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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

//節点uからでる枝の構造体の定義
typedef struct slobj_{
    struct slobj_ *next;
    int v; //uから枝が伸びている節点v
    int d; //枝(u,v)の重み
}* slobj;

//slobj型のリストの定義
typedef struct{
    slobj head;
    slobj tail;
}* slist;

//グラフの構造体の定義
typedef struct{
    int n; //節点数
    slist* E; //枝リストの配列E[0]~E[n-1],未ソート単方向
}* graph;

//vにn,dにxを持つslobj型のリスト要素のメモリ確保+作成する関数
slobj slobj_new(int n,int x){
    slobj p;
    NEW(p,1);
    p->v=n;
    p->d=x;
    p->next=NULL;
    return p;
}

//slist型のメモリを1個確保し、空リスト1つを作る関数
slist slist_new(void){
    slist L;
    NEW(L,1);
    L->head=NULL;
    L->tail=NULL;
    return L;
}

//各slistを空リストにした,節点nのgraph型のメモリを確保する関数(Eは0カウント)
graph graph_new(int n){
    graph G;
    NEW(G,1);
    
    G->n=n;
    
    NEW(G->E,G->n);
    int i;
    for(i=0;i<G->n;i++){
        G->E[i]=slist_new();
    }
    
    return G;
}

//リストLの末尾に要素rを追加する関数
void slist_append_tail(slist L,slobj r){
    if(L->head==NULL){
        L->head=r;
        L->tail=r;
    }
    else{
        L->tail->next=r;
        L->tail=r;
    }
    
    return;
}

//節点i(0~n-1)から節点j(0~n-1)に、重さxの枝を追加する関数
void add_edge(graph G, int i, int j, int x){
    //枝データを追加
    slobj p=slobj_new(j,x);
    slist_append_tail(G->E[i],p);

    return;
}

//根(節点i)からの長さをdfs
void dfs(graph G, int i, int* d){
	slobj p=G->E[i]->head;
	while(p!=NULL){
		if(d[p->v]==-1){ //未探索なら
			d[p->v]=d[i]+p->d;
			dfs(G,p->v,d);
		}
		p=p->next;
	}
	return;
}
 
int main(void) {
	int N;
	scanf("%d", &N);

	graph G = graph_new(N);

	int u,v,w;
	for(int i=1;i<=N-1;i++){
		scanf("%d%d%d",&u,&v,&w);
		add_edge(G,u-1,v-1,w);
		add_edge(G,v-1,u-1,w);
	}

	//節点iの根(0)からの深さを格納(初期値-1)
	int d[N];
	for(int i=0;i<N;i++) d[i]=-1;
	d[0]=0;

	dfs(G,0,d);

	for(int i=0;i<N;i++){
		printf("%d\n",d[i]%2);
	}

	return 0;
}