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<p ALIGN="JUSTIFY"><font size="5" color="#FFFFFF"><b><font FACE="??ì?,SimSun" LANG="ZH-CN">⑶做矩阵乘法。将</font>A.data<font FACE="??ì?,SimSun" LANG="ZH-CN">中三元组的列值与</font>B.data<font FACE="??ì?,SimSun" LANG="ZH-CN">中三元组的行值相等的非零元素相乘，并将具有相同下标的乘积元素相加。</font></b></font></p>
<p ALIGN="JUSTIFY"><font FACE="??ì?,SimSun" LANG="ZH-CN" size="5" color="#FFFFFF"><b>算法如下：</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>SPMatrix
*MulSMatrix (SPMatrix *A, SPMatrix *B)</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>/*<font FACE="??ì?,SimSun" LANG="ZH-CN">稀疏矩阵</font>A(m<sub>1</sub>×
n<sub>1</sub>)<font FACE="??ì?,SimSun" LANG="ZH-CN">和</font>B(m<sub>2</sub>×
n<sub>2</sub>) <font FACE="??ì?,SimSun" LANG="ZH-CN">用三元组表存储，求</font>A×B
*/</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>{
SPMatrix *C; /* <font FACE="??ì?,SimSun" LANG="ZH-CN">乘积矩阵的指针</font>
*/</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>int
p,q,i,j,k,r;</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>datatype
temp[n+1];</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>int
num[B->nu+1],rpot[B->nu+1];</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>if
(A->nu!=B->mu) return NULL; /*A<font FACE="??ì?,SimSun" LANG="ZH-CN">的列与</font>B<font FACE="??ì?,SimSun" LANG="ZH-CN">的行不相等</font>*/</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>C=malloc(sizeof(SPMatrix));
/*<font FACE="??ì?,SimSun" LANG="ZH-CN">申请</font>C<font FACE="??ì?,SimSun" LANG="ZH-CN">矩阵的存储空间</font>*/</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>C->mu=A->mu;
C->nu=B->nu;</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>if
(A->tu*B->tu==0) {C->tu=0; return C; }</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>for
(i=1;i<= B->mu;i++) num[i]=0; /*<font FACE="??ì?,SimSun" LANG="ZH-CN">求矩阵</font>B<font FACE="??ì?,SimSun" LANG="ZH-CN">中每一行非零元素的个数</font>*/</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>for
(k=1;k<=B->tu;k++)<font FACE="??ì?,SimSun" LANG="ZH-CN"> </font></b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>{
i= B->data[k].i;</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>num[i]++;</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>}</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b><font FACE="??ì?,SimSun" LANG="ZH-CN"> </font>
rpot[1]=1; /*<font FACE="??ì?,SimSun" LANG="ZH-CN">求矩阵</font>B<font FACE="??ì?,SimSun" LANG="ZH-CN">中每一行第一个非零元素在</font>B.data<font FACE="??ì?,SimSun" LANG="ZH-CN">中的位置</font>*/</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>for
(i=2;i<=B->mu;i++)</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>rpot[i]=
rpot[i-1]+num[i-1];</b></font></p>
<font SIZE="3">
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"> </p>
</font>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>r=0;
<font FACE="??ì?,SimSun" LANG="ZH-CN"> </font>/*<font FACE="??ì?,SimSun" LANG="ZH-CN">当前</font>C<font FACE="??ì?,SimSun" LANG="ZH-CN">中非零元素的个数</font>*/</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>p=1;<font FACE="??ì?,SimSun" LANG="ZH-CN"> </font>/*<font FACE="??ì?,SimSun" LANG="ZH-CN">指示</font>A.data<font FACE="??ì?,SimSun" LANG="ZH-CN">中当前非零元素的位置</font>*/<font FACE="??ì?,SimSun" LANG="ZH-CN"> </font></b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>for
( i= 1;i<=A->mu; i++)</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>{
for (j=1;j<=B->nu;j++) temp[j]=0;<font FACE="??ì?,SimSun" LANG="ZH-CN"> </font>/*c<sub>ij</sub><font FACE="??ì?,SimSun" LANG="ZH-CN">的累加器初始化</font>*/</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>while
(A->data[p].i==i ) ./*<font FACE="??ì?,SimSun" LANG="ZH-CN">求第</font>i<font FACE="??ì?,SimSun" LANG="ZH-CN">行的</font>*/</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>{
k=A->data[p].j;<font FACE="??ì?,SimSun" LANG="ZH-CN"> </font>/*A<font FACE="??ì?,SimSun" LANG="ZH-CN">中当前非零元的列号</font>*/</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>if
(k<B->mu) t=rpot[k+1];<font FACE="??ì?,SimSun" LANG="ZH-CN"> </font></b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>else
