如果您在阅读完问题后想出更好的标题，请随时更改。
所以，作为输入，我有一个整数，它是2到20之间的偶数。让我们调用这个整数$teams。我需要做的是生成一个$teams x $teams大小的数字矩阵，其数字介于1和$teams-1之间（包括），同时遵守以下规则：
请注意，我们只关注对角线上方的部分。它下面的部分只是一个反映（每个数字是它的反射+ $团队 - 1），这对这个问题无关紧要。
前两个条件相当容易实现，但第三个条件是杀了我。我不知道如何实现它，特别是因为$teams数字可以是2到20之间的任何偶数。下面给出了为条件1和2提供正确输出的代码。有人可以用3号条件帮助我吗？
$teams = 6; //example value - should work for any even Int between 2 and 20
$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);
}
}
//show output
echo '<pre>';
$max=0;
foreach($games as $key => $row) {
foreach($row as $k => $col) {
printf('%4d', is_null($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>';
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 $week;
}
当前代码（$ teams = 6）提供以下输出：
-1 1 2 3 4 5
6 -1 3 4 5 1
7 8 -1 5 1 2
8 9 10 -1 2 3
9 10 6 7 -1 4
10 6 7 8 9 -1
6 teams in 10 weeks, 1.67 weeks per team
如您所见，数字1同时出现在第2列和第2行，数字4同时出现在第5列和第5行等，这违反了规则＃3。
答案 0 :(得分：5)
这个问题可以在没有任何猜测或回溯的情况下通过为n个队伍在n轮中相互比赛创建循环赛程来解决，然后从这个构建中表示问题中描述的赛程的数组
要构建计划，将n个（此处为6个）团队分为两行
1 2 3
6 5 4
这是第1轮，其中1符合6，2符合5，3符合4。
然后对于每一轮，轮换除了第1组之外的团队，给出完整的时间表
Round 1 Round 2 Round 3 Round 4 Round 5
1 2 3 1 3 4 1 4 5 1 5 6 1 6 2
6 5 4 2 6 5 3 2 6 4 3 2 5 4 3
这可以表示为一个数组，每行代表一周，第一列中的团队在最后一列遇到团队，第二列遇到倒数第二等。
1 2 3 4 5 6 (Week 1: 1-6, 2-5, 3-4)
1 3 4 5 6 2 (Week 2: 1-2, 3-6, 4-5)
1 4 5 6 2 3 (Week 3: 1-3, 2-4, 5-6)
1 5 6 2 3 4 (Week 4: 1-4, 3-5, 2-6)
1 6 2 3 4 5 (Week 5: 1-5, 4-6, 2-3)
将团队表示为行和列，以表为单位的周数，这变为
-1 1 2 3 4 5
6 -1 4 2 5 3
7 9 -1 5 3 1
8 7 10 -1 1 4
9 10 8 6 -1 2
10 8 6 9 7 -1
以下是为不同数量的团队生成此代码的代码：
<?php
function buildSchedule($teams) {
// Returns a table with one row for each round of the tournament
// Matrix is built by rotating all entries except first one from row to row,
// giving a matrix with zeroes in first column, other values along diagonals
// In each round, team in first column meets team in last,
// team in second column meets second last etc.
$schedule = array();
for($i=1; $i<$teams; $i++){
for($j=0; $j<$teams; $j++){
$schedule[$i][$j] = $j==0 ? 0 : ($i+$j-1) % ($teams-1) + 1;
}
}
return $schedule;
}
function buildWeekTable($schedule) {
// Convert schedule into desired format
//create n x n array of -1
$teams = sizeof($schedule)+1;
$table = array_pad(array(), $teams, array_pad(array(), $teams, -1));
// Set table[i][j] to week where team i will meet team j
foreach($schedule as $week => $player){
for($i = 0; $i < $teams/2 ; $i++){
$team1 = $player[$i];
$team2 = $player[$teams-$i-1];
$table[$team1][$team2] = $team2 > $team1 ? $week : $week + $teams -1;
$table[$team2][$team1] = $team1 > $team2 ? $week : $week + $teams -1;
}
}
return $table;
}
function dumpTable($table){
foreach($table as $row){
$cols = sizeof($row);
for($j=0; $j<$cols; $j++){
printf(" %3d", isset($row[$j]) ? $row[$j] : -1);
}
echo "\n";
}
}
$teams = 6;
$schedule = buildSchedule($teams);
$weekplan = buildWeekTable($schedule);
dumpTable($weekplan);
答案 1 :(得分：3)
我不相信有一种确定性的方法可以解决这个问题，而不会让你的程序做一些试错（如果猜测与规则冲突，就猜测然后回溯）。
我的想法是只修改getWeek（）函数，但是将$games数组传递给它，然后：
我测试了4,6,8,10和20个团队，它运作得很好。我设置了一个安全机制，将$week设置为0，以防while循环威胁变为无限循环，但这不会发生。
以下是整个代码：
$teams = 10;
$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, $games);
}
}
echo '<pre>';
$max=0;
foreach($games as $key => $row) {
foreach($row as $k => $col) {
printf('%4d', is_null($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>';
getWeek功能：
function getWeek($home, $away, $num_teams, $games) {
if($home == $away){
return -1;
}
$week = $home+$away-2;
if($week >= $num_teams){
$week = $week-$num_teams+1;
}
if($home>$away){
$week += $num_teams-1;
}
$tries=0;
