假设您有货物。它需要从 A 点到 B 点，从 B 点到 C 点，最后从 C 点到 D 点。您需要用最少的钱在五天内到达那里。每条支线有三个可能的托运人，每条支线都有自己不同的时间和成本：
Array
(
[leg0] => Array
(
[UPS] => Array
(
[days] => 1
[cost] => 5000
)
[FedEx] => Array
(
[days] => 2
[cost] => 3000
)
[Conway] => Array
(
[days] => 5
[cost] => 1000
)
)
[leg1] => Array
(
[UPS] => Array
(
[days] => 1
[cost] => 3000
)
[FedEx] => Array
(
[days] => 2
[cost] => 3000
)
[Conway] => Array
(
[days] => 3
[cost] => 1000
)
)
[leg2] => Array
(
[UPS] => Array
(
[days] => 1
[cost] => 4000
)
[FedEx] => Array
(
[days] => 1
[cost] => 3000
)
[Conway] => Array
(
[days] => 2
[cost] => 5000
)
)
)
您将如何以编程方式寻找最佳组合？
到目前为止，我最好的尝试（第三或第四种算法）是：
- 为每条腿找到最长的托运人
- 淘汰最“贵”的一个
- 为每条腿找到最便宜的托运人
- 计算总成本和天数
- 如果天数可以接受，则完成，否则，转到 1
在 PHP 中快速模拟（请注意，下面的测试数组可以流畅地工作，但是如果您使用上面的测试数组尝试它，它不会找到正确的组合）：
$shippers["leg1"] = array(
"UPS" => array("days" => 1, "cost" => 4000),
"Conway" => array("days" => 3, "cost" => 3200),
"FedEx" => array("days" => 8, "cost" => 1000)
);
$shippers["leg2"] = array(
"UPS" => array("days" => 1, "cost" => 3500),
"Conway" => array("days" => 2, "cost" => 2800),
"FedEx" => array("days" => 4, "cost" => 900)
);
$shippers["leg3"] = array(
"UPS" => array("days" => 1, "cost" => 3500),
"Conway" => array("days" => 2, "cost" => 2800),
"FedEx" => array("days" => 4, "cost" => 900)
);
$times = 0;
$totalDays = 9999999;
print "<h1>Shippers to Choose From:</h1><pre>";
print_r($shippers);
print "</pre><br />";
while($totalDays > $maxDays && $times < 500){
$totalDays = 0;
$times++;
$worstShipper = null;
$longestShippers = null;
$cheapestShippers = null;
foreach($shippers as $legName => $leg){
//find longest shipment for each leg (in terms of days)
unset($longestShippers[$legName]);
$longestDays = null;
if(count($leg) > 1){
foreach($leg as $shipperName => $shipper){
if(empty($longestDays) || $shipper["days"] > $longestDays){
$longestShippers[$legName]["days"] = $shipper["days"];
$longestShippers[$legName]["cost"] = $shipper["cost"];
$longestShippers[$legName]["name"] = $shipperName;
$longestDays = $shipper["days"];
}
}
}
}
foreach($longestShippers as $leg => $shipper){
$shipper["totalCost"] = $shipper["days"] * $shipper["cost"];
//print $shipper["totalCost"] . " <?> " . $worstShipper["totalCost"] . ";";
if(empty($worstShipper) || $shipper["totalCost"] > $worstShipper["totalCost"]){
$worstShipper = $shipper;
$worstShipperLeg = $leg;
}
}
//print "worst shipper is: shippers[$worstShipperLeg][{$worstShipper['name']}]" . $shippers[$worstShipperLeg][$worstShipper["name"]]["days"];
unset($shippers[$worstShipperLeg][$worstShipper["name"]]);
print "<h1>Next:</h1><pre>";
print_r($shippers);
print "</pre><br />";
foreach($shippers as $legName => $leg){
//find cheapest shipment for each leg (in terms of cost)
unset($cheapestShippers[$legName]);
$lowestCost = null;
foreach($leg as $shipperName => $shipper){
if(empty($lowestCost) || $shipper["cost"] < $lowestCost){
$cheapestShippers[$legName]["days"] = $shipper["days"];
$cheapestShippers[$legName]["cost"] = $shipper["cost"];
$cheapestShippers[$legName]["name"] = $shipperName;
$lowestCost = $shipper["cost"];
}
}
//recalculate days and see if we are under max days...
$totalDays += $cheapestShippers[$legName]['days'];
}
//print "<h2>totalDays: $totalDays</h2>";
}
print "<h1>Chosen Shippers:</h1><pre>";
print_r($cheapestShippers);
print "</pre>";
我想我实际上可能需要做一些事情，我逐个制作每个组合（带有一系列循环）并将每个组合的总“分数”相加，然后找到最好的一个......
编辑：澄清一下，这不是“家庭作业”（我不在学校）。这是我当前工作项目的一部分。
要求（一如既往）一直在不断变化。如果在我开始解决这个问题时给了我当前的限制，我将使用 A* 算法的一些变体（或 Dijkstra 或最短路径或单纯形或其他东西）。但是一切都在变化和变化，这将我带到了现在的位置。
所以我想这意味着我需要忘记我到目前为止所做的所有废话，只使用我知道我应该使用的东西，这是一种寻路算法。