# Kitty's Calculations on a Tree

### Problem Statement :

```Kitty has a tree, T , consisting of n nodes where each node is uniquely labeled from  1 to n . Her friend Alex gave her q sets, where each set contains k distinct nodes. Kitty needs to calculate the following expression on each set:

where:

{ u ,v } denotes an unordered pair of nodes belonging to the set.
dist(u , v) denotes the number of edges on the unique (shortest) path between nodes  and .
Given T and q sets of k  distinct nodes, calculate the expression for each set. For each set of nodes, print the value of the expression modulo 10^9 + 7  on a new line.

Input Format

The first line contains two space-separated integers, the respective values of n (the number of nodes in tree T ) and  q (the number of nodes in the query set).
Each of the n - 1  subsequent lines contains two space-separated integers, a and b, that describe an undirected edge between nodes  and .
The 2 * q subsequent lines define each set over two lines in the following format:

The first line contains an integer, k  , the size of the set.
The second line contains  k space-separated integers, the set's elements.

Output Format

Print q lines of output where each line i contains the expression for the ith query, modulo 10^9 + 7.```

### Solution :

```                            ```Solution in C :

In C ++ :

#include <bits/stdc++.h>

using namespace std;

const int MOD=1000000007;
int N, Q;
int P[18][200001];
int depth[200001];
int in[200001];
int out[200001];
int now;
int A[400001];
int mul[200001];
long long sum[200001];

void dfs(int u, int p)
{
P[0][u]=p;
for(int i=1; i<18; i++)
P[i][u]=P[i-1][P[i-1][u]];
in[u]=++now;
{
depth[v]=depth[u]+1;
dfs(v, u);
}
out[u]=now;
}

int lca(int u, int v)
{
if(depth[u]<depth[v])
swap(u, v);
for(int i=17; i>=0; i--) if(depth[P[i][u]]>=depth[v])
u=P[i][u];
if(u==v)
return u;
for(int i=17; i>=0; i--) if(P[i][u]!=P[i][v])
u=P[i][u], v=P[i][v];
return P[0][u];
}

int dfs2(int u, long long tot)
{
int ret=0;
sum[u]=u*mul[u];
{
ret=(ret+dfs2(v, tot))%MOD;
sum[u]+=sum[v];
}
ret=(ret+1LL*((tot-sum[v])%MOD)
*(sum[v]%MOD)%MOD
*(depth[v]-depth[u])%MOD)%MOD;
return ret;
}

int main()
{
scanf("%d%d", &N, &Q);
for(int i=0; i<N-1; i++)
{
int a, b;
scanf("%d%d", &a, &b);
}
dfs(1, 1);
while(Q--)
{
int K;
scanf("%d", &K);
long long tot=0;
for(int i=0; i<K; i++)
scanf("%d", A+i), mul[A[i]]=1, tot+=A[i];
sort(A, A+K, [](int a, int b) {
return in[a]<in[b];
});
for(int i=0; i<K-1; i++)
A[i+K]=lca(A[i], A[i+1]);
sort(A, A+2*K-1);
int M=unique(A, A+2*K-1)-A;
sort(A, A+M, [](int a, int b) {
return out[a]-in[a]>out[b]-in[b];
});
int root=A[0];
map<int, int> m;
m[in[root]]=root;
for(int i=1; i<M; i++)
{
int u=A[i];
auto it=m.upper_bound(in[u]);
assert(it!=m.begin());
--it;
int p=it->second;
//printf("%d -> %d\n", p, u);
m[in[u]]=u;
if(out[u]<out[p] && (!m.count(out[u]+1) || P[0][m[out[u]+1]]!=p))
m[out[u]+1]=p;
}
printf("%d\n", dfs2(root, tot));
for(int i=0; i<M; i++)
}
return 0;
}

In Java :

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

public class KittysCalc {

public static final long constant = 1000000007;

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int queries = sc.nextInt();
int[] parents = new int[n+1];
long[] children = new long[n+1];
boolean[] valuesSet = new boolean[n+1];
long valuesSum = 0;
long sum = 0;
int a, b;
for(int i = 0; i < n-1; i++) {
a = sc.nextInt();
b = sc.nextInt();
if(a < b) {
parents[b] = a;
} else {
parents[a] = b;
}
}
parents[1] = 0;

for(int i = 0; i < queries; i++) {
int k = sc.nextInt();
Arrays.fill(valuesSet, false);
Arrays.fill(children, 0);
valuesSum = 0;
for(int j = 0; j < k; j++) {
a = sc.nextInt();
valuesSum += a;
valuesSet[a] = true;
}
sum = 0;
for (int j = n; j > 0; j--) {
long c = children[j];
if (valuesSet[j]) {
c += j;
}
if (c > 0) {
long x = ((c % constant) * ((valuesSum - c) % constant)) % constant;
if (constant - sum < x) {
sum -= constant;
}
sum += x;
}
children[parents[j]] += c;
}
System.out.println(sum);
}
sc.close();
}

}

In Python3 :

#!/usr/bin/env python3

def put(d, a, b):
if a in d: d[a].append(b)
else: d[a] = [b]

def main():
for n in ns[::-1]:
r = [tt[s] for s in tree[n] if s != f[n]]
bst = {s: [gl[n], n, 0] for s in queries[n]}
if r:
o = max(range(len(r)), key=lambda a: len(r[a]))
if len(r[o]) > len(bst): r[o], bst = bst, r[o]
ry = {}
for ae in r:
for y, v in ae.items():
put(ry, y, v)
for y, r in ry.items():
eq, z, t = 0, 0, 0
if len(r) == 1 and y not in bst:
bst[y] = r[0]
continue
if y in bst: r.append(bst.pop(y))
for d, v, c in r:
eq += (d - gl[n]) * v + c
z += v
for d, v, c in r:
c += (d - gl[n]) * v
diff = (eq - c) * v
t += diff
returns[y] += t
bst[y] = (gl[n], z, eq)
tt[n] = bst

def locate():
q = [r]
level = 0
while q:
level += 1
tmp = []
ns.extend(q)
for n in q:
for s in tree[n]:
if s not in f:
f[s] = n
gl[s] = level
tmp.append(s)
q = tmp

tree = {}
tt = {}
n, q = map(int, input().split())
returns = [0] * q
for _ in range(n - 1):
a, b = map(int, input().split())
put(tree, a, b)
put(tree, b, a)
queries = {a: set() for a in tree}
for y in range(q):
input()
for x in map(int, input().split()): queries[x].add(y)
r = next(iter(tree))
ns = []
f = {r: None}
gl = {r: 0}
locate()
main()
for s in returns: print(s % (10**9 + 7))```
```

## Array-DS

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## 2D Array-DS

Given a 6*6 2D Array, arr: 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 An hourglass in A is a subset of values with indices falling in this pattern in arr's graphical representation: a b c d e f g There are 16 hourglasses in arr. An hourglass sum is the sum of an hourglass' values. Calculate the hourglass sum for every hourglass in arr, then print t

## Dynamic Array

Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.

## Left Rotation

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## Sparse Arrays

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## Array Manipulation

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