显然无法在Stack Overflow上将其称为问题，但我目前正在尝试了解如何在背包问题中以项目组的形式集成约束。在这种情况下，我的数学技能被证明是相当有限的，但是我非常积极地使这项工作按预期进行，并弄清楚每个方面的作用（按照这个顺序，因为事情在工作时更有意义）。
话虽如此，我在Rosetta Code找到了一个非常漂亮的实现，并清理了一些变量名，以帮助我从一个非常基本的角度更好地理解这一点。
不幸的是，我非常难以确定如何应用此逻辑来包含项目组。我的目的是建立幻想团队，提供我自己的价值和每位运动员的体重（分/薪水），但没有小组（在我的情况下位置），我无法这样做。
有人能指出我正确的方向吗？我正在审查其他语言的代码示例以及整个问题的其他描述，但我希望通过任何可能的方式实现这些组。
<?php
function knapSolveFast2($itemWeight, $itemValue, $i, $availWeight, &$memoItems, &$pickedItems)
{
global $numcalls;
$numcalls++;
// Return memo if we have one
if (isset($memoItems[$i][$availWeight]))
{
return array( $memoItems[$i][$availWeight], $memoItems['picked'][$i][$availWeight] );
}
else
{
// At end of decision branch
if ($i == 0)
{
if ($itemWeight[$i] <= $availWeight)
{ // Will this item fit?
$memoItems[$i][$availWeight] = $itemValue[$i]; // Memo this item
$memoItems['picked'][$i][$availWeight] = array($i); // and the picked item
return array($itemValue[$i],array($i)); // Return the value of this item and add it to the picked list
}
else
{
// Won't fit
$memoItems[$i][$availWeight] = 0; // Memo zero
$memoItems['picked'][$i][$availWeight] = array(); // and a blank array entry...
return array(0,array()); // Return nothing
}
}
// Not at end of decision branch..
// Get the result of the next branch (without this one)
list ($without_i,$without_PI) = knapSolveFast2($itemWeight, $itemValue, $i-1, $availWeight,$memoItems,$pickedItems);
if ($itemWeight[$i] > $availWeight)
{ // Does it return too many?
$memoItems[$i][$availWeight] = $without_i; // Memo without including this one
$memoItems['picked'][$i][$availWeight] = array(); // and a blank array entry...
return array($without_i,array()); // and return it
}
else
{
// Get the result of the next branch (WITH this one picked, so available weight is reduced)
list ($with_i,$with_PI) = knapSolveFast2($itemWeight, $itemValue, ($i-1), ($availWeight - $itemWeight[$i]),$memoItems,$pickedItems);
$with_i += $itemValue[$i]; // ..and add the value of this one..
// Get the greater of WITH or WITHOUT
if ($with_i > $without_i)
{
$res = $with_i;
$picked = $with_PI;
array_push($picked,$i);
}
else
{
$res = $without_i;
$picked = $without_PI;
}
$memoItems[$i][$availWeight] = $res; // Store it in the memo
$memoItems['picked'][$i][$availWeight] = $picked; // and store the picked item
return array ($res,$picked); // and then return it
}
}
}
$items = array("map","compass","water","sandwich","glucose","tin","banana","apple","cheese","beer","suntan cream","camera","t-shirt","trousers","umbrella","waterproof trousers","waterproof overclothes","note-case","sunglasses","towel","socks","book");
$weight = array(9,13,153,50,15,68,27,39,23,52,11,32,24,48,73,42,43,22,7,18,4,30);
$value = array(150,35,200,160,60,45,60,40,30,10,70,30,15,10,40,70,75,80,20,12,50,10);
## Initialize
$numcalls = 0;
$memoItems = array();
$selectedItems = array();
## Solve
list ($m4, $selectedItems) = knapSolveFast2($weight, $value, sizeof($value)-1, 400, $memoItems, $selectedItems);
# Display Result
echo "<b>Items:</b><br>" . join(", ", $items) . "<br>";
echo "<b>Max Value Found:</b><br>$m4 (in $numcalls calls)<br>";
echo "<b>Array Indices:</b><br>". join(",", $selectedItems) . "<br>";
echo "<b>Chosen Items:</b><br>";
echo "<table border cellspacing=0>";
echo "<tr><td>Item</td><td>Value</td><td>Weight</td></tr>";
$totalValue = 0;
$totalWeight = 0;
foreach($selectedItems as $key)
{
$totalValue += $value[$key];
$totalWeight += $weight[$key];
echo "<tr><td>" . $items[$key] . "</td><td>" . $value[$key] . "</td><td>".$weight[$key] . "</td></tr>";
}
echo "<tr><td align=right><b>Totals</b></td><td>$totalValue</td><td>$totalWeight</td></tr>";
echo "</table><hr>";
?>
答案 0 :(得分：3)
背包计划是传统的，但我认为它模糊了正在发生的事情。让我向您展示如何从蛮力解决方案中更直接地推导出DP。
在Python中（对不起;这是我选择的脚本语言），暴力解决方案看起来像这样。首先，有一个函数用于生成具有广度优先搜索的所有子集（这很重要）。
def all_subsets(S): # brute force
subsets_so_far = [()]
for x in S:
new_subsets = [subset + (x,) for subset in subsets_so_far]
subsets_so_far.extend(new_subsets)
return subsets_so_far
然后有一个函数返回True如果解决方案有效（在预算范围内并且有适当的位置分解） - 称之为is_valid_solution - 并且给定一个解决方案的函数返回总数玩家价值（total_player_value）。假设players是可用玩家的列表，那么最佳解决方案就是这样。
max(filter(is_valid_solution, all_subsets(players)), key=total_player_value)
现在，对于DP，我们向cull添加了一个函数all_subsets。
def best_subsets(S): # DP
subsets_so_far = [()]
for x in S:
new_subsets = [subset + (x,) for subset in subsets_so_far]
subsets_so_far.extend(new_subsets)
subsets_so_far = cull(subsets_so_far) ### This is new.
