Author: Matthias Gemuh
Date: 08:07:39 12/26/05
nEng = number of engines
int nTeams[2*nEng], nPairing[nEng][nEng], k, x, y, nNr=0;
for (int t = 0; t < nEng; t++) {
for (int i = 0; i < nEng; i++) {
x = nTeams[i]; y = nTeams[i+nEng];
if (x == y) continue;
if (nPairing[x][y] == 0) nPairing[x][y] = nPairing[y][x] = nNr++;
}
k = nTeams[0]; for (int j = 0; j < nEng; j++) nTeams[j-1] = nTeams[j];
nTeams[nEng-1] = k;
}
I figured out this algo above, but it is not perfect.
Please, give me the perfect algorithm.
Thanks,
Matthias.
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