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 Match Editorial
SRM 207
Tuesday, August 10, 2004

Match summary

In division 2, today, coders were faced with a relatively simple easy problem, but many fell to a couple of special cases. The medium problem involved sorting of Strings based on a separate criteria, and proved to be a bit difficulty for both divisions. The division 2 hard problem was relatively typical, and turned out to be a simpler version of the division 1 hard. At the end of the competition, KelvinYe won with the help of two challenges. Not far behind was csd98412, in second place, and rrenaud was a distant third.

In division 1, coders were faced with an easier set than they were used to. Though the string manipulation of the easy problem tripped many people up, the medium problem required little more than standard brute force and the hard problem, though tricky, was well within reach. nicka81, despite losing his easy problem, found two challenges which enabled him win comfortably. Ryan was a little less than 100 points behind in second, and kalinov was a close third.

# The Problems

TransportCounting
Used as: Division Two - Level One:
 Value 250 Submission Rate 178 / 193 (92.23%) Success Rate 125 / 178 (70.22%) High Score 35C4P3 for 246.51 points (3 mins 23 secs) Average Score 206.17 (for 125 correct submissions)

If a bus starts ahead of us, then we will pass it if and only if we end up ahead of it or next to it after time minutes. Since all of the speeds are constant, there is no way for us to pass a bus but then end up behind it. We can calculate our final position as speed*time, and calculate each of the buses final positions as velocityi*time+positioni.

There is, however, an exception to this rule. If a bus starts at position 0, it must be counted, regardless of its velocity. Therefore, we must count a bus if it ends at a position less than or equal to our end position, or if it starts at position 0.

RegularSeason
Used as: Division Two - Level Two:
 Value 500 Submission Rate 83 / 193 (43.01%) Success Rate 45 / 83 (54.22%) High Score mmoll for 403.56 points (14 mins 38 secs) Average Score 239.71 (for 45 correct submissions)
Used as: Division One - Level One:
 Value 250 Submission Rate 163 / 170 (95.88%) Success Rate 110 / 163 (67.48%) High Score Yarin for 234.83 points (7 mins 18 secs) Average Score 165.54 (for 110 correct submissions)

The first thing to do in this problem is to calculate the expected number of wins for each team. Some coders might have been a little scared of the math here, but it's actually quite simple. If a team has a 50% chance of winning a game, then it is expected to win 0.5 games. If it has a 50% chance of winning one game, and a 20% chance of winning another game, then it is expected to win 0.5 + 0.2 = 0.7 games. In other words, you just add up all of the probabilities. As the notes suggested, however, it isn't a good idea to add up 0.5 and 0.2 because you might end up a little bit off due to the inexactness of floating point representation. Instead, you should add up 50 and 20, and wait until the end to divide by 100.

Once you calculate how many games each team is expected to win, it is just a matter of sorting. To accomplish this, you could either write your own comparator, and use the built in sorting functions of your favorite language, or just implement your own simple O(n2) sorting algorithm. Either way, you needed to be careful to sort by the expected number of wins prior to rounding, not the rounded values, and then you had to break ties alphabetically by team name.

CaptureThemAll
Used as: Division Two - Level Three:
 Value 1000 Submission Rate 33 / 193 (17.10%) Success Rate 14 / 33 (42.42%) High Score KelvinYe for 821.78 points (13 mins 52 secs) Average Score 643.95 (for 14 correct submissions)

Knight movement on a chessboard is a classic example of a search problem. There are a lot of different ways to solve it, but the two most well-known approaches are the Breadth-First Search (BFS) and the Depth-First Search (DFS). In most cases, a BFS is faster and if you can afford the extra memory it requires, it is generally preferable.

In a BFS, you typically maintain a First-In First-Out queue, which you initialize to contain a single location - the starting point. You also maintain a mapping that keeps track of how far it is to each location. Initially, you set the distance to every point to infinity, except for the starting point, whose distance you set to 0. Then, you start removing locations from your queue. Each time you remove a location from the queue, you look at all of the locations that can be reached from that location. For each such location that has a distance of infinity, you add that location to the queue and update its distance. Eventually, all of the reachable locations will have a distance assigned to them, and the queue will be empty, at which point you are done.

In this problem, our mapping can be a simple 2D array, with an element for each spot on the chessboard. There are a lot of different ways to implement the queue, but in problems like this I always like to use a 2D array for that as well. The queue array will have 64 rows, and 2 columns - one column for X, and one for Y. We will also have two integers, head and tail, telling us which row is the head of the queue, and which one is the tail of the queue. Initially, we will set the first row of the queue to be our starting location, head to 0, and tail to 1:

```    int distances[8][8];
int queue[64][2];
queue[0][0] = startX;
queue[0][1] = startY;
for (int i = 0; i<8; i++){
for (int j = 0; j<8; j++){
distances[i][j] = -1;//Using -1 for infinity
}
}
distances[startX][startY] = 0;
int head = 0;
int tail = 1;
while(tail > head){
int x = queue[head][0];
int y = queue[head][1];
head++;
foreach ((x',y') reachable from (x,y)){
if(distances[x'][y'] == -1){
distances[x'][y'] = distances[x][y] + 1;
queue[tail][0] = x';
queue[tail][1] = y';
tail++;
}
}
}
```
Now, we can calculate the distance from any location on the board to any other location, so we figure out the distances from the knight's initial location to both the rook and to the queen, as well as the distance from the rook to the queen (which is equal to the distance from the queen to the rook). With these distances, we just consider the two possible orders in which we can capture the two pieces, and return the minimum distance to get them both.

