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 :



title-img 


                            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 Add[N * 4];
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) {
    Add[id] = 0;
    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) {
    if(Add[id]) {
        Add[id * 2] += Add[id];
        Add[id * 2 + 1] += Add[id];
        Add[id] = 0;
    }
    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) {
        Add[id] += val;
        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) {
        return Add[id];
    }
    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 BufferedReader br = new
     BufferedReader(new InputStreamReader(System.in));
    static Node[] nodes;
    static int n;
    public static void main(String[] args) throws IOException {
        n = Integer.valueOf(br.readLine());
        nodes = new Node[n];
        String[] line = br.readLine().split(" ");
        for (int i = 0; i < n; i++)
            nodes[i] = new Node(i, Integer.valueOf(line[i]));
        for (int i = 1; i < n; i++) {
            line = br.readLine().split(" ");
            int u = Integer.valueOf(line[0])-1;
            int v = Integer.valueOf(line[1])-1;
            nodes[u].adj.add(nodes[v]);
            nodes[v].adj.add(nodes[u]);
        }
        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;
    List<Node> adj = new ArrayList<Node>();
    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:
                seen.add(edge)
                q.append((edge, colors | {node_colors[edge]}))
    return value


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








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