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: Runtime Error)

using System;

namespace ABC126
{
	class Program
	{
		public static void Main(String[] args)
		{
			// 入力
			int N = int.Parse(Console.ReadLine());
			int[] u = new int[N - 1];
			int[] v = new int[N - 1];
			int[] w = new int[N - 1];
			for (int i = 0; i < N - 1; i++)
			{
				int[] a = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
				// 頂点の数字は配列番号と1ずれている事に注意
				u[i] = a[0] - 1;
				v[i] = a[1] - 1;
				w[i] = a[2];
			}
			// 計算
			// 木を構築して、先頭から距離を求めながら色分けをしていけば良い。
			// 木の性質として、どこでも先頭にできるというものがある。
			int[,] weightTable = new int[N, N];// 各頂点について、全ての頂点への距離マップ
			for (int i = 0; i < N; i++)
			{
				for (int j = 0; j < N; j++)
				{
					weightTable[i, j] = -1;// ひとまず全て-1に
				}
			}
			// 各辺の重みを記録
			for (int i = 0; i < N - 1; i++)
			{
				weightTable[u[i], v[i]] = w[i];
				weightTable[v[i], u[i]] = w[i];
			}
			// 0からの距離を求める
			Int64[] rootWeight = new Int64[N];
			for (int i = 0; i < N; i++)
			{
				rootWeight[i] = -1;
			}
			int[] indexVec = new int[N];// 0から順番に見ていって、どの辺を次に計算するか
			for (int i = 0; i < N; i++)
			{
				indexVec[i] = 0;
			}
			indexVec[0] = 0;
			rootWeight[0] = 0;
			int lastIndex = 1;
			for (int i = 0; i < N; i++)
			{
				int nowRoot = indexVec[i];
				if (nowRoot<0 || nowRoot>=N)
				{
					// 自明だと思うんだけどなぁ……
					continue;
				}
				// indexVec[i]をi=0から順に見ていく。
				for (int j = 0; j < N; j++)
				{
					if (weightTable[nowRoot,j]>=0 && rootWeight[j]<0)
					{
						// まだ探索していない木の行き先(j)を見つけた場合
						// 行き先の点の重みを計算
						rootWeight[j] = rootWeight[nowRoot] + weightTable[nowRoot, j];
						// 探索対象に追加
						if (lastIndex<N)
						{
							indexVec[lastIndex] = j;
							lastIndex++;
						}
					}
				}
			}
			// 出力
			for (int i = 0; i < N; i++)
			{
				if (rootWeight[i]%2==0)
				{
					Console.WriteLine(0);
				}
				else
				{
					Console.WriteLine(1);
				}
			}
		}
	}
}