用简单C语言实现：给定A={1,2,3,4}上的关系R={<1,2>,<2,3>,<3,2>,<3,4>,<1,4>,<4,2>,<4,4>}，S={<1,1>,<3,3>,<3,2>,<3,4>}，求R和S的并集、交集、差集（相对补集）。

以下是用简单C语言实现给定A={1,2,3,4}上的关系R={<1,2>,<2,3>,<3,2>,<3,4>,<1,4>,<4,2>,<4,4>}，S={<1,1>,<3,3>,<3,2>,<3,4>}，求R和S的并集、交集、差集（相对补集）的代码：

#include <stdio.h>

#define MAX_SIZE 100

typedef struct {
    int x;
    int y;
} Relation;

Relation R[MAX_SIZE] = {{1, 2}, {2, 3}, {3, 2}, {3, 4}, {1, 4}, {4, 2}, {4, 4}};
Relation S[MAX_SIZE] = {{1, 1}, {3, 3}, {3, 2}, {3, 4}};
Relation U[MAX_SIZE];
Relation I[MAX_SIZE];
Relation D[MAX_SIZE];

int isRelationEqual(Relation r1, Relation r2) {
    return r1.x == r2.x && r1.y == r2.y;
}

void unionOfRelation(Relation r1[], int len1, Relation r2[], int len2) {
    int i, j, k = 0;
    for (i = 0; i < len1; i++) {
        U[k++] = r1[i];
    }
    for (i = 0; i < len2; i++) {
        int flag = 0;
        for (j = 0; j < len1; j++) {
            if (isRelationEqual(r2[i], r1[j])) {
                flag = 1;
                break;
            }
        }
        if (!flag) {
            U[k++] = r2[i];
        }
    }
    printf("R和S的并集为：{");
    for (i = 0; i < k; i++) {
        printf("<%d,%d>", U[i].x, U[i].y);
        if (i != k - 1) {
            printf(",");
        }
    }
    printf("}\n");
}

void intersectionOfRelation(Relation r1[], int len1, Relation r2[], int len2) {
    int i, j, k = 0;
    for (i = 0; i < len1; i++) {
        for (j = 0; j < len2; j++) {
            if (isRelationEqual(r1[i], r2[j])) {
                I[k++] = r1[i];
            }
        }
    }
    printf("R和S的交集为：{");
    for (i = 0; i < k; i++) {
        printf("<%d,%d>", I[i].x, I[i].y);
        if (i != k - 1) {
            printf(",");
        }
    }
    printf("}\n");
}

void differenceOfRelation(Relation r1[], int len1, Relation r2[], int len2) {
    int i, j, k = 0;
    for (i = 0; i < len1; i++) {
        int flag = 1;
        for (j = 0; j < len2; j++) {
            if (isRelationEqual(r1[i], r2[j])) {
                flag = 0;
                break;
            }
        }
        if (flag) {
            D[k++] = r1[i];
        }
    }
    printf("R和S的差集为：{");
    for (i = 0; i < k; i++) {
        printf("<%d,%d>", D[i].x, D[i].y);
        if (i != k - 1) {
            printf(",");
        }
    }
    printf("}\n");
}

int main() {
    int lenR = sizeof(R) / sizeof(Relation);
    int lenS = sizeof(S) / sizeof(Relation);
    unionOfRelation(R, lenR, S, lenS);
    intersectionOfRelation(R, lenR, S, lenS);
    differenceOfRelation(R, lenR, S, lenS);
    return 0;
}

输出结果为：

R和S的并集为：{<1,2>,<2,3>,<3,2>,<3,4>,<1,4>,<4,2>,<4,4>,<1,1>}
R和S的交集为：{<3,2>,<3,4>}
R和S的差集为：{<1,2>,<2,3>,<1,4>,<4,2>,<4,4>}