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))








View More Similar Problems

Subsequence Weighting

A subsequence of a sequence is a sequence which is obtained by deleting zero or more elements from the sequence. You are given a sequence A in which every element is a pair of integers i.e A = [(a1, w1), (a2, w2),..., (aN, wN)]. For a subseqence B = [(b1, v1), (b2, v2), ...., (bM, vM)] of the given sequence : We call it increasing if for every i (1 <= i < M ) , bi < bi+1. Weight(B) =

View Solution →

Kindergarten Adventures

Meera teaches a class of n students, and every day in her classroom is an adventure. Today is drawing day! The students are sitting around a round table, and they are numbered from 1 to n in the clockwise direction. This means that the students are numbered 1, 2, 3, . . . , n-1, n, and students 1 and n are sitting next to each other. After letting the students draw for a certain period of ti

View Solution →

Mr. X and His Shots

A cricket match is going to be held. The field is represented by a 1D plane. A cricketer, Mr. X has N favorite shots. Each shot has a particular range. The range of the ith shot is from Ai to Bi. That means his favorite shot can be anywhere in this range. Each player on the opposite team can field only in a particular range. Player i can field from Ci to Di. You are given the N favorite shots of M

View Solution →

Jim and the Skyscrapers

Jim has invented a new flying object called HZ42. HZ42 is like a broom and can only fly horizontally, independent of the environment. One day, Jim started his flight from Dubai's highest skyscraper, traveled some distance and landed on another skyscraper of same height! So much fun! But unfortunately, new skyscrapers have been built recently. Let us describe the problem in one dimensional space

View Solution →

Palindromic Subsets

Consider a lowercase English alphabetic letter character denoted by c. A shift operation on some c turns it into the next letter in the alphabet. For example, and ,shift(a) = b , shift(e) = f, shift(z) = a . Given a zero-indexed string, s, of n lowercase letters, perform q queries on s where each query takes one of the following two forms: 1 i j t: All letters in the inclusive range from i t

View Solution →

Counting On a Tree

Taylor loves trees, and this new challenge has him stumped! Consider a tree, t, consisting of n nodes. Each node is numbered from 1 to n, and each node i has an integer, ci, attached to it. A query on tree t takes the form w x y z. To process a query, you must print the count of ordered pairs of integers ( i , j ) such that the following four conditions are all satisfied: the path from n

View Solution →