网络流相关（拓扑）CodeForces 269C：Flawed Flow

Emuskald considers himself a master of flow algorithms. Now he has completed his most ingenious program yet — it calculates the maximum flow in an undirected graph. The graph consists ofn vertices andm edges. Vertices are numbered from 1 ton. Vertices1 andn being the source and the sink respectively.

However, his max-flow algorithm seems to have a little flaw — it only finds the flow volume for each edge, but not its direction. Help him find for each edge the direction of the flow through this edges. Note, that the resulting flow should be correct maximum flow.

More formally. You are given an undirected graph. For each it's undirected edge (ai,bi) you are given the flow volumeci. You should direct all edges in such way that the following conditions hold:

1. for each vertexv(1 < v < n), sum ofci of incoming edges is equal to the sum ofci of outcoming edges;
2. vertex with number1 has no incoming edges;
3. the obtained directed graph does not have cycles.

Input

The first line of input contains two space-separated integersn andm (2 ≤ n ≤ 2·105,n - 1 ≤ m ≤ 2·105), the number of vertices and edges in the graph. The followingm lines contain three space-separated integersai,bi andci (1 ≤ ai, bi ≤ n,ai ≠ bi,1 ≤ ci ≤ 104), which means that there is an undirected edge fromai tobi with flow volumeci.

It is guaranteed that there are no two edges connecting the same vertices; the given graph is connected; a solution always exists.

Output

Outputm lines, each containing one integerdi, which should be 0 if the direction of thei-th edge isai → bi (the flow goes from vertexai to vertexbi) and should be 1 otherwise. The edges are numbered from 1 tom in the order they are given in the input.

If there are several solutions you can print any of them.

Sample Input

Input

3 3
3 2 10
1 2 10
3 1 5

Output

1
0
1

Input

4 5
1 2 10
1 3 10
2 3 5
4 2 15
3 4 5

Output

0
0
1
1
0
　　可以发现这里有拓扑性质，可以直接做，O（N）复杂度。

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 using namespace std;
 6 const int N=200010,M=400010;
 7 int cnt=1,fir[N],nxt[M],to[M],cap[M];
 8 int n,m,in[N],vis[N],ans[N];queueq;
 9 void addedge(int a,int b,int c){
10     nxt[++cnt]=fir[a];
11     to[fir[a]=cnt]=b;
12     cap[cnt]=c;
13 }
14 int main(){
15     scanf("%d%d",&n,&m);
16     for(int i=1,a,b,c;i<=m;i++){
17         scanf("%d%d%d",&a,&b,&c);
18         addedge(a,b,c);addedge(b,a,c);
19         in[a]+=c;in[b]+=c;
20     }
21     for(int i=2;i