[Postgraduate entrance examination data structure code question 8] Fast transposition of sparse matrix represented by triples

Title: Fast transposition of sparse matrices represented by triples

Test point: Compression and storage of matrices

Difficulty: ★★★

Sparse matrix compression storage

triple structure

//三元组结构 
typedef struct {
    int row;
    int col;
    int value;
} Term;

core code

// 转置函数，a为原矩阵，b存放转置后的矩阵 
void transpose(Term a[], Term b[]) {
    int rowTerms[100], startingPos[100];
    int numCols = a[0].col, numRows = a[0].row, numTerms = a[0].value;

    b[0].row = numCols; // 转置后的矩阵行数等于原矩阵的列数
    b[0].col = numRows; // 转置后的矩阵列数等于原矩阵的行数
    b[0].value = numTerms; // 转置后的矩阵非零元素个数等于原矩阵非零元素个数

    if (numTerms > 0) {
        // 统计每一列非零元素的个数
        for (int i = 0; i < numCols; i++)
            rowTerms[i] = 0;
        for (int i = 1; i <= numTerms; i++)
            rowTerms[a[i].col]++;

        // 计算转置后每一列非零元素的起始位置
        startingPos[0] = 1;
        for (int i = 1; i < numCols; i++)
            startingPos[i] = startingPos[i - 1] + rowTerms[i - 1];

        // 执行转置操作（核心） 
        for (int i = 1; i <= numTerms; i++) {
            int j = startingPos[a[i].col]++;
            b[j].row = a[i].col; // 转置后的元素行下标等于原矩阵元素列下标
            b[j].col = a[i].row; // 转置后的元素列下标等于原矩阵元素行下标
            b[j].value = a[i].value; // 转置后的元素值不变
        }
    }
}

Complete code

#include <stdio.h>

#define MAX_TERMS 100
//三元组结构 
typedef struct {
    int row;
    int col;
    int value;
} Term;

// 转置函数，a为原矩阵，b存放转置后的矩阵 
void transpose(Term a[], Term b[]) {
    int rowTerms[100], startingPos[100];
    int numCols = a[0].col, numRows = a[0].row, numTerms = a[0].value;

    b[0].row = numCols; // 转置后的矩阵行数等于原矩阵的列数
    b[0].col = numRows; // 转置后的矩阵列数等于原矩阵的行数
    b[0].value = numTerms; // 转置后的矩阵非零元素个数等于原矩阵非零元素个数

    if (numTerms > 0) {
        // 统计每一列非零元素的个数
        for (int i = 0; i < numCols; i++)
            rowTerms[i] = 0;
        for (int i = 1; i <= numTerms; i++)
            rowTerms[a[i].col]++;

        // 计算转置后每一列非零元素的起始位置
        startingPos[0] = 1;
        for (int i = 1; i < numCols; i++)
            startingPos[i] = startingPos[i - 1] + rowTerms[i - 1];

        // 执行转置操作（核心） 
        for (int i = 1; i <= numTerms; i++) {
            int j = startingPos[a[i].col]++;
            b[j].row = a[i].col; // 转置后的元素行下标等于原矩阵元素列下标
            b[j].col = a[i].row; // 转置后的元素列下标等于原矩阵元素行下标
            b[j].value = a[i].value; // 转置后的元素值不变
        }
    }
}

void printMatrix(Term a[]) {
    int numRows = a[0].row, numCols = a[0].col, numTerms = a[0].value;

    for (int i = 0; i <= numTerms; i++) {
        printf("%3d", a[i].value);
        printf("%3d", a[i].row);
        printf("%3d", a[i].col);
        printf("\n");
    }

    printf("\n");
}

int main() {
    Term a[MAX_TERMS];
    Term b[MAX_TERMS];

    // 初始化原始矩阵
    a[0].row = 6;
    a[0].col = 6;
    a[0].value = 8;

    a[1].row = 0;
    a[1].col = 0;
    a[1].value = 15;
    a[2].row = 0;
    a[2].col = 3;
    a[2].value = 22;
    a[3].row = 0;
    a[3].col = 5;
    a[3].value = -15;
    a[4].row = 1;
    a[4].col = 1;
    a[4].value = 11;
    a[5].row = 1;
    a[5].col = 2;
    a[5].value = 3;
    a[6].row = 2;
    a[6].col = 3;
    a[6].value = -6;
    a[7].row = 4;
    a[7].col = 0;
    a[7].value = 91;

    printf("Original matrix:\n");
    printMatrix(a);

    // 转置矩阵
    transpose(a, b);

    printf("Transposed matrix:\n");
    printMatrix(b);

    return 0;
}