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📁 浙江大学计算机学院数据结构课程的教学课件
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<div id="Layer1" style="position:absolute; width:711px; height:21px; z-index:1; top: 10px; background-color: #CCCCCC; layer-background-color: #CCCCCC; border: 1px none #000000; left: 26px"><b>|</b><font face="宋体" size="2"><a href="../text1/text1-0.htm">第一章</a></font><b>|</b><font face="宋体" size="2"><a href="../text2/text2-0.htm">第二章</a></font><b>|</b><font face="宋体" size="2"><a href="../text3/text3-0.htm">第三章</a></font><b>|</b><font face="宋体" size="2"><a href="../text4/text4-0.htm">第四章</a></font><b>|</b><font face="宋体" size="2"><a href="../text5/text5-0.htm">第五章</a></font><b>|</b><font face="宋体" size="2"><a href="../text6/text6-0.htm">第六章</a></font><b>|</b><font face="宋体" size="2"><a href="../text7/text7-0.htm">第七章</a></font><b>|</b><font face="宋体" size="2"><a href="../text8/text8-0.htm">第八章</a></font><b>|</b><font face="宋体" size="2"><a href="../text9/text9-0.htm">第九章</a></font><b>|</b><font face="宋体" size="2">第十章</font><b>|</b><font size="2" face="宋体"><a href="../textA/textA-0.htm">算法分析</a><b><font color="#000000">|</font></b> 
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<pre align="left"><font face="Arial, Helvetica, sans-serif" size="4" color="#000000"><b>
<font color="#FF0000">brute force algorithm</font>

<i>O(n) -- the cost of each of the binary search tree
N(n) --  the total number of distinct binary search trees</i>
<font color="#FF0000">complexity: </font>            
                   O( nN(n) )                  N(n) = O( <img src="image/4n.gif" width="76" height="20" align="middle">  )

<font color="#FF0000">efficient algorithm</font>
 Let   <i> a1 < a2 < … < an</i>
<i><font color="#0033CC">T ij </font></i><font color="#0033CC"> </font>  ----  an optimal binary search tree for  <i>ai+1, …, aj ,</i>   i < j.
<i><font color="#0033CC">T ii </font> </i> ----   an empty tree for <i>0 &lt;=  i  &lt;= n    </i>	
<i> <font color="#0033CC">cij </font></i>   ----  the cost of the search tree <i><font color="#0033CC">Tij </font></i>  ( cii  = 0 )
<i><font color="#0033CC"> r ij   </font></i>----  the root of <i><font color="#0033CC">T ij </font></i>      ( <i><font color="#0033CC">rii</font></i>  = 0 )
<i><font color="#0033CC"> wij</font></i>  ----  the  weight of <i><font color="#0033CC">T ij</font></i> ,        <i><font color="#0033CC">wij  </font></i> = <img src="image/qi.gif" width="175" height="52" align="middle">   ,        ( <i>wii</i>  = 0 )
 <i><font color="#0033CC">T 0n</font></i>    an optimal binary search tree for <i> a1, a2,…, an</i>
<i><font color="#0033CC">T ij </font></i><font color="#0033CC"> </font> ---  an optimal binary search tree for  <i>ai+1, …, aj </i>, <img src="image/text10-m.gif" width="144" height="63" align="middle">
 <i><font color="#0033CC">r ij </font></i>  = k  (<i> i < k  &lt;=  j  </i>)                        
  
  L   ---  the left subtree   <i> ( ai+1, …, ak-1  )</i>
  R   ---  the right subtree <i> ( ak, …, ak-1    )</i>
  weight (L ) =  weight <i>(<font color="#0033CC">T i, k-1 </font>) = <font color="#0033CC">w i,k-1</font></i>
  weight (R ) =  weight<i> (<font color="#0033CC">T kj </font>) = <font color="#0033CC">w kj</font></i>

<i><font color="#0033CC">cij   </font></i> =  pk   + cost(L ) + cost (R ) + weight (L ) + weight (R )

if  cost(L ) =<i><font color="#0033CC"> ci,k-1</font></i>    and  cost (R ) = <i><font color="#0033CC"> c kj</font></i> ,  then cij  is minimal

<i><font color="#0033CC">cij </font></i>   =  <i>pk   + ci,k-1  + c kj  + w i,k-1 + w kj    =  w ij  + ci,k-1  + c kj
                 </i><font color="#FF0000"><i>w ij  + ci,k-1  + c kj</i>   = <img src="image/min.gif" width="43" height="35" align="middle"> { <i>w ij  + ci,l-1  + c lj </i>}
<font color="#000000">or    </font>         <i>ci,k-1  + c kj </i>  = </font><font face="Arial, Helvetica, sans-serif" size="4" color="#000000"><b><font color="#FF0000"><img src="image/min.gif" width="43" height="35" align="middle"></font></b></font><font color="#FF0000"> {<i> ci,l-1  + c lj</i> }</font></b></font></pre>
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