编写算法，实现三元组表表示的两个稀疏矩阵的加法，c语言

以下是实现稀疏矩阵加法的 C 语言代码，其中使用三元组表表示稀疏矩阵：

#include <stdio.h>
#include <stdlib.h>

#define MAX_TERMS 101

typedef struct {
    int row, col, value;
} term;

void sparse_add(term a[], term b[], term c[]) {
    int i = 1, j = 1, k = 0;
    while (i <= a[0].value && j <= b[0].value) {
        if (a[i].row < b[j].row) {
            c[++k] = a[i++];
        } else if (a[i].row > b[j].row) {
            c[++k] = b[j++];
        } else {
            if (a[i].col < b[j].col) {
                c[++k] = a[i++];
            } else if (a[i].col > b[j].col) {
                c[++k] = b[j++];
            } else {
                c[++k].row = a[i].row;
                c[k].col = a[i].col;
                c[k].value = a[i++].value + b[j++].value;
            }
        }
    }
    while (i <= a[0].value) {
        c[++k] = a[i++];
    }
    while (j <= b[0].value) {
        c[++k] = b[j++];
    }
    c[0].row = a[0].row;
    c[0].col = a[0].col;
    c[0].value = k;
}

int main() {
    int m1, n1, m2, n2;
    printf("Enter the rows and columns of first matrix: ");
    scanf("%d%d", &m1, &n1);
    printf("Enter the rows and columns of second matrix: ");
    scanf("%d%d", &m2, &n2);
    if (n1 != m2) {
        printf("Matrices cannot be added!\n");
        return 0;
    }
    term a[MAX_TERMS], b[MAX_TERMS], c[MAX_TERMS];
    printf("Enter the first matrix:\n");
    a[0].row = m1;
    a[0].col = n1;
    int cnt = 0;
    for (int i = 1; i <= m1; i++) {
        for (int j = 1; j <= n1; j++) {
            int x;
            scanf("%d", &x);
            if (x != 0) {
                a[++cnt].row = i;
                a[cnt].col = j;
                a[cnt].value = x;
            }
        }
    }
    a[0].value = cnt;
    printf("Enter the second matrix:\n");
    b[0].row = m2;
    b[0].col = n2;
    cnt = 0;
    for (int i = 1; i <= m2; i++) {
        for (int j = 1; j <= n2; j++) {
            int x;
            scanf("%d", &x);
            if (x != 0) {
                b[++cnt].row = i;
                b[cnt].col = j;
                b[cnt].value = x;
            }
        }
    }
    b[0].value = cnt;
    sparse_add(a, b, c);
    printf("The result matrix:\n");
    for (int i = 1; i <= c[0].value; i++) {
        printf("%d %d %d\n", c[i].row, c[i].col, c[i].value);
    }
    return 0;
}

该算法的时间复杂度为 $O(m_1n_1+m_2n_2)$，其中 $m_1,n_1$ 和 $m_2,n_2$ 分别是两个稀疏矩阵的行列数。