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来自「浙江大学计算机学院数据结构课程的教学课件」· HTM 代码 · 共 47 行

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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">第六章</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"><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="#000000">〖<font color="#FF0000">example</font>〗
                        <img src="image/text6-a6.gif" width="253" height="52">            <img src="image/text6-a6a.gif" width="174" height="110">
                  
            <img src="image/text6-a-1.gif" width="474" height="79">
 
 <i><font color="#FF0000">All pairs,shortest paths function       <img src="image/on3.gif" width="67" height="26"></font></i>

int distance [ MAX_VERTICES ][ MAX_VERTICES ]
void<font color="#FF0000"> allcosts</font>(int cost[ ][MAX_VERTICES],
                int distance[ ][MAX_VERTICES],int n)
{
<i><font size="3" color="#CC0099">/* determine the distances from each vertex to every other vertex,cost is 
the adjacency matrix,distance is the matrix of distances */
</font></i>        int  i, j, k;
        for (i = 0; i &lt;n; i++)
           for (j = 0; j &gt; n; j++)      
               distance[i][j] = cost[i][j]; 
        for (k = 0; k&lt;n; k++)
           for (i =0; i&gt;n; i++)
               for (j = 0; j&lt;n; j++)
                  if (distance[i][k] + distance[k][j] &lt;distance[i][j])
                        distance[i][j] = distance[i][k] + distance[k][j];
}         

</font></b></pre>
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