Google Code Jam Archive — Qualification Round 2008 problems

The first task in the new Google Code Jam is about, no surprise, search engines, and also, the grandiose feat of saving the universe in a parsimonious way. However, putting all the fancies aside, the problem itself is easy.

List all queries one by one, and break them up into segments. Each segment will be an interval where we use one search engine, and when we cross from one segment to another we will switch the search engine we use. What can you say about each segment? Well, one thing for sure is:

Never ever ever have all S different search engines appear as queries in one segment. (*)

Why is this? Because if all S names appeared in one segment, then any search engine used for that segment will encounter at least one query that is the same as its name, thus exploding the universe!

Working in the opposite direction, (*) is all we need to achieve; as long as you can partition the list of queries into such segments, it corresponds to a plan of saving the universe. You don't even care about which engine is used for one segment; any engine not appearing as a query on that segment will do. However, you might sometimes pick the same engine for two consecutive segments, laughing at yourself when you realize it; why don't I just join the two segments into one? Because your task is to use as few segments as possible, it is obvious that you want to make each segment as long as possible.

This leads to the greedy solution: Starting from the first query, add one query at a time to the current segment until the names of all S search engines have been encountered. Then we continue this process in a new segment until all queries are processed.

Sample code in C++, where st is the set of queries in the current segment, q is the next query, and count is the number of switches.

st.clear();
count = 0;
for (int i=0; i<Q; i++) {
  getline(cin, q);
  if (st.find(q) == st.end()) {
    if (st.size() == S-1) {
      st.clear();
      count++;
    }
    st.insert(q);
  }
}

If st is a hashset, you expect the solution run in O(n) time. Note that this solution uses the fact that each query will be a search engine name and so we can ignore the list of names provided in the input.

Let us justify that the greedy approach always gives the optimal answer. Think of the procedure as Q steps, and we want to show that, for each i, there is (at least) one optimal choice which agrees with us on the first i steps. We do this inductively for i = 0, then i = 1, and so on. The proposition for i = Q when proven true will imply that our algorithm is correct.

So, the key points in the inductive step i:

1. If adding the next query will explode the universe, we must start a new segment. Any optimal choice agrees with us for the first (i-1) steps must also do that.
2. If adding the next query will not explode the universe, we do not start a new segment. We know there is an optimal solution R agreed with us for (i-1) steps. Even if in R a new segment is started at step i, we can modify it a little bit. Let R' be the plan that agrees with R, but instead of starting a new segment on the i-th step, we delay this to the (i+1)st. It is clear that R' will also make the universe safe, and has no more switches than R does. So, R' is also an optimal solution, and agrees with our choice for the first i steps.

The similar lines of justifications work for many greedy algorithms, including the well beloved minimum spanning trees.

This problem can be solved with a greedy strategy. The simplest way to do this is by scanning through a list of all the trips, sorted by departure time, and keeping track of the set of trains that will be available at each station, and when they will be ready to take a trip.

When we examine a trip, we see if there will be a train ready at the departure station by the departure time. If there is, then we remove that train from the list of available trains. If there is not, then our solution will need one new train added for the departure station. Then we compute when the train taking this trip will be ready at the other station for another trip, and add this train to the set of available trains at the other station. If a train leaves station A at 12:00 and arrives at station B at 13:00, with a 5-minute turnaround time, it will be available for a return journey from B to A at 13:05.

We need to be able to efficiently identify the earliest a train can leave from a station; and update this set of available trains by adding new trains or removing the earliest train. This can be done using a heap data structure for each station.

Sample Python code provided below that solves a test case for this problem:

def SolveCase(case_index, case):
  T, (tripsa, tripsb) = case
  trips = []
  for trip in tripsa:
    trips.append([trip[0], trip[1], 0])
  for trip in tripsb:
    trips.append([trip[0], trip[1], 1])

  trips.sort()

  start = [0, 0]
  trains = [[], []]

  for trip in trips:
    d = trip[2]
    if trains[d] and trains[d][0] <= trip[0]:
      # We're using the earliest train available, and
      # we have to delete it from this station's trains.
      heappop(trains[d])
    else:
      # No train was available for the current trip,
      # so we're adding one.
      start[d] += 1
    # We add an available train in the arriving station at the
    # time of arrival plus the turnaround time.
    heappush(trains[1 - d], trip[1] + T)

  print "Case #%d: %d %d" % (case_index, start[0], start[1])

Luckily Python has methods that implement the heap data structure operations. This solution takes O(n log n) time, where n is the total number of trips, because at each trip we do at most one insert or one delete operation from the heaps, and heap operations take O(log n) time.

This was the hardest problem of the set, as we can see from the submission statistics, but 652 people succeeded in solving it correctly.

