I have prepared this problem some time ago but recently I realized beside my Author solution, another solution works but I just couldn't prove it. I need help on proving the correctness of this unexpected solution.
The problem
I have prepared a problem as follow: There is this secret sequence $$$S$$$ consists of $$$N$$$ distinct numbers. Given a matrix $$$G$$$ of size $$$N \times N$$$, where $$$G_{i,j}=$$$'\t{Y}' if $$$S_i>S_j$$$. Otherwise, $$$G_{i,j}=$$$'\t{?}', meaning we have no information about the relation of $$$S_i$$$ and $$$S_j$$$, whether they are larger or smaller. You need to output a sequence $$$P = {P_1, P_2, ..., P_n}$$$, such that $$$S_{P_1} > S_{P_2} > ... > S_{P_n}$$$, or report that it is impossible to determined exactly the order.
Example
\begin{array}{|c|c|c|} \hline G & 1 & 2 & 3 & 4 \cr \hline 1 & ? & ? & ? & ? \cr \hline 2 & ? & ? & Y & Y \cr \hline 3 & Y & ? & ? & Y \cr \hline 4 & Y & ? & ? & ? \cr \hline \end{array} The above input yeilds the answer $$$P={2,3,4,1}$$$ which corresponds to $$$S_2 > S_3 > S_4 > S_1$$$. This is the only possible answer.
Constraints:
- $$$(1 \leq N \leq 400)$$$
- Time limit: 1s
The intended solution
This problem can be solved by turn it into the All Pairs Shortest Path problem. First, I create a cost matrix $$$C$$$ of size $$$N \times N$$$. $$$C_{i,j} = 0$$$ if $$$G_{i,j}$$$ equals to '\t{Y}', otherwise, $$$C_{i,j} = \infinite$$$. Then I run Floyd-Warshall on $$$C$$$, which produces a "complete" matrix $$$C$$$. After that, if $$$S_i$$$ is the largest element then row $$$C_{i}$$$ will have exactly $$$N-1$$$ zeros, second largest will have $$$N-2$$$ zeros, and so forth.
If the above is not possible, then there is no answer for that input.
The unexpected solution
The solution is as follow: cpp for (int round = 0; round < 2; round++) { for (int a = 0; a < n; a++) for (int b = 0; b < n; b++) if (G[a][b] == 'Y') for (int c = 0; c < n; c++) if (G[b][c] == 'Y') G[a][c] = 'Y'; } It needs at max 2 rounds to completely fill the $$$G$$$ matrix, after that I only need to counts the number of '\t{Y}' on each row to reconstruct the sequence $$$S$$$.
Observation
- For sequence $$$S={S_1 > S_2 > S_3 > ... > S_n}$$$ and the its reverse sequence, it only needs 1 round.
- The above requires only 1 run maybe because I traverse all tuples in the "correct" order. For other cases, sequence $$$S$$$ is a permutation of the above and because I go through all possible ordered tuples, there will be a moment that I go into the "correct traverse order" for the current sequence $$$S$$$, but have not finished it. So it needs another round to complete the traversing... Maybe.
