# Unique Colors

### Problem Statement :

```You are given an unrooted tree of n nodes numbered from 1 to n . Each node i has a color, ci.

Let d( i , j )  be the number of different colors in the path between node i and node j. For each node i, calculate the value of sum, defined as follows:

Your task is to print the value of sumi  for each node  1  <=  i  <= n.

Input Format

The first line contains a single integer, n, denoting the number of nodes.
The second line contains n space-separated integers, c1, c2 , c3 , . . . cn , where each ci  describes the color of node i.
Each of the n - 1 subsequent lines contains 2 space-separated integers, a and b, defining an undirected edge between nodes a and b.

Constraints

1  <=   n  <=  10^5
1  <=  ci  <=  10^5

Output Format

Print n lines, where the ith line contains a single integer denoting sumi.```

### Solution :

```                            ```Solution in C :

In   C++  :

#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> pi;
#define N 101000
typedef long long ll;
set<pi> S;
set<pi> ::iterator it;
int dis[N], rev[N], col[N], pa[N], fin[N];
ll ans[N];
vector<int> v[N];
vector<int> C[N];
int cnt;
void dfs(int x, int p) {
dis[x] = ++cnt;
rev[cnt] = x;
C[col[x]].push_back(x);
for(int i = 0; i < v[x].size(); i ++) {
int y = v[x][i];
if(y == p) continue;
pa[y] = x;
dfs(y, x);
}
fin[x] = cnt;
}
pi A[N];

void build(int st, int en, int id) {
if(st == en) {
return ;
}
int mid = (st + en) >> 1;
build(st, mid, id * 2);
build(mid + 1, en, id * 2 + 1);
return ;
}

int n;

void push_down(int id) {
}
return ;
}

void add(int l, int r, int st, int en, int id, int val) {
if(l > en || st > r) return ;
if(l <= st && en <= r) {
return ;
}
push_down(id);
int mid = (st + en) >> 1;
add(l, r, st, mid, id * 2, val);
add(l, r, mid + 1, en, id * 2 + 1, val);
}

ll calc(int l, int st, int en, int id) {
if(st == en) {
}
push_down(id);
int mid = (st + en) >> 1;
if(l <= mid) return calc(l, st, mid, id * 2);
return calc(l, mid + 1, en, id * 2 + 1);
}
pi B[N];

void doit(int y) {
int st = dis[y];
int en = fin[y];
it = S.lower_bound(pi(en + 1, 0));
if(it == S.begin()) {
add(st, en, 1, n, 1, -(en - st + 1));
S.insert(pi(st, en));
return ;
}
it --;
pi bb = *it;
if(bb.second < st) {
add(st, en, 1, n, 1, -(en - st + 1));
S.insert(pi(st, en));
return ;
}
int cnt = 0;
while(1) {
pi aa = *it;
if(aa.second < st) break;
B[cnt ++] = aa;
if(it == S.begin()) break;
it --;
}
for(int i = 0; i < cnt; i ++) A[i] = B[cnt - 1- i];
int num = en - st + 1;
for(int i = 0; i < cnt; i ++) num -= (A[i].second - A[i].first + 1);
for(int i = 1; i < cnt; i ++) {
if(A[i].first > A[i - 1].second + 1) {
add(A[i - 1].second + 1, A[i].first - 1, 1, n, 1, -num);
}
continue;
}
if(A[0].first >= st + 1) add(st, A[0].first - 1, 1, n, 1, -num);
if(A[cnt - 1].second <= en - 1) add(A[cnt - 1].second + 1, en, 1, n, 1, -num);
for(int i = 0; i < cnt; i ++) {
it = S.find(pi(A[i].first, A[i].second));
S.erase(it);
}
S.insert(pi(st, en));
}
int a[N];
int main() {
//freopen("1.in", "r", stdin);
scanf("%d", &n);
for(int i = 1; i <= n; i ++) scanf("%d", &col[i]);
for(int i = 1; i <= n; i ++) a[i - 1] = col[i];
sort(a, a + n);
int num = unique(a, a + n) - a;
for(int i = 1; i <= n; i ++) col[i] = lower_bound(a, a + num, col[i]) - a + 1;
for(int i = 1; i <= n; i ++) ans[i] = 1ll * num * n;
for(int i = 1; i < n; i ++) {
int x, y;
scanf("%d%d", &x, &y);
v[x].push_back(y);
v[y].push_back(x);
}
if(n == 1) {
puts("0");
return 0;
}
dfs(1, 0);
build(1, n, 1);
for(int i = 1; i <= num; i ++) {
S.clear();
for(int j = C[i].size() - 1; j >= 0; j --) {
int x = C[i][j];
for(int k = 0; k < v[x].size(); k ++) {
int y = v[x][k];
if(y == pa[x]) continue;
doit(y);
}
for(int k = 0; k < v[x].size(); k ++) {
int y = v[x][k];
if(y == pa[x]) continue;
S.erase(pi(dis[y], fin[y]));
}
S.insert(pi(dis[x], fin[x]));
}
doit(1);
}
for(int i = 1; i <= n; i ++) ans[i] += calc(i, 1, n, 1);
for(int i = 1; i <= n; i ++) printf("%lld\n", ans[dis[i]]);
return 0;
}

In   Java  :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
static Node[] nodes;
static int n;
public static void main(String[] args) throws IOException {
nodes = new Node[n];
for (int i = 0; i < n; i++)
nodes[i] = new Node(i, Integer.valueOf(line[i]));
for (int i = 1; i < n; i++) {
int u = Integer.valueOf(line[0])-1;
int v = Integer.valueOf(line[1])-1;
}
for (int i = 0; i < n; i++) {
System.out.println(f(i, i, -1, new BitSet(n), new BitSet(n), 0));
}
}
static long f(int start, int index,
int prev, BitSet visited, BitSet colors, int bitCount) {
if (visited.get(index))
return bitCount;
int color = nodes[index].color;
boolean increased = false;
if (!colors.get(color)) {
colors.set(color);
bitCount++;
increased = true;
}

long sum = bitCount;
for (Node node : nodes[index].adj) {
if (node.id != prev)
sum += f(start, node.id, index, visited, colors, bitCount);
}

if (increased) {
colors.clear(color);
bitCount--;
}
visited.clear(index);
return sum;
}
}
class Node {
int id, color;
public Node(int id, int color) {
this.id = id; this.color = color;
}
}

In    Python 3  :

import collections

n = int(input())
node_colors = input().split()
edges = {i: [] for i in range(n)}
for _ in range(n - 1):
one, two = [int(i) - 1 for i in input().split()]
edges[one].append(two)
edges[two].append(one)

def bfs_sum(i):
value = 0
seen = {i}
q = collections.deque([(i, {node_colors[i]})])
while q:
t, colors = q.popleft()

value += len(colors)

for edge in edges[t]:
if edge not in seen:
q.append((edge, colors | {node_colors[edge]}))
return value

for i in range(n):
print(bfs_sum(i))```
```

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