4

鉴于：

1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8 
1 2 3 4 5 6 7 8

我想将二维数组（struct MATRIX）拆分为给定块大小 CS 的 struct MATRIX 数组：假设 cs 为 2，答案是

Seg[0]:
1 2 
1 2 
1 2
Seg[1]:
3 4 
3 4 
3 4
....
Seg[3]:
7 8
7 8
7 8

这是我的矩阵结构：

typedef struct MATRIX {
    int nrow;
    int ncol;
    int **element;
} MATRIX;

这是将它们分开的功能：

void SegmentMatrix(MATRIX input,MATRIX* segs,int Chunksize, int p) {
    int i,j,r;

    //Allocate segs
    for (i = 0; i<p;i++)
    {
        CreateMatrix(&(segs[i]),input.nrow ,Chunksize,0);
    }

    //Now Copy the elements from input to the segs
    //where seg0 takes from 0 to cs cols of a, and all their rows, and seg1 takes from cs to 2cs ...
    printf("Stats:\n\t P: %d\t CS: %d\n",p,Chunksize);
    for (r = 0; r<p; r++) {
        for (i = 0; i<input.nrow;i++) {
            for (j = r*Chunksize; j<r*Chunksize+Chunksize-1; j++) {
                 //I tried (&(segs[r]))->element... Doesn't work, produces wrong data
                 segs[r].element[i][j] = input.element[i][j];

        }
    }
    PRINTM(segs[r]);
    }


}

请注意，PRINTM 基本上打印矩阵，它通过检查 segs[r].nrow 和 ncol 来了解限制，并且 CreateMatrix 从内部获取以下输入（&matrix、行数、列数、填充类型）和 mallocs。

filltype: 
0- generates zeroth matrix
1- generates identity
else A[i][j] = j; for simplicity

问题是，如果我打印矩阵 Segs[i]，它们都以 CreateMatrix 给出的默认值归结，而不是新添加的值。

澄清：好的，所以如果你们检查 SegmentMatrix 函数中的最后一个 PRINTM，它会输出矩阵，就好像没有发生 for 循环一样，我可以删除 for 循环并得到相同的输出.. 我做了什么吗此行错误（取自 SegmentMatrix）

Segs[r].element[i][j] = input.element[i][j];
4

2 回答 2

5

我不明白为什么和你用ChunkSizeand乘法来操作r（无论如何都未初始化），我建议简化代码（经验法则：如果它看起来很乱，那就太复杂了）。您只需要一个 3 维数组来存储块数组，然后将模算术加整数除法插入到相应块的相应列中：

/* the variable-sized dimension of the `chunks' argument is w / chsz elements big
 * (it's the number of chunks)
 */
void split(int h, int w, int mat[h][w], int chsz, int chunks[][h][chsz])
{
    /* go through each row */
    for (int i = 0; i < h; i++) {
        /* and in each row, go through each column */
        for (int j = 0; j < w; j++) {
            /* and for each column, find which chunk it goes in
             * (that's j / chsz), and put it into the proper row
             * (which is j % chsz)
             */
            chunks[j / chsz][i][j % chsz] = mat[i][j];
        }
    }
}

演示，又名如何称呼它：

int main(int agrc, char *argv[])
{
    const size_t w = 8;
    const size_t h = 3;
    const size_t c = 2;

    int mat[h][w] = {
        { 1, 2, 3, 4, 5, 6, 7, 8 },
        { 1, 2, 3, 4, 5, 6, 7, 8 },
        { 1, 2, 3, 4, 5, 6, 7, 8 }
    };

    int chunks[w / c][h][c];

    split(h, w, mat, c, chunks);

    for (int i = 0; i < w / c; i++) {
        for (int j = 0; j < h; j++) {
            for (int k = 0; k < c; k++) {
                printf("%3d ", chunks[i][j][k]);
            }
            printf("\n");
        }
        printf("\n\n");
    }

    return 0;
}
于 2013-05-13T19:14:45.443 回答
2

问题不清楚。所以我认为他只想知道如何实现这一目标。所以我写了这个简单的伪代码。否则接受我的道歉：

matrix[i] matrix
//matrixes total column size should be bigger big 2d array column size
first condition check: sum(matrix[i].colsize)>=big2d.colsize
//in this simple code raw sizes must be equal
second condition: for all i matrix[i].rawsize=big2d.rawsize
//if columns sizes will be equal the algorithm could be simplified , does not mean optimized
 //splitting big2d into matrixes
for (int br=0;br<big2d.rawsize;br++){
i=0;//store matrix index
int previndex=0;//store offset for next matrix
  for(int bc=0;bc<big2d.colsize;bc++){

      matrix[i].val[bc-previndex][br]=big2d.val[bc][br]; //assign (bc,br) 

      if(bc-previndex==matrix[i].colsize-1){
             i++; //move to next matrix;//if we not have next matrix then break;
            previndex=bc+1; 
          }
     /*if it be for equal chunks matrixes offset can be calculated this way too
         matrix[bc/chunk].val[bc%chunk][br]=big2d.val[bc][br];
      */
  }//loop columns
}//loop raws
于 2013-05-13T18:59:47.087 回答