我正在尝试为足球联赛制作一个夹具列表。我已设法生成实际的夹具列表，没有任何问题，但我现在正试图生成一个“周”，游戏将被播放。这只需要“1到n”，没有其他日期信息，只需一周号。
我在MySQL中创建的表格如下：
Home Away GameID WeekID
Team 2 Team 1 1 0
Team 3 Team 1 2 0
Team 3 Team 2 3 0
Team 4 Team 1 4 0
Team 4 Team 2 5 0
Team 4 Team 3 6 0
Team 1 Team 2 7 0
Team 1 Team 3 8 0
Team 2 Team 3 9 0
Team 1 Team 4 10 0
Team 2 Team 4 11 0
Team 3 Team 4 12 0
每支球队都必须在主场和客场比赛，因此在一些比赛中会出现重复。但是，如前所述，我需要做的是为游戏分配一个周数，一个团队每周只能玩一次。
我想要创建的是：
Home Away GameID WeekID
Team 2 Team 1 1 1
Team 3 Team 1 2 2
Team 3 Team 2 3 3
Team 4 Team 1 4 4
Team 4 Team 2 5 5
Team 4 Team 3 6 6
Team 1 Team 2 7 6
Team 1 Team 3 8 5
Team 2 Team 3 9 4
Team 1 Team 4 10 3
Team 2 Team 4 11 2
Team 3 Team 4 12 1
对此有任何帮助将不胜感激。
答案 0 :(得分：4)
以下解决方案正常工作，但我认为它不会像非二人权力的团队那样有效地填补时间表。对于所有情况，代码的效率也是n ^ 2 [可能是n ^ 3？]，因此我希望您不需要一次安排超过几百个团队。 ：P
<?php
$teams = $_GET['t'];
$games = array(); //2D array tracking which week teams will be playing
$weeks = array(); //2D array tracking which teams are playing in a given week
// initialize
for( $i=0; $i<$teams; $i++ ) {
$games[$i] = array();
for( $j=0; $j<$teams; $j++ ) {
if( $i == $j ) { $games[$i][$j] = -1; } //you can't play with yourself ;D
else { $games[$i][$j] = NULL; }
}
}
// do the work
for( $w=1, $noblanks=false; !$noblanks; $w++) {
if( !isset($weeks[$w]) ) { $weeks[$w] = array(); }
$noblanks = true; //begin assuming there are no blank spots in the matrix
for( $i=0; $i<$teams; $i++ ) {
for( $j=0; $j<$teams; $j++ ) {
if( $i == $j ) { continue; } //you can't play with yourself ;D
if( is_null($games[$i][$j]) ) {
if( !isset($weeks[$w][$i]) && !isset($weeks[$w][$j]) ) {
$games[$i][$j] = $w; //game between team i and j in week w
$weeks[$w][$i] = true; //mark that team i has game in week w
$weeks[$w][$j] = true; //mark that team j has game in week w
} else { $noblanks = false; } //this cell is blank, and will be left blank.
}
}
}
}
// display
echo '<pre>';
foreach($games as $row) {
foreach($row as $col) {
printf('%4d', is_null($col) ? -2 : $col);
}
echo "\n";
}
printf("%d teams in %d weeks\n", $teams, count($weeks));
echo '</pre>';
示例输出：
-1 1 2 3
4 -1 3 2
5 6 -1 1
6 5 4 -1
4 teams in 6 weeks
-1 1 2 3 4 5 6
7 -1 3 2 5 4 8
8 6 -1 1 7 9 4
9 10 5 -1 6 7 11
10 9 11 8 -1 1 2
11 12 10 13 3 -1 14
12 13 15 16 17 18 -1
7 teams in 18 weeks
我已经找到了一种对所有情况都更“周效率”的方法，除非团队数量是2的幂。基本上，需要的周数变为2 * number_of_teams。
使用我的'pen-and-paper'方法，我注意到在矩阵对角线条纹数字非常理想，在我步行回家的路上，我想到了一种方法，你可以只输入2个团队ID和数量团队，它会让你回到那个游戏应该发生的那一周。
<?php
function getweek($home, $away, $num_teams) {
if($home == $away) { return -1; }
$week = $home+$away-2;
if( $week > ($num_teams) ) {
$week = $week-$num_teams;
}
if( $home>$away ) {
$week += $num_teams;
}
return $week;
}
$teams = $_GET['t'];
$games = array(); //2D array tracking which week teams will be playing
// do the work
for( $i=1; $i<=$teams; $i++ ) {
$games[$i] = array();
for( $j=1; $j<=$teams; $j++ ) {
$games[$i][$j] = getweek($i, $j, $teams);
}
}
// display
echo '<pre>';
$max=0;
foreach($games as $row) {
foreach($row as $col) {
printf('%4d', is_null($col) ? -2 : $col);
if( $col > $max ) { $max=$col; }
}
echo "\n";
}
printf("%d teams in %d weeks, %.2f weeks per team\n", $teams, $max, $max/$teams);
echo '</pre>';
示例输出：
-1 1 2 3
5 -1 3 4
6 7 -1 1
7 8 5 -1
4 teams in 8 weeks, 2.00 weeks per team
-1 1 2 3 4 5 6
8 -1 3 4 5 6 7
9 10 -1 5 6 7 1
10 11 12 -1 7 1 2
11 12 13 14 -1 2 3
12 13 14 8 9 -1 4
13 14 8 9 10 11 -1
7 teams in 14 weeks, 2.00 weeks per team
我修改了getWeek()功能，适用于任意数量的团队。请参阅下面的新功能。约万
function getWeek($home, $away, $num_teams) {
if($home == $away){
return -1;
}
$week = $home+$away-2;
if($week >= $num_teams){
$week = $week-$num_teams+1;
}
if($home>$away){
$week += $num_teams-1;
}
return $week;
}