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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Diagnostics;

class Program
{
    static void Main(string[] args)
    {
        int n = int.Parse(Console.ReadLine());

        var to = new List<List<int>>();
        for (int i = 0; i < n; i++)
            to.Add(new List<int>());
        var cost = new List<List<int>>();
        for (int i = 0; i < n; i++)
            cost.Add(new List<int>());

        for (int i = 1; i < n; i++)
        {
            string[] cin = Console.ReadLine().Split();
            int a = int.Parse(cin[0]);
            int b = int.Parse(cin[1]);
            int w = int.Parse(cin[2]);
            // -1してインデックス0スタートに
            a--;
            b--;
            // cin[0]から出るi番目の辺について
            // toとcostを記録
            to[a].Add(b);
            cost[a].Add(w);
            // 無向木なので
            to[b].Add(a);
            cost[b].Add(w);
        }

        // 木構造の全点間距離dist出す
        var dist = new long[n][];
        for (int i = 0; i < n; i++)
        {
            dist[i] = Dijkstra(n, i, to, cost);
        }
        var ans = new bool[n + 1];

        var even = new List<Tuple<int, int>>();
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                // i→j間が偶数のリスト作る
                if (dist[i][j] != long.MaxValue && dist[i][j] != 0 && dist[i][j] % 2 == 0)
                {
                    even.Add(new Tuple<int, int>(i, j));
                }
            }
        }

        // ビット全探索で偶数距離が同じ色に塗られてる組み合わせを見つける
        for (int i = 1; i < 1 << n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if ((1 & i >> j) == 1) ans[j] = true;
                else ans[j] = false;
            }
            bool flag = false;
            for (int j = 0; j < even.Count; j++)
            {
                if (ans[even[j].Item1] != ans[even[j].Item2])
                {
                    flag = false;
                    break;
                }
                else
                {
                    flag = true;
                }
            }
            if (flag) break;
        }

        // [高速化]出力多いのでまとめてFlushする
        Console.SetOut(new System.IO.StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false });
        for (int i = 0; i < n; i++)
        {
            if (ans[i]) Console.WriteLine("1");
            else Console.WriteLine("0");
        }
        Console.Out.Flush();
    }


    // 木構造の距離分からんからダイクストラ
    static long[] Dijkstra(int n, int s, List<List<int>> to, List<List<int>> cost)
    {
        var dist = new long[n];
        for (int i = 0; i < n; i++)
        {
            dist[i] = long.MaxValue;
        }
        dist[s] = 0;

        var fix = new bool[n];

        while (true)
        {
            int p = int.MaxValue;
            long q = long.MaxValue;
            for (int i = 0; i < n; i++)
            {
                if (!fix[i] && dist[i] <= q)
                {
                    p = i;
                    q = dist[i];
                }
            }
            if (p == int.MaxValue) break;
            fix[p] = true;
            for (int i = 0; i < to[p].Count; i++)
            {
                if (!fix[to[p][i]]) dist[i] = Math.Min(dist[i], dist[p] + cost[p][i]);
            }
        }
        return dist;
    }
}

// chokudai's Scanner
class Scanner
{
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner()
    {
        s = new string[0];
        i = 0;
    }

    public string next()
    {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        if (s.Length == 0) return next();
        i = 0;
        return s[i++];
    }

    public int nextInt()
    {
        return int.Parse(next());
    }

    public int[] ArrayInt(int N, int add = 0)
    {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextInt() + add;
        }
        return Array;
    }

    public long nextLong()
    {
        return long.Parse(next());
    }

    public long[] ArrayLong(int N, long add = 0)
    {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextLong() + add;
        }
        return Array;
    }

    public double nextDouble()
    {
        return double.Parse(next());
    }

    public double[] ArrayDouble(int N, double add = 0)
    {
        double[] Array = new double[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextDouble() + add;
        }
        return Array;
    }
}