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;
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
Tree: Postorder Traversal
Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the
View Solution →Tree: Inorder Traversal
In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func
View Solution →Tree: Height of a Binary Tree
The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary
View Solution →Tree : Top View
Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.
View Solution →Tree: Level Order Traversal
Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F
View Solution →Binary Search Tree : Insertion
You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <
View Solution →