BOTTOM - The Bottom of a Graph

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 $V \times V$, its elements being called edges. Then $G = (V, E)$ is called a directed graph.

Let $n$ be a positive integer, and let $p = (e_1, \ldots, e_n)$ be a sequence of length $n$ of edges $e_i \in E$ such that $e_i = (v_i, v_{i+1})$ for a sequence of vertices ($v_1, \ldots, v_{n+1}$). Then $p$ is called a path from vertex $v_1$ to vertex $v_{n+1}$ in $G$ and we say that $v_{n+1}$ is reachable from $v_1$, writing $(v_1 \to v_{n+1})$.

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., $\mathrm{bottom}(G) = \{v \in V \mid \forall w \in V : (v \to w) \Rightarrow (w \to v) \}$. 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 $V = \{1, \ldots, v\}$. You may assume that $1 \le v \le 5000$. That is followed by a non-negative integer $e$ and, thereafter, $e$ pairs of vertex identifiers $v_1, w_1, \ldots, v_e, w_e$ with the meaning that $(v_i, w_i) \in E$. 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

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

hide comments
2016-06-29 17:28:04
@ayush because if a vertex u is directed towards another vertex v (i.e. u->v), where u and v belong to different SCCs, then
1. u definitely cannot belong to a sink ( or bottom)
2. every vertex belonging to the same SCC as u cannot belong to a sink because we know that u is connected to every other vertex in that SCC, and hence we can reach v from each of those vertices.
for example, in your case,
1. 2->3 => 2 cannot be a sink
2. since 1->2 and 2->1 => (2,1) forms an SCC, and if we can reach 3 from 2, we can definitely reach 3 from 1 (via 2)
Hope it helped!!
2016-06-29 16:09:49 ayush
if a vertex belongs to one component and has a neighbour of other component (by component i mean a SCC group), why the whole SCC group of that vertex is discarded, as given by test case:
3 3
1 2 2 1 2 3
0
output:
3
Please explain.Thanks in advance.
2016-06-26 21:56:03
Finally! I got ac.. But I was getting WA when I was using low[] values as id for an SCC (yeah! u guessed right, I was using Tarjan's Algo for finding SCC). I was assuming that low value will be different for each SCC. Can anyone explain where I went wrong!!!
2016-06-26 19:04:21 Abhishek Naik
@kshubham02: Your comment was really very helpful. Got my code accepted! Thank you! :)

Last edit: 2016-06-26 19:04:57
2016-06-25 20:10:46
@deerishi If u're suggesting this case:
Input:
1 1
1 1
my code gives
Output:
1
and I guess this is right!!!
2016-06-25 01:19:56 deerishi
Nice Problem. For those getting WA consider the case of a single vertex with a self loop.
2016-06-23 22:37:38
Please help me! I'm constantly getting WA. I've tried all the given testcases (including those in comments!!!). Help me find bugs in my last submission...
2016-06-12 22:37:30
@Saumye Malhotra it looks like you haven't understood the question well. There is no concept of 'Multiple bottoms'. Please look at my reply to Abhishek Naik's comment. Hope it is helpful. Cheers :)
2016-06-12 19:39:34
@Abhishek Naik You've got it wrong. In case 2, there's only one edge, going from 1->2. So, for 2, there are no nodes reachable from it. Hence it is a sink. For 1, there is one node reachable from it (2). However, from 2 you cannot reach back to 1. Hence, 1 is not a sink.
In TC1, from 1 you can reach only 3 and from 3 you can reach 1. Hence 1 is a sink.
From 2 you can reach 3 but from 3 you cannot reach 2. Hence 2 is not a sink.
From 3 you can reach only 1 and from 1 you can reach back to 3. Hence 3 is also a sink.
2016-04-22 15:31:43 Abhishek Naik
I didn't understand the 2nd test case. There is an edge from 1->2 and implicitly from 2->2. Going by this implication, in the first test case, we should also have had 1->1, 2->2 and 3->3, in addition to 1->3 and 3->1. Could someone please explain this to me?
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