Problem:
Score : 400 points

Problem Statement

We have a tree with N vertices numbered 1 to N.
The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i.
Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied:

For any two vertices painted in the same color, the distance between them is an even number.

Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq u_i < v_i \leq N

1 \leq w_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
u_1 v_1 w_1
u_2 v_2 w_2
.
.
.
u_{N - 1} v_{N - 1} w_{N - 1}

Output

Print a coloring of the vertices that satisfies the condition, in N lines.
The i-th line should contain 0 if Vertex i is painted white and 1 if it is painted black.

If there are multiple colorings that satisfy the condition, any of them will be accepted.

Sample Input 1

3
1 2 2
2 3 1

Sample Output 1

0
0
1

Sample Input 2

5
2 5 2
2 3 10
1 3 8
3 4 2

Sample Output 2

1
0
1
0
1

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <deque>
#include <stack>
#include <functional>
#include <algorithm>
#include <cmath>
#include <iomanip>
#include <limits>
#include <utility>

using namespace std;

#define ll long long

#define MOD 1000000007
#define PI acos(-1)
#define pb push_back

#define endl '\n'

#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl

#define pllll pair<ll, ll>
#define plls pair<ll, string>
#define psll pair<string, ll>

#define For(i, n) for(int i = 0; i < n; i++)

#define phoge print("hoge")
#define phugo print("hugo")

#define vll vector<ll>
#define v2ll vec2<ll>
#define v3ll vec3<ll>

template<typename T>
using vec = vector<T>;

template<typename T>
using vec2 = vector<vector<T>>;

template<typename T>
using vec3 = vector<vector<vector<ll>>>;

template<typename T>
using uset = unordered_set<T>;

template<typename T, typename S>
using umap = unordered_map<T, S>;

bool inRange(ll a, ll x, ll b) {
	if (a > b) {
		swap(a, b);
	}
	return (a <= x && x <= b);
}

bool between(ll a, ll x, ll b) {
	if (a > b) {
		swap(a, b);
	}
	return (a < x && x < b);
}

template<typename T> void sort(T &a, T &b) {
	if(a > b){
		swap(a, b);
	}
}

template<typename T> 
void input(T *data, int n) {
	For(i, n) {
		cin >> data[i];
	}
}

template<typename T, typename S>
void input(T *data1, S *data2, int n){
	For(i, n){
		cin >> data1[i] >> data2[i];
	}
}

template<typename T> 
void input(vector<T> &vec, int n) {
	T temp;
	For(i, n) {
		cin >> temp;
		vec.push_back(temp);
	}
}

template<typename T, typename S> 
void input(vector<T> &vec1, vector<S> &vec2, int n){
	T temp1;
	S temp2;
	For(i, n){
		cin >> temp1 >> temp2;
		vec1.push_back(temp1);
		vec2.push_back(temp2);
	}
}

template<typename T> 
void sort(vector<T> &ary) {
	sort(ary.begin(), ary.end());
}

template<typename T> 
void rsort(vector<T> &ary) {
	sort(ary.rbegin(), ary.rend());
}

void rstring(string &str) {
	string temp = str;

	For(i, (int)str.size()) {
		str[i] = temp[(str.size() - 1) - i];
	}
}

template<typename T>
T print(T x){
	cout << x << endl;
	return x;
}

template<typename T> 
void print(T *x, int n) {
	For(i, n) {
		cout << x[i] << " ";
	}
	cout << endl;
}

template<typename T> 
void print(vector<T> &x) {
	for(auto t : x){
		cout << t << " ";
	}
	cout << endl;
}

template<typename T, typename S> 
void print(pair<T, S> &p){
	out2(p.first, p.second);
}

template<typename T>
T gcd(T n, T m) {
	if (n % m == 0) {
		return m;
	} else {
		return gcd<T>(m, n % m);
	}
}

void lsum(vll &a, vll &sum){
	sum[0] = a[0];
	for (int i = 1; i < (int)a.size() ; i++) {
		sum[i] = sum[i - 1] + a[i];
	}
}

void rsum(vll &a, vll &sum){
	sum[(int)a.size() - 1] = a[(int)a.size() - 1];
	for (int i = (int)a.size() - 2; i >= 0; i--) {
		sum[i] = sum[i + 1] + a[i];	
	}
}

struct UnionFind {
	vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

	UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
		for (int i = 0; i < N; i++) par[i] = i;
	}
	UnionFind() {}

	int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}

	void unite(int x, int y) { // xとyの木を併合
		int rx = root(x); //xの根をrx
		int ry = root(y); //yの根をry
		if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
		par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
	}

	bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}

	int size() {
		int count = 0;
		for (int i = 0; i < (int)par.size(); i++) {
			if (par[i] == i) {
				count++;
			}
		}
		return count;
	}
};

template<class Abel> struct PotentialUnionFind {
	vector<int> par;
	vector<int> rank;
	vector<Abel> diff_weight;
 
