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

Queue using Two Stacks

A queue is an abstract data type that maintains the order in which elements were added to it, allowing the oldest elements to be removed from the front and new elements to be added to the rear. This is called a First-In-First-Out (FIFO) data structure because the first element added to the queue (i.e., the one that has been waiting the longest) is always the first one to be removed. A basic que

View Solution →

Castle on the Grid

You are given a square grid with some cells open (.) and some blocked (X). Your playing piece can move along any row or column until it reaches the edge of the grid or a blocked cell. Given a grid, a start and a goal, determine the minmum number of moves to get to the goal. Function Description Complete the minimumMoves function in the editor. minimumMoves has the following parameter(s):

View Solution →

Down to Zero II

You are given Q queries. Each query consists of a single number N. You can perform any of the 2 operations N on in each move: 1: If we take 2 integers a and b where , N = a * b , then we can change N = max( a, b ) 2: Decrease the value of N by 1. Determine the minimum number of moves required to reduce the value of N to 0. Input Format The first line contains the integer Q.

View Solution →

Truck Tour

Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr

View Solution →

Queries with Fixed Length

Consider an -integer sequence, . We perform a query on by using an integer, , to calculate the result of the following expression: In other words, if we let , then you need to calculate . Given and queries, return a list of answers to each query. Example The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is . The secon

View Solution →

QHEAP1

This question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: " 1 v " - Add an element to the heap. " 2 v " - Delete the element from the heap. "3" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element

View Solution →