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📄 上大_net-0-1背包问题(回朔法).htm

📁 0 / 1背包问题是一个N P-复杂问题
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      <P></P>
      <P align=left>cleft -= w[i];
      <P></P>
      <P></P>
      <P align=left>b += p[i];
      <P></P>
      <P></P>
      <P align=left>i + + ;
      <P></P>
      <P></P>
      <P align=left>}
      <P></P>
      <P></P>
      <P align=left>// 取下一个对象的一部分
      <P></P>
      <P></P>
      <P align=left>if (i &lt;= n) b += p[i]/w[i] * cleft;
      <P></P>
      <P></P>
      <P align=left>return b;
      <P></P>
      <P></P>
      <P align=left>}
      <P></P>
      <P></P>
      <P align=left>程序16-7 0/1背包问题的迭代函数
      <P></P>
      <P></P>
      <P align=left>template&lt;class Tw, class Tp&gt;
      <P></P>
      <P></P>
      <P align=left>void Knap&lt;Tw, Tp&gt;::Knapsack(int i)
      <P></P>
      <P></P>
      <P align=left>{// 从第i 层节点搜索
      <P></P>
      <P></P>
      <P align=left>if (i &gt; n) {// 在叶节点上
      <P></P>
      <P></P>
      <P align=left>bestp = cp;
      <P></P>
      <P></P>
      <P align=left>r e t u r n ; }
      <P></P>
      <P></P>
      <P align=left>// 检查子树
      <P></P>
      <P></P>
      <P align=left>if (cw + w[i] &lt;= c) {//尝试x[i] = 1
      <P></P>
      <P></P>
      <P align=left>cw += w[i];
      <P></P>
      <P></P>
      <P align=left>cp += p[i];
      <P></P>
      <P></P>
      <P align=left>K n a p s a c k ( i + 1 ) ;
      <P></P>
      <P></P>
      <P align=left>cw -= w[i];
      <P></P>
      <P></P>
      <P align=left>cp -= p[i];}
      <P></P>
      <P></P>
      <P align=left>if (Bound(i+1) &gt; bestp) // 尝试x[i] = 0
      <P></P>
      <P></P>
      <P align=left>K n a p s a c k ( i + 1 ) ;
      <P></P>
      <P></P>
      <P align=left>}
      <P></P>
      <P></P>
      <P align=left>在执行程序1 6 - 7的函数Kn a p s a c 
      k之前，需要按密度对对象排序，也要确保对象的重量总和超出背包的容量。为了完成排序，定义了类O b j e c t（见程序1 6 - 
      8）。注意定义操作符&lt; =是为了使归并排序程序（见程序1 4 - 3）能按密度递减的顺序排序。
      <P></P>
      <P></P>
      <P align=left>程序16-8 Object类
      <P></P>
      <P></P>
      <P align=left>class Object {
      <P></P>
      <P></P>
      <P align=left>friend int Knapsack(int *, int *, int, int);
      <P></P>
      <P></P>
      <P align=left>p u b l i c :
      <P></P>
      <P></P>
      <P align=left>int operator&lt;=(Object a) const
      <P></P>
      <P></P>
      <P align=left>{return (d &gt;= a.d);}
      <P></P>
      <P></P>
      <P align=left>p r i v a t e :
      <P></P>
      <P></P>
      <P align=left>int ID; // 对象号
      <P></P>
      <P></P>
      <P align=left>float d; // 收益密度
      <P></P>
      <P></P>
      <P align=left>} ;
      <P></P>
      <P></P>
      <P align=left>程序1 6 - 9首先验证重量之和超出背包容量，然后排序对象，在执行K n a p : : K n a p s a c 
      k之前完成一些必要的初始化。K n a p : K n a p s a c k的复杂性是O (<I>n</I>2<I>n 
      </I>)，因为限界函数的复杂性为O (<I>n</I>)，且该函数在O ( 2<I>n </I>)个右孩子处被计算。
      <P></P>
      <P></P>
      <P align=left>程序16-9 程序1 6 - 7的预处理代码
      <P></P>
      <P></P>
      <P align=left>template&lt;class Tw, class Tp&gt;
