学步园

先讲一个最基本的矩阵变换函数以后的各种变换都要用到：

/*由于元素为double类型故当与0或1比较式应当近似比较：*/<br /> #define N 0.000001<br /> #define Ng (N + 1)<br /> #define Nl (1 - N)</p> <p>lbian change_data(data *D )<br /> {<br /> lbian count;<br /> lbian i,j,k,l;<br /> lbian hang,lie;<br /> datatype temp;<br /> datatype num;</p> <p> hang = D -> hang;<br /> lie = D -> lie;<br /> count = 0;</p> <p> for(i = 1; i < lie ;i ++) { /*从1到lie-1列扫描 */<br /> for(j = i+1;j <= hang ;j ++) { /*第i列则从第i+1行开始变换使第i行以下的全变为零*/<br /> if( ((D->top)[j][i] < N) && ((D->top)[j][i] > -N)) /*遇到top[i][j]是0则不用变换*/<br /> continue;<br /> for( k = i ;k <= hang ;k ++) {<br /> if(((D->top)[k][i] > N) || ((D->top)[k][i] < -N))<br /> break;<br /> } /*在第i列开始（包括第i列）找到第一个非0的元素*/<br /> if( k != i && k <= hang ) {<br /> for( l=1 ; l <= lie ; l ++) {<br /> num = (D -> top)[i][l];<br /> (D -> top)[i][l] = (D -> top)[k][l];<br /> (D -> top)[k][l] = num;<br /> }<br /> count ++; /*找到一个第k行第i列(k != i)非零<br /> 则交换第k行和第i行并 count ++*/<br /> }<br /> if( k > hang ) /*说明第i列的从第i+1行到最末行已经全是0*/<br /> continue;</p> <p> temp = -(datatype)(D -> top)[j][i] / (D -> top)[i][i];<br /> for( k = 1;k <= lie ;k ++)<br /> (D -> top)[j][k] += (datatype)(temp* (D -> top)[i][k]);<br /> /*将第j行加上temp与第i行的乘积可使top[j][i]为0*/<br /> }<br /> }<br /> return count; /*共进行了count次交换行变换*/<br /> }</p> <p>/*这个<span class='wp_keywordlink'><a href="http://www.xuebuyuan.com/category/%E7%AE%97%E6%B3%95" title="算法" target="_blank">算法</a></span>比较复杂（奉劝未学过线性代数的不要看）根据行列式的上三角变换原则进行变换的，<br /> 对于矩阵也可以变换：<br /> 将一个矩阵尽可能使“主对角线”以下为零（对于一个有hang行，lie列的矩阵设min = hang < lie ? hang :lie ;<br /> 则这个算法可以将top[1][1]到top[min][min]这个矩阵变为上三角形矩阵，例如：<br /> --------------------------------------------------------------------------------------------<br /> 4 5 6 变为： 4 5 6<br /> 1 2 3 0 0.75 1.5<br /> --------------------------------------------------------------------------------------------<br /> 1 2 变为： 1 2<br /> 3 4 0 -2<br /> 5 6 0 -4<br /> --------------------------------------------------------------------------------------------<br /> 3 2 0 5 0<br /> 3 -2 3 6 -1<br /> 2 0 1 5 -3<br /> 1 6 -4 -1 4<br /> 变为：<br /> 3 2 0 5 0<br /> 0 -4 3 1 -1<br /> 0 0 0 4/3 -8/3<br /> 0 0 0 -4/3 8/3<br /> -------------------------------------------------------------------------------------------*/

以下的三个函数都必须为目的矩阵申请好内存：

/*------------------------------两矩阵和或差,A和B的和放在c中--------------------------------<br /> --------------------------------如果flag>0则求和否则求差-----------------------------------*/<br /> /*------------------------------两矩阵和 ,A和B的和放在c中-----------------------------------------*/<br /> int32 data_plus(data *C,data *A,data *B,int8 flag)<br /> {</p> <p> lbian i,j;<br /> if((A->hang != B->hang) || (A->lie!=B->lie)) {<br /> printf("条件不符不能求和! /n");<br /> return 0;<br /> }<br /> for(i=1 ; i<=A->hang; i ++)<br /> for(j=1 ; j <= A->lie;j ++) {<br /> if(flag>0)<br /> (C->top)[i][j] = (A->top)[i][j] + (B->top)[i][j];<br /> else<br /> (C->top)[i][j] = (A->top)[i][j] - (B->top)[i][j];<br /> }<br /> return 1;<br /> } </p> <p>/*------------------------------两矩阵积 ,A和B的乘积放在c中-----------------------------------------*/<br /> int data_mul(data *C,data *A,data *B)<br /> {<br /> lbian i, j , k;<br /> if(A->lie!=B->hang){<br /> printf("条件不符不能求积! /n");<br /> return 0;<br /> }<br /> for ( i =1 ;i <= C -> hang ; i ++ )<br /> for ( j =1 ; j <= C ->lie ; j ++)<br /> (C -> top)[i][j] = 0;<br /> for(i = 1;i <= A->hang;i ++)<br /> for(j = 1;j <= B->lie;j++)<br /> for(k=1;k<=A->lie;k++)<br /> C -> top [i][j] += (A -> top)[i][k] * (B->top)[k][j];<br /> return 1;<br /> } </p> <p>/*----------------------------------拷贝两个矩阵的内容 -------------------------------------------------*/<br /> void data_cpy ( data *des , data *src )<br /> {<br /> lbian hang = src ->hang;<br /> lbian lie = src ->lie ;<br /> lbian i,j;<br /> for (i = 1; i <= hang;i ++)<br /> for (j = 1; j <= lie; j ++)<br /> (des->top)[i][j] = (src->top)[i][j];<br /> }<br />