The first step in solving this problem involves a standard trick. Instead of looking for the region that will hit the fly's circular body, we look for the region where the center of the fly must avoid. The problem is thus transformed into an equivalent one where we thicken the size of the racquet to t + f and the radius of the strings from r to r + f. And we consider the fly to be simply a single point, only focusing on its center.

Now the main problem is to find the area of the holes (the areas between the strings, or between the strings and the racquet), and divide it by the area of the disc corresponding to the original racquet. It is easy to see this gives the probability that the racquet does not hit the fly.

A simplifying observation is that we can just look at the first quadrant of the circle, because the racquet is symmetrical.

Now we can go through each square gap in the racquet and check a few cases. Each case consists in computing the area of intersection between the square and the circle representing the inside of the racquet. The case where the square is totally inside or totally outside the circle is easy. The cases left when one corner, two or three corners of the square are inside the circle and the other corners are outside. There are no other cases because we just look at the intersection of a quarter of the circle with small squares.

Each of these cases can be solved by splitting the intersection area into triangles and a circular segment. Here's some code that does this:

double circle_segment(double rad, double th) {
  return rad*rad*(th - sin(th))/2;
}

double rad = R-t-f;
double ar = 0.0;
for (double x1 = r+f; x1 < R-t-f; x1 += g+2*r)
for (double y1 = r+f; y1 < R-t-f; y1 += g+2*r) {
  double x2 = x1 + g - 2*f;
  double y2 = y1 + g - 2*f;
  if (x2 <= x1 || y2 <= y1) continue;
  if (x1*x1 + y1*y1 >= rad*rad) continue;
  if (x2*x2 + y2*y2 <= rad*rad) {
    // All points are inside circle.
    ar += (x2-x1)*(y2-y1);
  } else if (x1*x1 + y2*y2 >= rad*rad &&
             x2*x2 + y1*y1 >= rad*rad) {
    // Only (x1,y1) inside circle.
    ar += circle_segment(rad, acos(x1/rad) - asin(y1/rad)) +
          (sqrt(rad*rad - x1*x1)-y1) *
          (sqrt(rad*rad - y1*y1)-x1) / 2;
  } else if (x1*x1 + y2*y2 >= rad*rad) {
    // (x1,y1) and (x2,y1) inside circle.
    ar += circle_segment(rad, acos(x1/rad) - acos(x2/rad)) +
          (x2-x1) * (sqrt(rad*rad - x1*x1)-y1 +
                     sqrt(rad*rad - x2*x2)-y1) / 2;
  } else if (x2*x2 + y1*y1 >= rad*rad) {
    // (x1,y1) and (x1,y2) inside circle.
    ar += circle_segment(rad, asin(y2/rad) - asin(y1/rad)) +
          (y2-y1) * (sqrt(rad*rad - y1*y1)-x1 +
                     sqrt(rad*rad - y2*y2)-x1) / 2;
  } else {
    // All except (x2,y2) inside circle.
    ar += circle_segment(rad, asin(y2/rad) - acos(x2/rad)) +
          (x2-x1)*(y2-y1) -
          (y2-sqrt(rad*rad - x2*x2)) *
          (x2-sqrt(rad*rad - y2*y2)) / 2;
  }
}
printf("Case #%d: %.6lf\n", prob++, 1.0 - ar / (PI*R*R/4));

This solution takes O(S²) time, where S is the number of vertical strings of the racquet. It's not hard to come up with an O(S) solution because there are at most 4S border squares which can be found efficiently, but the previous solution was fast enough.

Instead of solving the problem exactly, an iterative solution which approximates the area to the needed precision would have also worked. One such solution uses divide and conquer by splitting the square into four smaller squares and then checking the simple cases where the squares are totally inside or totally outside the square. In the cases where the circle and square intersect just recurse if the current square is larger than some chosen precision. An observation is that we can divide every length by the radius of the racquet because it gets canceled in the division between the area of the gaps in the racquet and the disc area. This observation helps the iterative solution since we can make the number of iterations smaller. Here's some sample code:

double intersection(double x1, double y1,
                    double x2, double y2) {
  // the normalized radius is 1
  if (x1*x1 + y1*y1 > 1) {
    return 0;
  }
  if (x2*x2 + y2*y2 < 1) {
    return (x2-x1) * (y2-y1);
  }
  // EPS2 = 1e-6 * 1e-6
  if ((x2-x1)*(y2-y1) < EPS2) {
    // this trick helps in doing 10 or 100 times less
    // iterations than we would need to get the same
    // precision if we just return 0;
    return (x2-x1) * (y2-y1) / 2;
  }
 
  double mx = (x1 + x2) / 2;
  double my = (y1 + y2) / 2;
 
  return intersection(x1, y1, mx, my) +
    intersection(mx, y1, x2, my) +
    intersection(x1, my, mx, y2) +
    intersection(mx, my, x2, y2);
}