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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">第七章</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"><a href="../text10/text10-0.htm">第十章</a></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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<b><font face="Arial, Helvetica, sans-serif" size="4" color="#FF0000"><i>n</i></font><font face="Arial, Helvetica, sans-serif" size="4" color="#000000"> records  --> <i><font color="#FF0000"> m </font></i>  runs  --&gt; [   <img src="image/log2m.gif" width="68" height="24" align="middle"> ] + 1  level  --> k-way merge
--> merge [  <img src="image/logkm.gif" width="69" height="24" align="middle"> ]   passes  

merge k-runs ( <font color="#0033CC"><i>s<font size="2">1</font>,s<font size="2">2</font>,s<font size="2">3</font>,,  …,sk<font size="2"> </font></i></font> ) --> O((<i>k-1) </i>  <img src="image/sums.gif" width="46" height="29" align="middle">) =n(k-1) computing time 
 <font color="#FF0000">total computing time : </font><i>n(k-1)</i> <img src="image/logkm.gif" width="69" height="24" align="middle">    computing number

 <i>n(k-1)</i> <img src="image/logkm.gif" width="69" height="24" align="middle">  = <i> n(k-1)</i> <img src="image/log2m.gif" width="68" height="24" align="middle">  / <img src="image/log2k.gif" width="61" height="24" align="middle">  = <i>n</i> <img src="image/log2m.gif" width="68" height="24" align="middle">  * <i>(k-1) / </i> <img src="image/log2k.gif" width="61" height="24" align="middle">

<i><font color="#FF0000">As k increase , the reduction in input/output time will be offset
 by the resulting increase in CPU time needed to perform the
 k-way merge. </font></i>

<font color="#0033CC">the total internal processing time per level</font>  =  <font color="#FF0000">O(n</font><img src="image/log2k.gif" width="61" height="24" align="middle"><font color="#FF0000"> )</font>
<font color="#0033CC">the total internal processing time :</font>

                   <font color="#FF0000">   O( <i>n </i></font><img src="image/log2k.gif" width="61" height="24" align="middle"> <font color="#FF0000"> </font><img src="image/logkm.gif" width="69" height="24" align="middle"><font color="#FF0000">   ) = O (<i> n</i> </font><img src="image/log2m.gif" width="68" height="24" align="middle"><font color="#FF0000">  )</font></font></b></pre>
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