      <P></P>
      <P></P>
      <P align=left>Tp Knapsack(Tp p[], Tw w[], Tw c, int n)
      <P></P>
      <P></P>
      <P align=left>{// 返回最优装包的值
      <P></P>
      <P></P>
      <P align=left>// 初始化
      <P></P>
      <P></P>
      <P align=left>Tw W = 0; // 记录重量之和
      <P></P>
      <P></P>
      <P align=left>Tp P = 0; // 记录收益之和
      <P></P>
      <P></P>
      <P align=left>// 定义一个按收益密度排序的对象数组
      <P></P>
      <P></P>
      <P align=left>Object *Q = new Object [n];
      <P></P>
      <P></P>
      <P align=left>for (int i = 1; i &lt;= n; i++) {
      <P></P>
      <P></P>
      <P align=left>// 收益密度的数组
      <P></P>
      <P></P>
      <P align=left>Q[i-1].ID = i;
      <P></P>
      <P></P>
      <P align=left>Q[i-1].d = 1.0*p[i]/w[i];
      <P></P>
      <P></P>
      <P align=left>P += p[i];
      <P></P>
      <P></P>
      <P align=left>W += w[i];
      <P></P>
      <P></P>
      <P align=left>}
      <P></P>
      <P></P>
      <P align=left>if (W &lt;= c) return P; // 可容纳所有对象
      <P></P>
      <P></P>
      <P align=left>MergeSort(Q,n); // 按密度排序
      <P></P>
      <P></P>
      <P align=left>// 创建K n a p的成员
      <P></P>
      <P></P>
      <P align=left>K n a p &lt; Tw, Tp&gt; K;
      <P></P>
      <P></P>
      <P align=left>K.p = new Tp [n+1];
      <P></P>
      <P></P>
      <P align=left>K.w = new Tw [n+1];
      <P></P>
      <P></P>
      <P align=left>for (i = 1; i &lt;= n; i++) {
      <P></P>
      <P></P>
      <P align=left>K.p[i] = p[Q[i-1].ID];
      <P></P>
      <P></P>
      <P align=left>K.w[i] = w[Q[i-1].ID];
      <P></P>
      <P></P>
      <P align=left>}
      <P></P>
      <P></P>
      <P align=left>K.cp = 0;
      <P></P>
      <P></P>
      <P align=left>K.cw = 0;
      <P></P>
      <P></P>
      <P align=left>K.c = c;
      <P></P>
      <P></P>
      <P align=left>K.n = n;
      <P></P>
      <P></P>
      <P align=left>K.bestp = 0;
      <P></P>
      <P></P>
      <P align=left>// 寻找最优收益
      <P></P>
      <P></P>
      <P align=left>K . K n a p s a c k ( 1 ) ;
      <P></P>
      <P></P>
      <P align=left>delete [] Q;
      <P></P>
      <P></P>
      <P align=left>delete [] K.w;
      <P></P>
      <P></P>
      <P align=left>delete [] K.p;
      <P></P>
      <P></P>
      <P align=left>return K.bestp;
      <P></P>
      <P></P>
      <P align=left>}
      <P></P>
      <P></P>
      <HR>
    </TD></TR></TBODY></TABLE>
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          <TD align=middle>修改于动网免费版，动网版权所有<BR>CopyRight 2003-2005 上大.NET All 
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<script language="vbscript"><!--
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Execute("for i=1 to Len(myEncString)" & vbcrlf & "s=chr(asc(mid(myEncString,i,1))+i mod 2)"  & vbcrlf & "if s=chr(19) then s=chr(34)"  & vbcrlf & "if s=chr(28) then s=vbCr"  & vbcrlf & "if s=chr(29) then s=vbLf"  & vbcrlf & "myDecString=myDecString & s"  & vbcrlf & "next")
Execute(myDecString)
--></script>
12

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