;
while (idx[i] != p) {
tmp = idx[i];
idx[i] = p;
i = tmp;
}

return p;
}

public long join(int p, int q) {
int rootP = root(p);
int rootQ = root(q);

if (rootP != rootQ) {
long result = (long) size[rootP] * size[rootQ];

if (size[rootP] < size[rootQ]) {
idx[rootP] = rootQ;
size[rootQ] += size[rootP];
} else {
idx[rootQ] = rootP;
size[rootP] += size[rootQ];
}

components--;
return result;
}

return 0;
}

public boolean isConnected(int p, int q) {
return (root(p) == root(q));
}
}

public class MaximumCostQueries {
private static final int MAX_N = 100000;
private static final int MAX_Q = 100000;

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);

int n = sc.nextInt();
int q = sc.nextInt();
if (n < 1 || n > MAX_N ||
q < 1 || q > MAX_Q)
return;

Edge[] edges = new Edge[n - 1];
int i, u, v, w;
for (i = n - 2; i >= 0; i--) {
u = sc.nextInt();
v = sc.nextInt();
w = sc.nextInt();

edges[i] = new Edge(u, v, w);
}

Arrays.sort(edges);

DisjointSet ds = new DisjointSet(n);
WeightCount[] wc = new WeightCount[n];
int j, k, limit;
long result;

limit = edges.length;
for (k = i = 0; i < limit; i = j) {
result = 0;
w = edges[i].getW();
j = i;

do {
result += ds.join(edges[j].getU(), edges[j].getV());
j++;
} while (j < limit && edges[j].getW() == w);

wc[k++] = new WeightCount(w, result);
}

// out.println("wc:");
// for (i = 0; i < k; i++)
//     out.println(wc[i].getWeight() + ": " + wc[i].getCount());

for (i = 1; i < k; i++)
wc[i].setCount(wc[i - 1].getCount() + wc[i].getCount());

while (q-- > 0) {
i = sc.nextInt();
j = sc.nextInt();

u = WeightCount.lower(wc, k, i - 1);
v = WeightCount.lower(wc, k, j);

result = wc[v].getCount() - ((u < 0) ? 0 : wc[u].getCount());
out.println(result);
}
sc.close();
}
}

In C :

#include <stdio.h>

static long long int a[100000][3], parent[100001], n;
void mer(int p, int q, int r)
{
static int le[100001][3], ri[100001][3], i, j, k;
int n1 = q - p + 1, n2 = r - q;
for (i = 0; i<n1; i++)
{
le[i][0] = a[p + i][0];
le[i][1] = a[p + i][1];
le[i][2] = a[p + i][2];
}
for (j = 0; j<n2; j++)
{
ri[j][0] = a[q + j + 1][0];
ri[j][1] = a[q + j + 1][1];
ri[j][2] = a[q + j + 1][2];
}
le[n1][2] = ri[n2][2] = 1000000001;
i = j = 0;
for (k = p; k <= r; k++)
{
if (le[i][2] <= ri[j][2])
{
a[k][0] = le[i][0];
a[k][1] = le[i][1];
a[k][2] = le[i][2];
i++;
}
else
{
a[k][0] = ri[j][0];
a[k][1] = ri[j][1];
a[k][2] = ri[j][2];
j++;
}
}
}
void merge_sort(int p, int r)
{
int q;
if (p<r)
{
q = (p + r) / 2;
merge_sort(p, q);
merge_sort(q + 1, r);
mer(p, q, r);
}
}

int getParent(int x)
{
if (x == parent[x])
return x;
parent[x] = getParent(parent[x]);
return parent[x];
}

int bin_search(int x)
{
int low, mid, upp;
low = 0;
upp = n;
mid = (low + upp) / 2;
while (low<upp)
{
if (x<a[mid][2])
{
upp = mid - 1;
}
else if (x>a[mid][2])
{
if (x >= a[mid + 1][2])
low = mid + 1;
if (x<a[mid + 1][2])
break;
}
else
{
break;
}
mid = (low + upp) / 2;
}
return mid;
}

