如何为两支球队相互比赛找到最佳解决方案？

Question

I am given a table of teams A and B where for each pair of 2 players there is number. The rows represent players of players of team A and columns of players of the team B. If a number is positive, it means that the player A is better than the plyaer from the B team and vice versa if negative.

For example:

-710 415 527 -641 175 48
-447 -799 253 626 304 895
509 -523 -758 -678 -689 92
24 -318 -61 -9 174 255
487 408 696 861 -394 -67

Both teams know this table. Now, what is done is that the team A reports 5 players, the team B can look at them and choose the best 5 players for them. If we want to compere the teams we sum up the numbers on the given positions from the table knowing that each team has a captain who is counted twice (as if a team had 6 players and the captain is there twice), if the sum is positive, the team A is better.

The input are numbers a (the number of rows/players A) and b (columns/players B) and the table like this:

6
6
-54 -927 428 -510 911 93
-710 415 527 -641 175 48
-447 -799 253 626 304 895
509 -523 -758 -678 -689 92
24 -318 -61 -9 174 255
487 408 696 861 -394 -67

The output should be 1282.

So, what I did was that I put the numbers into a matrix like this:

a, b = int(input()), int(input())

matrix = [list(map(int,input().split())) for _ in range(a)]

I used a MinHeap and a MaxHeap for this. I put the rows into the MaxHeap because A team wants the biggest, then I get 5 best A players from it as follows:

for player, values in enumerate(matrix):
    maxheap.enqueue(sum(values), player)

playersA = []
overallA = 0

for i in range(5):
    ov, pl  = maxheap.remove_max()
    if i == 0: # it is a captain
        playersA.append(pl)
        overallA += ov
        
    playersA.append(pl)
    overallA += ov

The B team knowing the A players the uses the MinHeap to find its best 5 players:

for i in range(b):
    player = []
    ov = 0
    for j in range(a): #take out a column of a matrix
        player.append(matrix[j][i])


    for rival in playersA: #counting only players already chosen by A
        ov += player[rival]

    minheap.enqueue(ov,i)

playersB = []
overallB = 0

for i in range(5):
    ov, pl = minheap.remove_min()
    if i == 0:
        playersB.append(pl)
        overallB += ov
        
    playersB.append(pl)
    overallB += ov

Having the players, then I count the sum from the matrix:

out = 0
for a in playersA:
    for b in playersB:
        out += matrix[a][b]
print(out)

However, this solution doesn't give the right solutions always. For example, it does for the input:

10
10
-802 -781 826 997 -403 243 -533 -694 195 182
103 182 -14 130 953 -900 43 334 -724 716
-350 506 184 691 -785 742 -303 -682 186 -520
25 -815 475 -407 -78 509 -512 714 898 243
758 -743 -504 -160 855 -792 -177 747 188 -190
333 -439 529 795 -500 112 625 -2 -994 282
824 498 -899 158 453 644 117 598 432 310
-799 594 933 -15 47 -687 68 480 -933 -631
741 400 979 -52 -78 -744 -573 -170 882 -610
-376 -928 -324 658 -538 811 -724 848 344 -308

But it doesn't for

11
11
279 475 -894 -641 -716 687 253 -451 580 -727 -509
880 -778 -867 -527 816 -458 -136 -517 217 58 740
360 -841 492 -3 940 754 -584 715 -389 438 -887
-739 664 972 838 -974 -802 799 258 628 3 815
952 -404 -273 -323 -948 674 687 233 62 -339 352
285 -535 -812 -452 -335 -452 -799 -902 691 195 -837
-78 56 459 -178 631 -348 481 608 -131 -575 732
-212 -826 -547 440 -399 -994 486 -382 -509 483 -786
-94 -983 785 -8 445 -462 -138 804 749 890 -890
-184 872 -341 776 447 -573 405 462 -76 -69 906
-617 704 292 287 464 -711 354 428 444 -42 45

So the question is: Can it be done like this or is there another fast algorithm ( O(n ** 2 ) / O(n ** 3) etc.), or I just gave to try all the possible combinations using brute force in O(n!) time complexity?

Answer 1

There is a way to do that with a polynomial complexity.

To show you why your solution doesn't work, let's consider an other simpler problem. Let's say each team only choose 2 players and there is no captain.

Let's also take a simple score matrix:

1 1 1 2 1
1 1 1 1 1
0 3 0 2 0
0 0 0 0 4
0 0 0 0 4

Here you can see that team A has no chance to win (as there are no negative numbers), but still they are going to try their best. Who should they pick?

Using your algorithm, team A should pick their best players and their ranking would be:

pa0 < pa1 = pa2 < pa3 = pa4

If they choose pa3 and pa4, who both have a score of 4 (which is bad, but not as bad as pa0's score of 6), team B will win by 8 (they will choose pb4 and an other player who doesn't matter).

On the other hand, if team A chose pa0 and pa1 (who are worse than pa3 and pa4 by your metric), the best team B can get is winning by 5 (if they choose pb3 and any other player)

Basically, your approximation fails to take into consideration that team B can only choose two players and thus can't take advantage of the pa0+pa1 weakness while it can easily exploit pa3+pa4's one.

A better solution would be for team A to evaluate each player's score only by taking into account their 2 worst scores (or 5 if 5 players are to be selected): this would make the ranking as such:

pa2 < pa3 = pa4 < pa0 < pa1

Still it would be an approximation: some combinations like pa2+pa3 are actually not as bad as they sound as, once again, the weaknesses are spread enough that team B can't exploit them all (although for this very example the approximation yields the best result).

What we really need to pick is not the two best players, but the best combination of two players, and sadly there is no way I know of other than trying all the $s!/(k!(sk)!)$ combinations of k players among s (the size of the team). It is not so bad, though, as for k=2 that's only $s*(s-1)/2$ and for k=5 that's $s*(s-1) (s-2) (s-3)*(s-4)/5!$, which is still polynomial in complexity despite being in O(s^5). Adding a captain to the mix only multiplies the number of combinations by k. It also requires a twist on how to calculate the score but you should be able to find that.

Now that team A have selected their players, team B have the easy job to select theirs. This is way simpler as here each player can be chosen individually.

---

example of how this last algorithm should work with the score matrix provided in the beginning.

team A has 10 possible combinations: pa0+pa1, pa0+pa2, pa0+pa3, pa0+pa4, pa1+pa2, pa1+pa3, pa1+pa4, pa2+pa3, pa2+pa4, pa3+pa4. Their respective scores are: 5, 8, 7, 7, 7, 6, 6, 7, 7, 8.

The best combination is pa0+pa1, so that's what they send to team B.

Team B calculate each of its player's score against pa0+pa1: pb0:2, pb1:2, pb2:2, pb3:3, pb4:2. pb3 is the best, all the others are equals, thus team B sends pb3+pb4 (for example), and the "answer" is 5.