In this

**HackerEarth Holiday decorations problem solution**Recently, the first model of decoration was assembled from multi-colored glowing light bulbs at the holiday jewelry factory. The prototype of the decoration was assembled as follows:First, two light bulbs were taken and connected with a wire.

Then, a light bulb was taken for N - 2 times and it was connected with a wire to one of the previously added to garland light bulbs.

The result was a decoration of N-colored light bulbs. The factory has K bulbs of different colors. When the prototype was ready, it was handed over to the Beauty Department. In this department, it was decided to consider the beauty of jewelry the number of pairs of light bulbs of the same color, connected by a wire.

Employees of the Beauty Department M repaint some of the light bulbs in one of the K colors for some reason known only to them. All they need to produce the perfect jewelry is to quickly determine the beauty of the product, after repainting the next light bulb.

The staff of the beauty department asks you, the best programmer, to write a program that will determine the beauty of jewelry according to the given prototype of jewelry and the sequence of repainting light bulbs.

## HackerEarth Holiday decorations problem solution.

`#include <bits/stdc++.h>`

#define ll long long

using namespace std;

const int MAX_N = 3e5 + 1;

int n, m, k;

int color[MAX_N];

vector<int> graph[MAX_N], v1[MAX_N], v2[MAX_N];

int S;

bool big[MAX_N];

unordered_map<int, int> M[MAX_N];

int get(int v, int col) {

if (big[v]) {

return M[v][col];

}

int ans = 0;

for (int to : graph[v]) {

if (color[to] == col) {

ans++;

}

}

return ans;

}

void update(int v, int col) {

for (int to : v2[v]) {

if(!--M[to][color[v]]){

M[to].erase(color[v]);

}

M[to][col]++;

}

color[v] = col;

}

int main() {

ios_base::sync_with_stdio(false);

cin.tie(NULL);

cout.tie(NULL);

cin >> n;

cin >> m;

cin >> k;

for(int i = 1; i <= n; i++) {

cin >> color[i];

}

graph[1].push_back(2);

graph[2].push_back(1);

for(int i = 3; i <= n; i++) {

int parent;

cin >> parent;

graph[parent].push_back(i);

graph[i].push_back(parent);

}

S = 550;

for(int i = 1; i <= n; i++) {

big[i] = (int) graph[i].size() >= S;

}

for(int i = 1; i <= n; i++) {

for(int to : graph[i]) {

if(big[to]) {

v2[i].push_back(to);

} else {

v1[i].push_back(to);

}

}

}

for(int i = 1; i <= n; i++) {

if(big[i]) {

for(int to : graph[i]) {

M[i][color[to]]++;

}

}

}

ll ans = 0;

for(int i = 1; i <= n; i++) {

ans += get(i, color[i]);

}

ans /= 2;

while(m--) {

int pos, col;

cin >> pos;

cin >> col;

ans -= get(pos, color[pos]);

update(pos, col);

ans += get(pos, color[pos]);

cout << ans << '\n';

}

}

## 0 Comments