有N种物品和一个容量为V的背包。还有一个阈值T。
第i种物品最多有n[i]件可用,每件费用是c[i],价值是w[i]。
求解将哪些物品装入背包可使这些物品的费用总和不超过背包容量,且价值总和大于且最接近T。
附完全背包代码
/**
*背包问题描述:一个承受最大重量为W的背包,现在有n个物品,每个物品重量为t, 每个物品的价值为v。
* 要使得这个背包重量最大(但不能超过W),同时又需要背包的价值最大。
*思路:定义一个二维数组,一维为物品数量(表示每个物品),二维是重量(不超过最大,这里是10),下面数组a,
*动态规划原理思想,max(opt(i-1,w),wi+opt(i-1,w-wi)) 当中最大值,
*opt(i-1,w-wi)指上一个最优解
*/
//这是我根据动态规划原理写的
// max(opt(i-1,w),wi+opt(i-1,w-wi))
//背包可以装最大的重量
$w=10;
//这里有四件物品,每件物品的重量
$dx=array(3,4,5);
//每件物品的价值
$qz=array(4,5,6);
//定义一个数组
$a=array();
//初始化
for($i=0;$i for ($j=0;$j //opt(i-1,w),wi+opt(i-1,w-wi)
for ($j=1;$j for($i=1;$i $a[$j][$i]=$a[$j-1][$i];
//不大于最大的w=10
if($dx[$j-1] if(!isset($a[$j-1][$i-$dx[$j-1]])) continue;
//wi+opt(i-1,wi)
$tmp = $a[$j-1][$i-$dx[$j-1]]+$qz[$j-1];
//opt(i-1,w),wi+opt(i-1,w-wi) => 进行比较
if($tmp>$a[$j][$i]){
$a[$j][$i]=$tmp;
}
}
}
}
//打印这个数组,输出最右角的值是可以最大价值的
for ($j=0;$j for ($i=0;$i echo $a[$j][$i]." ";
} echo "
";
}
?>
回复讨论(解决方案)
你遇到了什么问题呢?你的结果应该是对的
不过我喜欢这么写
$w = 10;$ar = array( array('w' => 3, 'v' => 4), array('w' => 4, 'v' => 5), array('w' => 5, 'v' => 6),);foreach($ar as $k=>$v) { $v['k'][] = $k; $res[] = $v;}$p = 0;for(;$p<count($res); $p++) { $r = $res[$p]; foreach($ar as $i=>$v) { if(in_array($i, $res[$p]['k'])) continue; if($r['w'] + $v['w'] <= $w) { $res[] = array( 'w' => $r['w'] + $v['w'], 'v' => $r['v'] + $v['v'], 'k' => array_merge($r['k'], array($i)), ); } }}foreach($res as $v) $t[] = $v['v'];array_multisort($t, SORT_DESC, $res); print_r($res);
Array( [0] => Array ( [w] => 9 [v] => 11 [k] => Array ( [0] => 1 [1] => 2 ) ) [1] => Array ( [w] => 9 [v] => 11 [k] => Array ( [0] => 2 [1] => 1 ) ) [2] => Array ( [w] => 8 [v] => 10 [k] => Array ( [0] => 0 [1] => 2 ) ) [3] => Array ( [w] => 8 [v] => 10 [k] => Array ( [0] => 2 [1] => 0 ) ) [4] => Array ( [w] => 7 [v] => 9 [k] => Array ( [0] => 0 [1] => 1 ) ) [5] => Array ( [w] => 7 [v] => 9 [k] => Array ( [0] => 1 [1] => 0 ) ) [6] => Array ( [w] => 5 [v] => 6 [k] => Array ( [0] => 2 ) ) [7] => Array ( [w] => 4 [v] => 5 [k] => Array ( [0] => 1 ) ) [8] => Array ( [w] => 3 [v] => 4 [k] => Array ( [0] => 0 ) ))
目前这个是 装满了w=10的背包的最大价值。
我是想找出:在尽量装满背包的情况下,总价值接近给定的一个阈值T 的组合
$f = 9;print_r(array_filter($res, function($a) use ($f) { return $a['v'] > 0.9*$f && $a['v']<1.1*$f;}));
太强了,感谢版主大人!非常牛B,结果完全达到了
这个算法你写的太简洁了,有点看不懂,再请教您一下,怎么过滤掉重复的组合啊? 比如 0 3 6 ,3 0 6,6 0 3
....array_multisort($t, SORT_DESC, $res); //排序(已有的)foreach($res as $i=>$v) if(isset($res[$i-1]) && ! array_diff($res[$i-1]['k'], $v['k'])) unset($res[$i]); //去重$res = array_values($res); //规格化
我这个算法并未完全按照动态规划去做
而是记录了所有可能的组合,最后用排序做检查
太感谢了,多谢您的指教

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