(Analysis by Brian Dean)

This is a nice problem where the coding is straightforward once you have figured out the right structure to search for as an answer.

The abstract structure we are dealing with here is a tree --- a set of $n$ nodes connected by $n-1$ edges, where every node is reachable from every other node and there is no cycle. Trees are everywhere in computer science and you can expect to see them often in higher divisions.

Let's call a node with only incoming directed edges a "sink", since things can flow into it but not out. The short answer to the problem is that we need to see if our tree has a single unique sink, in which case this is the answer. Let's see if we can argue this:

1. If all nodes in the tree can reach node $x$, then we claim node $x$ must be the unique sink. All nodes aside from $x$ need at least one outgoing directed edge to be able to escape from them, so they aren't sinks. Further, node $x$ cannot have any outgoing edges, since if there was an edge $x \rightarrow y$ then $y$ couldn't reach $x$.

2. If $x$ is the unique sink, then all nodes in the tree can reach $x$. Suppose some node $y$ cannot reach $x$. We know $y$ has an outgoing edge since it isn't a sink, so let's follow such an edge. This lands us on another node (say, $z$) which if not a sink must also have an outgoing edge, so let's follow such an edge. If we keep following outgoing edges until we no longer can, we inevitably must get stuck at a sink, since this is the only node with no outgoing edges (and we can't go around in cycles since a tree has no cycles). This means we have reached $x$, since $x$ is the only sink.

My code for solving this problem is below.

#include <iostream>
#include <fstream>
using namespace std;

int N, incoming[101], outgoing[101];

int main(void)
{
ifstream fin ("factory.in");
fin >> N;
for (int i=0; i<N-1; i++) {
int a, b;
fin >> a >> b;
outgoing[a]++;
incoming[b]++;
}

ofstream fout ("factory.out");