给定以下示例数组,如何找到所有可用时间的排列,以满足amountNeeded?换句话说,follow数组应该产生以下结果:

  

2008年4月14日08:00至08:10使用资源10和13

     

使用资源10和13

于2008-05-14 08:10至08:20提供
print("Array(

    [amountNeeded] => 2
    [resources] => Array
        (
            [0] => Array
                (
                    [resourceID] => 10
                    [blocks] => Array
                        (
                            [0] => Array
                                (
                                    [availDate] => 2008-05-14
                                    [startTime] => 08:00
                                    [endTime] => 08:10
                                )

                            [1] => Array
                                (
                                    [availDate] => 2008-05-14
                                    [startTime] => 08:10
                                    [endTime] => 08:20
                                )

                            [2] => Array
                                (
                                    [availDate] => 2008-05-14
                                    [startTime] => 08:20
                                    [endTime] => 08:30
                                )
                    ...
            [1] => Array
                (
                    [resourceID] => 13
                    [blocks] => Array
                        (
                            [0] => Array
                                (
                                    [availDate] => 2008-05-14
                                    [startTime] => 08:00
                                    [endTime] => 08:10
                                )

                            [1] => Array
                                (
                                    [availDate] => 2008-05-14
                                    [startTime] => 08:10
                                    [endTime] => 08:20
                                )

                            [2] => Array
                                (
                                    [availDate] => 2008-05-14
                                    [startTime] => 08:30
                                    [endTime] => 08:40
                                )
                    ...
");
有帮助吗?

解决方案

您所寻找的与排列无关。您正在考虑重叠时间段,我看到两种方法:

  1. 将所有时间段预处理到时间轴,然后查询时间线,或
  2. 并行扫描所有资源块。
  3. 第一个选项需要更多内存但更容易理解;第二种可能是资源不足,但编程要复杂得多。两者都将受益于最小化要考虑的数据集,即限制目标时间段。

    选项#1如下(在OO PHP5中实现):

    <?php
    
    class Resource {
        protected $name;    // resource ID
        protected $start;   // start timestamp
        protected $finish;  // end timestamp
        // resource available while $start <= current time < $end
    
        function __construct($n, $sd, $st, $ed, $et) {
            $this->name = $n;
            $this->start = strtotime("$sd $st");
            $this->finish = strtotime("$ed $et");
        }
    
        function getID() { return $this->name; }
        function getStart() { return $this->start; }
        function getEnd() { return $this->finish; }
    }
    
    class Timeline {
        protected $times;       // ordered list of start-times;
        protected $resources;   // resources available in each timeslot
        protected $offs;        // iterator offset
    
        function __construct() {
            $this->times = array();
            $this->resources = array();
            $this->offs = 0;
        }
    
        // binary search, insert if not found, return index
        private function time_ins($time) {
            // array is empty?
            if (count($this->times) == 0) {
                $this->times[0]= $time;
                $this->resources[0] = array();
                return 0;
            }
    
            $min = $lo = 0;
            $max = $hi = count($this->times)-1;
            // binary search
            while($lo <= $hi) {
                $mid = ($lo+$hi) >> 1;
    
                if ($this->times[$mid] == $time) {
                    // already exists - return index
                    return $mid;
                }
                elseif ($this->times[$mid] < $time) {
                    // if value exists, is in upper half of array
                    $lo = $mid+1;
    
                    if ($lo > $max || $this->times[$lo] > $time) {
                        // $lo points to first value greater than $time
                        // insert new value at $lo
                        array_splice($this->times, $lo, 0, $time);
                        $t = isset($this->resources[$lo-1]) ? $this->resources[$lo-1] : array();
                        array_splice($this->resources, $lo, 0, $t);
                        return $lo;
                    }
                }
                else {
                    // if value exists, is in lower half of array
                    $hi = $mid-1;
    
                    if ($hi < $min || $this->times[$hi] < $time) {
                        // $hi points to first value less than $time
                        // insert new value at $hi+1
                        array_splice($this->times, $hi+1, 0, $time);
                        $t = isset($this->resources[$hi+1]) ? $this->resources[$hi+1] : array();
                        array_splice($this->resources, $hi+1, 0, $t);
                        return $hi+1;
                    }
                }
            }
        }
    
        function Add( $start, $end, $id ) {
            $s = $this->time_ins($start);
            $e = $this->time_ins($end);
    
            for($i = $s; $i < $e; ++$i)
                $this->resources[$i][]= $id;
        }
    
        function reset()    { $offs = 0; }
        function isValid()  { return ($this->offs+1 < count($this->times)); }   // omit last time (is end-time only)
        function next()     { $this->offs++; }
        function resCount() { return count( $this->resources[ $this->offs ] ); }
        function getStart() { return $this->times[$this->offs]; }
        function getEnd()   { return $this->times[$this->offs + 1]; }
        function getRes()   { return $this->resources[$this->offs]; }
    }
    
    
    $res = array();
    $res[]= new Resource('10', '2008-05-14', '08:00', '2008-05-14', '08:10');
    $res[]= new Resource('10', '2008-05-14', '08:10', '2008-05-14', '08:20');
    $res[]= new Resource('10', '2008-05-14', '08:20', '2008-05-14', '08:30');
    $res[]= new Resource('13', '2008-05-14', '08:00', '2008-05-14', '08:10');
    $res[]= new Resource('13', '2008-05-14', '08:10', '2008-05-14', '08:20');
    $res[]= new Resource('13', '2008-05-14', '08:30', '2008-05-14', '08:40');
    
    $tl = new Timeline();
    foreach($res as $R)
        $tl->Add( $R->getStart(), $R->getEnd(), $R->getID() );
    
    $needed = 2;
    $_pre = "\n<p>";
    $_post = "</p>";
    for( $tl->reset(); $tl->isValid(); $tl->next() ) {
        $cnt = $tl->resCount();
    
        if ($cnt >= $needed) {
            $st = date("Y-m-d H:i", $tl->getStart());
            $fn = date("Y-m-d H:i", $tl->getEnd());
            $res = join(', ', $tl->getRes());
    
            echo ($cnt == $needed
                ? "{$_pre}Available from $st to $fn using resources ($res){$_post}"
                : "{$_pre}Available from $st to $fn using any $needed of resources ($res){$_post}"
            );
        }
    }
    
    ?>
    

其他提示

它被称为排列,用一个好词。有关通用实施,请查看此博客文章

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