Processing math: 100%

BOTTOM - The Bottom of a Graph

no tags 

We will use the following (standard) definitions from graph theory. Let V be a nonempty and finite set, its elements being called vertices (or nodes). Let E be a subset of the Cartesian product V×V, its elements being called edges. Then G=(V,E) is called a directed graph.

Let n be a positive integer, and let p=(e1,,en) be a sequence of length n of edges eiE such that ei=(vi,vi+1) for a sequence of vertices (v1,,vn+1). Then p is called a path from vertex v1 to vertex vn+1 in G and we say that vn+1 is reachable from v1, writing (v1vn+1).

Here are some new definitions. A node v in a graph G=(V,E) is called a sink, if for every node w in G that is reachable from v, v is also reachable from w. The bottom of a graph is the subset of all nodes that are sinks, i.e., bottom(G)={vVwV:(vw)(wv)}. You have to calculate the bottom of certain graphs.

Input Specification

The input contains several test cases, each of which corresponds to a directed graph G. Each test case starts with an integer number v, denoting the number of vertices of G=(V,E), where the vertices will be identified by the integer numbers in the set V={1,,v}. You may assume that 1v5000. That is followed by a non-negative integer e and, thereafter, e pairs of vertex identifiers v1,w1,,ve,we with the meaning that (vi,wi)E. There are no edges other than specified by these pairs. The last test case is followed by a zero.

Output Specification

For each test case output the bottom of the specified graph on a single line. To this end, print the numbers of all nodes that are sinks in sorted order separated by a single space character. If the bottom is empty, print an empty line.

Sample Input

3 3
1 3 2 3 3 1
2 1
1 2
0

Sample Output

1 3
2

hide comments
badry atef: 2015-08-05 15:26:40

a very nice SCC problem :)

Medo: 2015-08-05 13:19:05

It's impossible for a graph to have no sink nodes. Don't account for that case if you are getting a WA.

Ankit Kumar: 2015-07-16 21:07:13

O(V+E) solution -> AC
Best DFS prob solved by me on SPOJ till now :)

i_am_looser: 2015-06-06 21:27:19

Good question..... learnt something useful

Arafat dad Khan: 2015-06-03 17:57:47

Great problem for learning new things

Arya08: 2015-05-29 20:39:00

best dfs problem. solved by me(till now )

ehacker: 2015-05-26 22:53:03

My testcases are correct even the ones which are mentioned in the comments, don't know, i am using a do while loop and output the value as i calculated for that case????

eli: 2015-03-11 16:18:36

2 can't reach 1 so it works.

Archit Jain: 2014-12-26 20:13:55

seems easy but difficult to implement

|RAMSDEN|: 2014-12-25 21:09:22

A silly mistake costed me many WAs :p


Added by:Wanderley Guimarăes
Date:2007-09-21
Time limit:1s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ERL JS-RHINO
Resource:University of Ulm Local Contest 2003