	PotentialUnionFind(int n = 1, Abel SUM_UNITY = 0) {
		init(n, SUM_UNITY);
	}
 
	void init(int n = 1, Abel SUM_UNITY = 0) {
		par.resize(n); rank.resize(n); diff_weight.resize(n);
		for (int i = 0; i < n; ++i) par[i] = i, rank[i] = 0, diff_weight[i] = SUM_UNITY;
	}
 
	int root(int x) {
		if (par[x] == x) {
			return x;
		}
		else {
			int r = root(par[x]);
			diff_weight[x] += diff_weight[par[x]];
			return par[x] = r;
		}
	}
 
	Abel weight(int x) {
		root(x);
		return diff_weight[x];
	}
 
	bool issame(int x, int y) {
		return root(x) == root(y);
	}
 
	bool merge(int x, int y, Abel w) {
		w += weight(x); w -= weight(y);
		x = root(x); y = root(y);
		if (x == y) return false;
		if (rank[x] < rank[y]) swap(x, y), w = -w;
		if (rank[x] == rank[y]) ++rank[x];
		par[y] = x;
		diff_weight[y] = w;
		return true;
	}
 
	Abel diff(int x, int y) {
		return weight(y) - weight(x);
	}
};


template<typename T, typename S>
struct _CountMap {
	T data;

	void add(S key, int n) {
		auto itr = data.find(key);
		if (itr == data.end()) {
			data.insert(pair<S, int>(key, n));
		} else {
			itr->second += n;
		}
	}

	auto top() {
		return data.begin();
	}

	int val(S key) {
		return data[key];
	}

	int size() {
		return data.size();
	}

	int sub(S key, int n) {
		auto elm = data.find(key);
		int ret = min(elm->second, n);
		if (ret == elm->second) {
			data.erase(elm);
		}
		elm->second -= ret;
		return ret;
	}

	int& operator [](S key) {
		auto itr = data.find(key);
		if (itr == data.end()) {
			data.insert(pair<S, int>(key, 0));
		}
		return itr->second;
	}
};

struct Area {
	vector<ll> data;

	Area(ll range): data(range + 1) {}

	void add(pair<ll, ll> newdata) {
		data[newdata.first]++;
		data[newdata.second]--;
	}

	ll num() {
		ll sum = 0;
		ll ret = 0;
		For(i, (int)data.size()) {
			sum += data[i];
			if (sum != 0) {
				ret++;
				for (i++; i < (int)data.size() && (sum += data[i]) != 0; i++);
			}
		}
		return ret;
	}
};

struct Interval{
	vector<pllll> data;

	void add(ll s, ll f){
		data.push_back(pllll(f, s));
	}

	ll num(){
		sort(data);

		ll ret = 0;
		ll lastf = -1;
		for(auto t : data){
			if(t.first == lastf){
				continue;
			}
			if(t.second >= lastf){
				lastf = t.first;
				ret++;
			}
		}

		return ret;
	}
};

template<typename T>
using CountMap = _CountMap<map<T, int>, T>;

template<typename T>
using CountMapr = _CountMap<multimap<T, int, greater<T>>, T>;

void factoring(ll n, CountMap<ll> &p){
	for(ll i = 2; i + i <= n; i++){
		while(n % i == 0){
			p.add(i, 1);
			n /= i;
		}
	}
}

template<typename T, typename S>
using rmap = multimap<T, S, greater<T>>;

template<typename T>
using rset = multiset<T, greater<T>>;

vector<ll> fac(2001); //n!(mod M)
vector<ll> ifac(2001); //k!^{M-2} (mod M)
//a,bの範囲的にこれだけ配列を用意していけば十分


ll mpow(ll x, ll n){ //x^n(mod M) ←普通にpow(x,n)では溢れてしまうため，随時mod計算
    ll ans = 1;
    while(n != 0){
        if(n&1) ans = ans*x % 1000000007;
        x = x*x % 1000000007;
        n = n >> 1;
    }
    return ans;
}

ll comb(ll a, ll b){ //aCbをmod計算
    if(a == 0 && b == 0)return 1;
    if(a < b || a < 0)return 0;
    ll tmp = ifac[a-b]* ifac[b] % 1000000007;
    return tmp * fac[a] % 1000000007;
}

ll syakutori(vll &data, ll key){
	ll ans = 0;
	ll sum = 0;
	ll cur = 0;
	For(i, data.size()){

		while(sum <= key){
			if(cur == data.size()){
				break;
			}
			else{
				sum += data[cur];
				cur++;
			}
		}

		ans += cur - i - 1 + ((sum <= key) ? 1 : 0);
		sum -= data[i];

	}

	return ans;
}

void dfs(vec2<pllll> &line, vll &ans, vll passed, ll longth, ll now){

	if(longth % 2 == 0){
		ans[now] = 1;
	}
	else{
		ans[now] = 0;
	}

	if(now != 0 && line[now].size() == 1) return;

	for(auto t : line[now]){
		if(passed[t.first]) continue;
		passed[t.first] = 1;
		dfs(line, ans, passed, longth + t.second, t.first);
	}

}

int main() {

	ll n, m, k, a, b, c, x, y, z, h, w, sum = 0;
	string s;

	cin >> n;
	PotentialUnionFind<int> puf(n);
	For(i, n - 1){
		cin >> a >> b >> c;
		puf.merge(a - 1, b - 1, c);
	}

	For(i, n){
		if(puf.weight(i) % 2 == 0){
			print(1);
		}
		else{
			print(0);
		}
	}

	return 0;
}