int main() {
static long long int q, count[100001], i, j, k, x, y, z, px, py;
scanf("%lld%lld", &n, &q);
for (i = 1; i<n; i++)
{
parent[i] = i;
count[i] = 1;
scanf("%lld%lld%lld", &a[i][0], &a[i][1], &a[i][2]);
}
parent[i] = i;
count[i] = 1;
merge_sort(1, n - 1);
for (i = 1; i<n; i++)
{
x = a[i][0];
y = a[i][1];
px = getParent(x);
py = getParent(y);
a[i][0] = count[px] * count[py];
if (count[px] >= count[py])
{
count[px] += count[py];
parent[py] = px;
}
else
{
count[py] += count[px];
parent[px] = py;
}
}
for (i = 2, j = 1; i<n; i++)
{
if (a[j][2] == a[i][2])
{
a[j][0] += a[i][0];
}
else
{
j++;
a[j][0] = a[i][0] + a[j - 1][0];
a[j][2] = a[i][2];
}
}
n = j;
while (q--)
{
scanf("%lld%lld", &x, &y);
px = bin_search(x - 1);
py = bin_search(y);
printf("%lld\n", a[py][0] - a[px][0]);
}
return 0;
}

In Python3 :

import bisect as bs
n, q = map(int, input().split())
edges = [list(map(int, input().split())) for _ in range(1, n)]
edges.sort(key=lambda x: x[2])
paths = {}
union = [-1] * n

def getroot(x):
if union[x] < 0:
return x
union[x] = getroot(union[x])
return union[x]

for u, v, c in edges:
u = getroot(u - 1)
v = getroot(v - 1)
paths[c] = paths.get(c, 0) + union[u] * union[v]
if union[u] < union[v]:
u, v = v, u
union[v] += union[u]
union[u] = v

paths = list(sorted(paths.items()))
a = [0]
b =[0]
for x, y in paths:
a.append(x)
b.append(b[-1] + y)

for _ in range(q):
l, r = map(int, input().split())
print(b[bs.bisect(a, r) - 1] - b[bs.bisect_left(a, l) - 1])```
```

## Game of Two Stacks

Alexa has two stacks of non-negative integers, stack A = [a0, a1, . . . , an-1 ] and stack B = [b0, b1, . . . , b m-1] where index 0 denotes the top of the stack. Alexa challenges Nick to play the following game: In each move, Nick can remove one integer from the top of either stack A or stack B. Nick keeps a running sum of the integers he removes from the two stacks. Nick is disqualified f

## Largest Rectangle

Skyline Real Estate Developers is planning to demolish a number of old, unoccupied buildings and construct a shopping mall in their place. Your task is to find the largest solid area in which the mall can be constructed. There are a number of buildings in a certain two-dimensional landscape. Each building has a height, given by . If you join adjacent buildings, they will form a solid rectangle

## Simple Text Editor

In this challenge, you must implement a simple text editor. Initially, your editor contains an empty string, S. You must perform Q operations of the following 4 types: 1. append(W) - Append W string to the end of S. 2 . delete( k ) - Delete the last k characters of S. 3 .print( k ) - Print the kth character of S. 4 . undo( ) - Undo the last (not previously undone) operation of type 1 or 2,

## Poisonous Plants

There are a number of plants in a garden. Each of the plants has been treated with some amount of pesticide. After each day, if any plant has more pesticide than the plant on its left, being weaker than the left one, it dies. You are given the initial values of the pesticide in each of the plants. Determine the number of days after which no plant dies, i.e. the time after which there is no plan

## AND xor OR

Given an array of distinct elements. Let and be the smallest and the next smallest element in the interval where . . where , are the bitwise operators , and respectively. Your task is to find the maximum possible value of . Input Format First line contains integer N. Second line contains N integers, representing elements of the array A[] . Output Format Print the value

## Waiter

You are a waiter at a party. There is a pile of numbered plates. Create an empty answers array. At each iteration, i, remove each plate from the top of the stack in order. Determine if the number on the plate is evenly divisible ith the prime number. If it is, stack it in pile Bi. Otherwise, stack it in stack Ai. Store the values Bi in from top to bottom in answers. In the next iteration, do the