TCSocks
Used as: Division One - Level Two:
 Value 500 Submission Rate 80 / 170 (47.06%) Success Rate 62 / 80 (77.50%) High Score Yarin for 393.46 points (15 mins 41 secs) Average Score 275.34 (for 62 correct submissions)

A quick glance at the constraints for this problem reveals that there are at most 10 cities. Since you have to start and finish at city 0, there are only 9! possible routes you could take - few enough to consider them all.

The first thing that I did was to parse the input into something more usable. I parsed cost and time into 2D int arrays, and then I made a new 2D array, prev, so I could quickly look up now many competing salesmen had visited a city at a given time. I set it up so that prev[i][j] told me how many people had visited city j by time i, noting that I would never need to look up times greater than about 100.

Then, it was just a matter of implementing a brute force recursive algorithm. The basic idea here is that the recursive function considers heading to each of the cities that haven't been visited next:

```    recurse(boolean[] visited, int cur, int tm){
int maxProfit = (money[cur]>>prev[tm][cur])-cost[cur][0];
foreach (city i != 0){
if(!visited[i]){
visited[i] = true;
int profit = recurse(visited,i,tm+time[cur][i]);
maxProfit = max(maxProfit, profit +
(money[i]>>prev[tm][cur]) - cost[cur][i]);
visited[i] = false;
}
}
return maxProfit;
}
```
Note that >> is the shift right operator, and does the same thing as dividing by powers of 2. As a challenge, try to implement an algorithm that runs under the time limit even when there are 20 cities.

GetThemAll
Used as: Division One - Level Three:
 Value 1000 Submission Rate 16 / 170 (9.41%) Success Rate 8 / 16 (50.00%) High Score nicka81 for 720.24 points (19 mins 21 secs) Average Score 569.29 (for 8 correct submissions)

This problem is very similar to the medium in that it could also be solved using a brute force approach. In fact, it is an instance of the famous traveling salesman problem, and can be solved similarly. However, the difficulty here lies in finding the distances between two locations in terms of knights moves. Without any loss of generality, lets say that the knight starts at (0,0) and must go to (x,y), where both x and y are positive. We can establish a lower bound on the number of moves by observing that we never move more than 2 at a time, so it takes at least ceil(x/2) moves and at least ceil(y/2) moves. Also, we only move a total of 3 moves at a time, so it takes at least ceil((x+y)/3) moves. However, these are only approximations, and we'll need to be a bit more clever to get the actual numbers. Luckily, a mostly greedy approach will work. We can use a greedy algorithm to get close to our destination, and then use a BFS (as in the div 2 hard) to finish up in as few moves as possible. This works because when the knight is far away from its destination, it is easy to predict the direction that most of its moves will be in. For example, if the knight is going to (247,201), we can approximate that about 2/3 of its moves will be in the (+2,+1) direction, while 1/3 of its moves will be in the (+1,+2) direction. There will be a few moves that are not in either of those directions, but not very many, so we can figure them out with our BFS at the end.

Now, there are a bunch of different approaches we can take to the greedy part of the problem, and I'll only outline a couple of them. The first is to always move in the direction that takes us closest to the line between the two points. This approach follows the general principle that the shortest path between two points is a straight line, but it requires a bit more work than a simpler approach with a similar idea. Here, we work backwards from (x,y), and always take the move (-2,-1) or (-1,-2) that brings us closer to the diagonal x==y. If we get to the x or y axis while following this approach, we know that we can move 4*d along the axis in 2*d moves. Eventually, we will get reasonably close to the origin, and then we use our BFS. How close should we get before we switch to a BFS though? Switching to a BFS as soon as both coordinates get to be less than 30 will certainly work, and 10 will work also. You could easily run your BFS all the way out to 1000 or so though, if you were really nervous.

Another approach is to consider a few special cases where the number of moves required can be computed directly. Let's say that y is greater than x. We could make floor(y/2) moves in the (+1,+2) direction and y mod 2 moves in the (+2,+1) direction. This will bring us an x position of floor(y/2) + (y mod 2) * 2. If this position is greater than x, by g and g mod 2 == 0, then we can convert g/2 of the (+1,+2) moves into (-1,+2) moves, and we would end up at the right place with a minimum number of moves. Alternatively, if floor(y/2) + (y mod 2) * 2 is less than x by g, and g mod 3 == 0 we can convert g/3 of the (+1,+2) moves into (+2,+1) moves and then add g/3 more (+2,+1) moves. This will increase x by g and keep y the same. By combining this closed form approach with a BFS, we can figure out the distance between any two points in constant time.

It turns out that, when the destination is far away, we can always get to one of these special cases in at most one move. From here, we can actually eliminate the BFS portion of the code, as marian did in room 3 (though he looked up the formula). He failed only due to an input with no pieces, and though it isn't obvious that his f() function will calculate the distance correctly, it does.

By lbackstrom
TopCoder Member