CF 269 E Flawed Flow

CF 269 E Flawed Flow

Emuskald considers himself a master of flow algorithms. Now he has completed his most ingenious program yet cipher 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 too much 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? V? N), 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? ¡U?N? ¡U? 2 Tib limit 105,N? -? 1? ¡U?M? ¡U? 2 rows between 105), the number of vertices and edges in the graph. The followingMLines contain three space-separated integersAI,BIAndCI(1? ¡U?AI,?BI? ¡U?N,AI? Too many?BI, 1? ¡U?CI? ¡U? 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 isAI? ~Ú?BI(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 :. <symbol · blank "http://www.bkjia.com/kf/ware/vc/" target = "_ blank" class = "keylink"> vcD4KPHA + Crj4s/bN + MLnwfe1xM7ez/keys/yoaPK18/IyejL + keys/keys + keys + CjxwPgo8cHJlIGNsYXNzPQ = "brush: java; "> # include # Include # Include # Include 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) 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 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; // Ç Â ¼ ± ß Æ flow ¼ µ ¶ ½ · ½ Ï ò 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 ;}