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PHP回溯法解决0-1背包问题实例分析_php技巧

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Release: 2016-05-16 20:19:38
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本文实例讲述了PHP回溯法解决0-1背包问题的方法。分享给大家供大家参考。具体分析如下:

这段代码是根据《软件设计师》教程的伪代码写的;
最麻烦的不是伪代码改成php,而是数组下标从0开始,及相应的下标判断问题;
带着调试输出一块写上

<&#63;php
  $v_arr = array(11,21,31,33,43,53,55,65);
  $w_arr = array(1,11,21,23,33,43,45,55);
  $n = count($w_arr );
  //测试输出
  var_dump(bknap1(110));
//var_dump(bound(139,89,7,110));
  function bound($v,$w,$k,$W_total){
    global $v_arr,$w_arr,$n;
    $b = $v;
    $c = $w;
//var_dump($W_total);var_dump($n);var_dump($k);var_dump($v);var_dump($w);
//die;
    for($i=$k+1;$i<$n;$i++){
      $c = $c + $w_arr[$i];
      //var_dump($W_total);var_dump($c);
      if($c<$W_total)
        $b += $v_arr[$i];
      else{
//var_dump((1-($c-$W_total)/$w_arr[$i])*$v_arr[$i]);
        $b = $b+(1-($c-$W_total)/$w_arr[$i])*$v_arr[$i];
        return $b; 
      }
    }
    /*var_dump('------bound head');
    var_dump($k);
    var_dump($b);
    var_dump('------bound end');*/
    return $b; 
  }
  function bknap1($W_total){
    global $v_arr,$w_arr,$n;
    $cw = $cp = 0;
    $k = 0;
    $fp = -1;
    while(true){
      while($k<$n && $cw+$w_arr[$k]<=$W_total){
        $cw += $w_arr[$k];
        $cp += $v_arr[$k];
        $Y_arr[$k] = 1;
        $k +=1;
      }
//var_dump($cw);var_dump($cp);var_dump($Y_arr);var_dump($k);var_dump($n);
      if($k==$n){
        $fp = $cp;
        $fw = $cw;
        $k = $n-1;
        $X_arr = $Y_arr;
//bound($cp,$cw,$k,$W_total);
//var_dump(bound($cp,$cw,$k,$W_total),$fp,$k);die;
//var_dump($fp);var_dump($fw);var_dump($Y_arr);var_dump($k);var_dump($n);
      }else{
        $Y_arr[$k] = 0;
      }
//var_dump($Y_arr);var_dump($k);var_dump($n);//die;
//var_dump(bound($cp,$cw,$k,$W_total),$fp);die;
      while(bound($cp,$cw,$k,$W_total)<=$fp)
      {
        while($k>=0 && $Y_arr[$k]!=1){
          $k -= 1;
        }
        if($k<0)
        {
          return $X_arr;
        }
        var_dump($k);
        $Y_arr[$k] = 0;
        $cw -= $w_arr[$k];
        $cp -= $v_arr[$k];
      }
      $k += 1;
    }
  }
&#63;>
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希望本文所述对大家的php程序设计有所帮助。

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