return subsets_so_far
cull做的是抛弃我们寻求最佳解决方案时明显不会错过的部分解决方案。如果部分解决方案已经超出预算，或者在一个位置已经有太多玩家，则可以安全地丢弃它。让is_valid_partial_solution成为测试这些条件的函数（它可能看起来很像is_valid_solution）。到目前为止，我们有这个。
def cull(subsets): # INCOMPLETE!
return filter(is_valid_partial_solution, subsets)
另一个重要的测试是，某些部分解决方案比其他解决方案更好。如果两个部分解决方案具有相同的位置分解（例如，两个前向和一个中心）并且成本相同，那么我们只需要保留更有价值的一个。让cost_and_position_breakdown采用解决方案并生成一个对指定属性进行编码的字符串。
def cull(subsets):
best_subset = {} # empty dictionary/map
for subset in filter(is_valid_partial_solution, subsets):
key = cost_and_position_breakdown(subset)
if (key not in best_subset or
total_value(subset) > total_value(best_subset[key])):
best_subset[key] = subset
return best_subset.values()
就是这样。这里有很多优化要做（例如，抛弃部分解决方案，其中有更便宜和更有价值的部分解决方案;修改数据结构，以便我们不总是从头开始计算价值和位置细分并减少存储成本），但可以逐步解决。
答案 1 :(得分：0)
在PHP中组合递归函数的一个潜在的小优势是变量按值传递（意味着复制）而不是引用，这可以节省一两步。
也许您可以通过包含示例输入和输出来更好地阐明您要查找的内容。这是一个从给定组中进行组合的示例 - 我不确定这是否是您的意图......我使访问部分结果的部分允许使用较少值的组合，如果他们的体重较低 - 所有这些都可以改变为以你想要的具体方式修剪。
function make_teams($players, $position_limits, $weights, $values, $max_weight){
$player_counts = array_map(function($x){
return count($x);
}, $players);
$positions = array_map(function($x){
$positions[] = [];
},$position_limits);
$num_positions = count($positions);
$combinations = [];
$hash = [];
$stack = [[$positions,0,0,0,0,0]];
while (!empty($stack)){
$params = array_pop($stack);
$positions = $params[0];
$i = $params[1];
$j = $params[2];
$len = $params[3];
$weight = $params[4];
$value = $params[5];
// too heavy
if ($weight > $max_weight){
continue;
// the variable, $positions, is accumulating so you can access the partial result
} else if ($j == 0 && $i > 0){
// remember weight and value after each position is chosen
if (!isset($hash[$i])){
$hash[$i] = [$weight,$value];
// end thread if current value is lower for similar weight
} else if ($weight >= $hash[$i][0] && $value < $hash[$i][1]){
continue;
// remember better weight and value
} else if ($weight <= $hash[$i][0] && $value > $hash[$i][1]){
$hash[$i] = [$weight,$value];
}
}
// all positions have been filled
if ($i == $num_positions){
$positions[] = $weight;
$positions[] = $value;
if (!empty($combinations)){
$last = &$combinations[count($combinations) - 1];
if ($weight < $last[$num_positions] && $value > $last[$num_positions + 1]){
$last = $positions;
} else {
$combinations[] = $positions;
}
} else {
$combinations[] = $positions;
}
// current position is filled
} else if (count($positions[$i]) == $position_limits[$i]){
$stack[] = [$positions,$i + 1,0,$len,$weight,$value];
// otherwise create two new threads: one with player $j added to
// position $i, the other thread skipping player $j
} else {
if ($j < $player_counts[$i] - 1){
$stack[] = [$positions,$i,$j + 1,$len,$weight,$value];
}
if ($j < $player_counts[$i]){
$positions[$i][] = $players[$i][$j];
$stack[] = [$positions,$i,$j + 1,$len + 1
,$weight + $weights[$i][$j],$value + $values[$i][$j]];
}
}
}
return $combinations;
}
输出：
$players = [[1,2],[3,4,5],[6,7]];
$position_limits = [1,2,1];
$weights = [[2000000,1000000],[10000000,1000500,12000000],[5000000,1234567]];
$values = [[33,5],[78,23,10],[11,101]];
$max_weight = 20000000;
echo json_encode(make_teams($players, $position_limits, $weights, $values, $max_weight));
/*
[[[1],[3,4],[7],14235067,235],[[2],[3,4],[7],13235067,207]]
*/