CF 269 E Flawed Flow, 269 flawed

CF 269 E Flawed Flow, 269 flawed

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 consistsNVertices andMEdges. Vertices are numbered from 1N. Vertices 1 andNBeing 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 shocould 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 shoshould direct all edges in such way that the following conditions hold:

1. For each vertexV(1 hour VLatency <latencyN), SumCIOf incoming edges is equal to the sumCIOf outcoming edges;
2. Vertex with number 1 has no incoming edges;
3. The obtained directed graph does not have cycles.

Input

The first line of input contains two space-separated integersNAndM(2 cores ≤ CoresNLimit ≤ limit 2 · 105,NExecutor-cores 1 ≤ CoresMLimit ≤ limit 2 · 105), the number of vertices and edges in the graph. The followingMLines contain three space-separated integersAI,BIAndCI(1 digit ≤ DigitAI, Bytes,BILimit ≤ limitN,AI  =BI, 1 limit ≤ limitCILimit ≤1_104), which means that there is an undirected edge fromAIToBIWith 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

OutputMLines, each containing one integerDI, Which shocould be 0 if the direction ofI-Th edge isAINavigation → navigationBI(The flow goes from vertexAITo vertexBI) And shoshould be 1 otherwise. The edges are numbered from 1MIn the order they are given in the input.

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

Sample test (s) input

3 33 2 101 2 103 1 5

Output

101

Input

4 51 2 101 3 102 3 54 2 153 4 5

Output

00110

Note

In the first test case, 10 flow units pass through path, and 5 flow units pass directly from source to sink :.

The following figure shows the traffic of the network flow in an undirected graph. First, set flow [I] [0] for all vertices to indicate the outbound direction.

It can be thought that if a point meets the following requirements: flow [I] [[0] = flow [I] [1], the direction can be determined.

BFS practice: from known to continuous BFS, to a point: flow [I] [[0] = flow [I] [1], add to the queue ..

# Include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <queue> typedef long LL; using namespace std; # define REPF (I, a, B) for (int I = a; I <= B; ++ I) # define REP (I, n) for (int I = 0; I <n; ++ I) # define CLEAR (a, x) memset (a, x, sizeof a) const int maxn = 2*(1e5 + 100); struct node {int u, v, w; int num, next;} e [maxn <1]; int vis [maxn], head [maxn]; int pre [maxn ], S [maxn]; int flow [maxn] [2]; int n, m, cnt; void add (int x, int y, int z, int n) {e [cnt]. u = x; e [cnt]. v = y; e [cnt]. w = z; e [cnt]. num = n; e [cnt]. next = head [x]; head [x] = cnt ++;} void BFS () {CLEAR (vis, 0); queue <int> q; q. push (1); vis [1] = 1; while (! Q. empty () {int st = q. front (); q. pop (); for (int I = head [st]; I! =-1; I = e [I]. next) {int v = e [I]. v; int num = e [I]. num; int w = e [I]. w; if (vis [v]) continue; s [num] = st; // flow [v] [1] + = w; flow [v] [0]-= w; if (v! = N & flow [v] [1] = flow [v] [0]) {vis [v] = 1; q. push (v) ;}}} int main () {int x, y, z; std: ios: sync_with_stdio (false ); while (cin> n> m) {CLEAR (head,-1); CLEAR (flow, 0); cnt = 0; REPF (I, 1, m) {cin> x> y> z; pre [I] = x; add (x, y, z, I); add (y, x, z, i); flow [x] [0] + = z; flow [y] [0] + = z;} BFS (); REPF (I, 1, m) {if (pre [I] = s [I]) puts ("0"); else puts ("1") ;}} return 0 ;}