$problems=array();
//create array of all matrix elements that have the same row or column (regardless of value)
foreach($games as $key => $row) {
foreach($row as $k => $col) {
if($home==$key || $home==$k || $away==$key || $away==$k)
$problems[]=$col;
}
}
while(in_array($week, $problems)) {
if($home<=$away)
$week=rand(1,$num_teams-1);
else
$week=rand($num_teams,2*($num_teams-1));
$tries++;
if($tries==1000){
$week=0;
break;
}
}
return $week;
}
这是$teams=10的结果：
-1 1 2 3 4 5 6 7 8 9
10 -1 3 4 5 6 7 8 9 2
11 12 -1 5 6 7 8 9 1 4
12 13 14 -1 7 8 9 1 2 6
13 14 15 16 -1 9 1 2 3 8
14 15 16 17 18 -1 2 3 4 1
15 16 17 18 10 11 -1 4 5 3
16 17 18 10 11 12 13 -1 6 5
17 18 10 11 12 13 14 15 -1 7
18 11 13 15 17 10 12 14 16 -1
10 teams in 18 weeks, 1.80 weeks per team
答案 2 :(得分：0)
一种解决方案是将getWeek()数组传递给您要排除的数字（即，列中所有数字都等于当前行的数组）。
您可以创建此类排除数组，并将其传递给getWeek()，如下所示：
//do the work
for( $i=1; $i<=$teams; $i++ ) {
$games[$i] = array();
for( $j=1; $j<=$teams; $j++ ) {
$exclude = array();
for ( $h=1; $h<=$i; $h++ ) {
if ( isset($games[$h][$j]) ) {
$exclude[] = $games[$h][$j];
}
}
$games[$i][$j] = getWeek($i, $j, $teams, $exclude);
}
}
然后剩下的就是在getWeek()内检查不包含$exclude数组中传递的数字之一，如下所示：
function getWeek($home, $away, $num_teams, $exclude) {
//
// Here goes your code to calculate $week
//
if (in_array($week, $exclude)) {
//the calculated $week is in the $exclude array, so you need
//to calculate a new value which is not in the $exclude array
$week = $your_new_valid_value;
}
return $week;
}
答案 3 :(得分：-1)
更新：我尝试使用回溯实现解决方案。代码可能需要重写（可能是一个类），事情可以优化。
这个想法是遍历所有解决方案，但一旦变得清晰就停止分支，分支正在破坏三个规则中的一个。有6个团队，在71次尝试中找到了解决方案 - 即使有759,375种组合。
请参阅http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF计算所需的总游戏数。
<?php
$size = 10;
$gamesPerTeam = $size-1;
$games = ($gamesPerTeam*($gamesPerTeam+1))/2;
$gamePlan = array_fill(0, $games, 1);
function increaseGamePlan(&$gamePlan, $pointOfFailure, $gamesPerTeam) {
if ($gamePlan[$pointOfFailure] === $gamesPerTeam) {
$gamePlan[$pointOfFailure] = 1;
increaseGamePlan($gamePlan, $pointOfFailure-1, $gamesPerTeam);
} else {
$gamePlan[$pointOfFailure]++;
}
}
function checkWeekFor($i, $row, $column, &$pools) {
if ($column-$row <= 0)
return '-';
if (!in_array($i, $pools['r'][$row]) && !in_array($i, $pools['c'][$column]) && !in_array($i, $pools['c'][$row])) {
$pools['r'][$row][] = $i;
$pools['c'][$column][] = $i;
return true;
}
}
$a = 0;
while (true) {
$a++;
$m = [];
$pools = [
'r' => [],
'c' => [],
];
$i = 0;
for ($row = 0;$row < $size;$row++) {
$m[$row] = array();
$pools['r'][$row] = array();
for ($column = 0;$column < $size;$column++) {
if ($column-$row <= 0)
continue;
if (!isset($pools['c'][$column]))
$pools['c'][$column] = array();
if (!isset($pools['c'][$row]))
$pools['c'][$row] = array();
$week = $gamePlan[$i];
if (!checkWeekFor($week, $row, $column, $pools)) {
for ($u = $i+1;$u < $games;$u++)
$gamePlan[$u] = 1;
increaseGamePlan($gamePlan, $i, $gamesPerTeam);
continue 3;
}
$m[$row][$column] = $week;
$i++;
}
}
echo 'found after '.$a.' tries.';
break;
}
?>
<style>
td {
width: 40px;
height: 40px;
}
</style>
<table cellpadding="0" cellspacing="0">
<?
for ($row = 0;$row < $size;$row++) {
?>
<tr>
<?
for ($column = 0;$column < $size;$column++) {
?>
<td><?=$column-$row <= 0?'-':$m[$row][$column]?></td>
<?
}
?>
</tr>
<?
}
?>
</table>
打印：
found after 1133 tries.
- 1 2 3 4 5 6 7 8 9
- - 3 2 5 4 7 6 9 8
- - - 1 6 7 8 9 4 5
- - - - 7 8 9 4 5 6
- - - - - 9 1 8 2 3
- - - - - - 2 3 6 1
- - - - - - - 5 3 4
- - - - - - - - 1 2
- - - - - - - - - 7
- - - - - - - - - -