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<font color="#FF0033"><b><font face="宋体" size="4">最小价周游路线问题 = 归约矩阵C'的最小代价周游路线问题</font></b></font><b><font face="宋体" size="4" color="#000000">
<font color="#FF0033">方法</font>：
①对每一个节点建立一个相应的归约代价矩阵。并定义下界值<font color="#FF0033">r</font>.
根节点下界值 r = <img src="IMAGE/c.gif" width="13" height="23" align="middle">(root)
原始归约代价矩阵 (25)
扩展节点下界值
<font color="#FF0033"><i><font face="Arial, Helvetica, sans-serif">r = <img src="IMAGE/c.gif" width="13" height="23" align="middle">( S ) = <img src="IMAGE/c.gif" width="13" height="23" align="middle">( R ) + Ar (i, j) + rs</font></i></font>
<i><font face="Arial, Helvetica, sans-serif"><img src="IMAGE/c.gif" width="13" height="23" align="middle">( R ) </font></i> 父节点的下界值
<i><font face="Arial, Helvetica, sans-serif">Ar (i, j)</font></i> 归约代价矩阵的 Cij 代价
<font face="Arial, Helvetica, sans-serif"><i>rs </i></font> 新矩阵归约后的r.
② 选取活动节点表中具有最小<i><font face="Arial, Helvetica, sans-serif"><img src="IMAGE/c.gif" width="13" height="23" align="bottom"></font></i>的点为当前扩展点。
新矩阵归约：
· 选定一节点i扩展到k时，
<i><font face="Arial, Helvetica, sans-serif">Ar(i, k) </font></i>的 i 行， k列的点不可取, 相应元素量为∞
· <i><font face="Arial, Helvetica, sans-serif">Ar(k, 1)</font></i>为∞
· 再对新矩阵归约，求rs.
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