t=B->tu+1; /*<font FACE="??ì?,SimSun" LANG="ZH-CN">确定</font>B<font FACE="??ì?,SimSun" LANG="ZH-CN">中第</font>k<font FACE="??ì?,SimSun" LANG="ZH-CN">行的非零元素在</font>B.data<font FACE="??ì?,SimSun" LANG="ZH-CN">中的下限位置</font>*/</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>for
(q=rpot[k]; q<t; q++;)<font FACE="??ì?,SimSun" LANG="ZH-CN"> </font>/* B<font FACE="??ì?,SimSun" LANG="ZH-CN">中第</font>k<font FACE="??ì?,SimSun" LANG="ZH-CN">行的每一个非零元素</font>*/</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>{
j=B->data[q].j;<font FACE="??ì?,SimSun" LANG="ZH-CN"> </font></b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>temp[j]+=A->data[p].v
* B->data[q].v</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>}</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>p++;</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>}
/* while */</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>for
(j=1;j<=B->nu;j++)</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>if
(temp[j] )</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>{
r++;;</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>C->data[r]={i,j,temp[j]
};</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>}</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>}
/*for i*/</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>C->tu=r;</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>return
C;</b></font></p>
<p ALIGN="justify" style="margin-top: 0; margin-bottom: 0"><font size="5" color="#FFFFFF"><b>}/*
MulSMatrix */</b></font></p>
<p ALIGN="JUSTIFY"><font size="5" color="#FFFFFF"><b><font FACE="??ì?,SimSun" LANG="ZH-CN">算法</font>5.3
<font FACE="??ì?,SimSun" LANG="ZH-CN">稀疏矩阵的乘积</font></b></font></p>
<p> <font color="#FFFFFF"><b><font SIZE="5"><font FACE="??ì?,SimSun" LANG="ZH-CN">分析上述算法的时间性能如下：</font></font></b></font></p>
<p style="margin-top: 0; margin-bottom: 0"><font face="??ì?,SimSun" lang="ZH-CN" size="5" color="#FFFFFF"><b> </b></font><font color="#FFFFFF"><b><font SIZE="5"><font FACE="??ì?,SimSun" LANG="ZH-CN">1、求</font>num<font FACE="??ì?,SimSun" LANG="ZH-CN">的时间复杂度为</font>O(B->nu+B->tu)<font FACE="??ì?,SimSun" LANG="ZH-CN">；</font></font></b></font></p>
<p style="margin-top: 0; margin-bottom: 0"><font face="??ì?,SimSun" lang="ZH-CN" size="5" color="#FFFFFF"><b> </b></font><font color="#FFFFFF"><b><font SIZE="5"><font FACE="??ì?,SimSun" LANG="ZH-CN">2、求</font>rpot
<font FACE="??ì?,SimSun" LANG="ZH-CN">时间复杂度为</font>O(B->mu)<font FACE="??ì?,SimSun" LANG="ZH-CN">；</font></font></b></font></p>
<p style="margin-top: 0; margin-bottom: 0"><font face="??ì?,SimSun" lang="ZH-CN" size="5" color="#FFFFFF"><b> </b></font><font color="#FFFFFF"><b><font SIZE="5"><font FACE="??ì?,SimSun" LANG="ZH-CN">3、求</font>temp<font FACE="??ì?,SimSun" LANG="ZH-CN">时间复杂度为</font>O(A->mu*B->nu)<font FACE="??ì?,SimSun" LANG="ZH-CN">；</font></font></b></font></p>
<p style="margin-top: 0; margin-bottom: 0"><font face="??ì?,SimSun" lang="ZH-CN" size="5" color="#FFFFFF"><b> </b></font><font color="#FFFFFF"><b><font SIZE="5"><font FACE="??ì?,SimSun" LANG="ZH-CN">4、求</font>C<font FACE="??ì?,SimSun" LANG="ZH-CN">的所有非零元素的时间复杂度为</font>O(</font><font size="3">A->tu*B->tu/B->mu</font><font SIZE="5">)<font FACE="??ì?,SimSun" LANG="ZH-CN">；</font></font></b></font></p>
<p style="margin-top: 0; margin-bottom: 0"><font face="??ì?,SimSun" lang="ZH-CN" size="5" color="#FFFFFF"><b> </b></font><font color="#FFFFFF"><b><font SIZE="5"><font FACE="??ì?,SimSun" LANG="ZH-CN">5、压缩存储时间复杂度为</font>O(A->mu*B->nu)<font FACE="??ì?,SimSun" LANG="ZH-CN">；</font></font></b></font></p>
<p><font color="#FFFFFF"><b><font SIZE="5"><font FACE="??ì?,SimSun" LANG="ZH-CN">所以总的时间复杂度为</font>O(</font><font size="4">A->mu*B->nu+(A->tu*B->tu)/B->nu</font><font SIZE="5">)<font FACE="??ì?,SimSun" LANG="ZH-CN">。</font></font></b></font></p>
<p align="center"><b><a href="ds5.3.HTM"><font size="5" color="#FFFF00">返回</font></a></b></p>
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