Analysis: Wormholes, by Brian Dean

The process of solving this problem is described in the video above; the final code is shown below.

#include <iostream>
#include <fstream>
using namespace std;
#define MAX_N 12

int N, X[MAX_N+1], Y[MAX_N+1];
int partner[MAX_N+1];
int next_on_right[MAX_N+1];

bool cycle_exists(void)
{
  for (int start=1; start<=N; start++) {
    // does there exist a cylce starting from start
    int pos = start;
    for (int count=0; count<N; count++)
      pos = next_on_right[partner[pos]];
    if (pos != 0) return true;
  }
  return false;
}

// count all solutions
int solve(void) 
{
  // find first unpaired wormhole
  int i, total=0;
  for (i=1; i<=N; i++) 
    if (partner[i] == 0) break;

  // everyone paired?
  if (i > N) {
    if (cycle_exists()) return 1;
    else return 0;
  }

  // try pairing i with all possible other wormholes j
  for (int j=i+1; j<=N; j++)
    if (partner[j] == 0) {
      // try pairing i & j, let recursion continue to 
      // generate the rest of the solution
      partner[i] = j;
      partner[j] = i;
      total += solve();
      partner[i] = partner[j] = 0;
    }
  return total;
}

int main(void)
{
  ifstream fin("wormhole.in");
  fin >> N;
  for (int i=1; i<=N; i++) fin >> X[i] >> Y[i];
  fin.close();
  
  for (int i=1; i<=N; i++) // set next_on_right[i]...
    for (int j=1; j<=N; j++)
      if (X[j] > X[i] && Y[i] == Y[j]) // j right of i...
	if (next_on_right[i] == 0 ||
	    X[j]-X[i] < X[next_on_right[i]]-X[i])
	  next_on_right[i] = j;

  ofstream fout("wormhole.out");
  fout << solve() << "\n";
  fout.close();
  return 0;
}