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 :



title-img


                            Solution in C :

In C ++ :




#include <bits/stdc++.h>

using namespace std;

const int MOD=1000000007;
int N, Q;
vector<int> adj[200001];
vector<int> adj2[200001];
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;
    for(auto& v: adj[u]) if(v!=p)
    {
        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];
    for(auto& v: adj2[u])
    {
        ret=(ret+dfs2(v, tot))%MOD;
        sum[u]+=sum[v];
    }
    for(auto& v: adj2[u])
        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);
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
    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;
            adj2[p].push_back(u);
            //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++)
            adj2[A[i]].clear(), mul[A[i]]=0;
    }
    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))
                        








View More Similar Problems

Binary Search Tree : Lowest Common Ancestor

You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b

View Solution →

Swap Nodes [Algo]

A binary tree is a tree which is characterized by one of the following properties: It can be empty (null). It contains a root node only. It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees. In-order traversal is performed as Traverse the left subtree. Visit root. Traverse the right subtree. For this in-order traversal, start from

View Solution →

Kitty's Calculations on a Tree

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 a

View Solution →

Is This a Binary Search Tree?

For the purposes of this challenge, we define a binary tree to be a binary search tree with the following ordering requirements: The data value of every node in a node's left subtree is less than the data value of that node. The data value of every node in a node's right subtree is greater than the data value of that node. Given the root node of a binary tree, can you determine if it's also a

View Solution →

Square-Ten Tree

The square-ten tree decomposition of an array is defined as follows: The lowest () level of the square-ten tree consists of single array elements in their natural order. The level (starting from ) of the square-ten tree consists of subsequent array subsegments of length in their natural order. Thus, the level contains subsegments of length , the level contains subsegments of length , the

View Solution →

Balanced Forest

Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to a

View Solution →