输入一个集合的二元关系，判定其是否满足自反性、反自反性、对称性、反对称性、传递性。并求出自反、对称和传递闭包。
大二上学期时的写的代码，C++语言实现。
#include<iostream>
#include<string>
using namespace std;class Relation//构造函数
{
private:int R[11][2];//存储关系Rint R1[11][2], R2[11][2], R3[11][2];//分别存储自反，对称，传递闭包int m,n;//分别存储二元关系R中的最大值和最小值int o;//存储二元关系个数int M[10][10];//存储转换后的矩阵
public:Relation()//构造函数{n = 10;m = -1;o = PutInR();memset(M,0,sizeof(M));}int PutInR();//输入关系Rvoid Convent();//转换为矩阵void Give();//用R初始化R1,R2bool Reflex();//判断自反性bool UnReflex();//判断反自反性bool Symme();//判断对称性bool AnSymme();//判断反对称性bool Delivery();//判断传递性void Decide();//性质判定void ReflexB();//求自反闭包void SymmeB();//求对称闭包void DeliveryB();//求传递闭包
};//输入关系R -- 以‘-1’结尾
int Relation::PutInR()
{int i = 0, k = 0, l = 0;cout << "请输入关系R： "<<endl;for (i; i <= 10; i++){cin >> R[i][0];cin >> R[i][1];if (R[i][1] == -1)break;//结束标志为-1l = R[i][0] > R[i][1] ? R[i][0] : R[i][1];k = R[i][0] < R[i][1] ? R[i][0] : R[i][1];m = m > l ? m : l;n = n > k ? k : n;}return i;
}//转换为矩阵
void Relation::Convent()
{int i = 0;while (R[i][0] != -1){M[R[i][0]][R[i][1]] = 1;//相应矩阵位置赋1i++;}
}//判断自反性
bool Relation::Reflex()
{for (int i = n; i <= m;i++){if (M[i][i] == 0)return false;//对角线上有元素为0，则不满足}cout << "具有自反性" << endl;return true;
}//判断反自反性
bool Relation::UnReflex()
{for (int i = n; i <= m; i++){if (M[i][i] == 1)return false;//对角线上有元素为1，则不满足}cout << "具有反自反性" << endl;return true;
}//判断对称性
bool Relation::Symme()
{for (int i = n; i <= m; i++){for (int j = i; j <= m; j++){if (M[i][j] == 1 && M[j][i] != 1)return false;//不关与对角线对称，则不满足}}cout << "具有对称性" << endl;return true;
}//判断反对称性
bool Relation::AnSymme()
{for (int i = n; i <= m; i++){for (int j = n; j <= m; j++){if (M[i][j] == 1 && M[j][i] == 1)return false;//关于对角线对称，则不满足if (M[i][j] == 0 && M[j][i] == 0)return false;}}cout << "具有反对称性" << endl;return true;
}//判断传递性
bool Relation::Delivery()
{int k[10] = { 0 }, l[10] = { 0 };for (int i = n; i <= m; i++)//以对角线上每个元素循环{int e = 0, f = 0;for (int j = n; j <= m; j++){if (M[i][j] == 1 && i != j) { k[e++] = j; }//找出第i行的非0元素，列下标记录在a数组中if (M[j][i] == 1 && i != j) { l[f++] = j; }//找出第i列的非0元素，行下标记录在b数组中}for (int c = k[0]; c <= k[--e]; c++)//行上的非0元素{for (int d = l[0]; d <= k[--f]; d++)//列上的非0元素{if (M[i][c] == 1 && M[d][i] != 1|| M[i][c] != 1 && M[d][i] == 1)return false;}}}cout << "具有传递性" << endl;return true;
}//性质判定
void Relation::Decide()
{Convent();Reflex();UnReflex();Symme();AnSymme();Delivery();
}//用R初始化R1,R2
void Relation::Give()
{memcpy(R1, R, 22);memcpy(R1, R, 22);
}//求自反闭包
void Relation::ReflexB()
{int g = o;for (int i = n; i <= m ; i++){if (M[i][i] == 0)//如果缺少对角线上的元素{//连接在R1后面R1[g][0] = i;R1[g][1] = i;g++;}}cout << "自反闭包是：" << endl;for (int j = 0; j < g ; j++){cout <<'<'<<R1[j][0]<<','<<R1[j][1]<<'>';}cout << endl;
}//求对称闭包
void Relation::SymmeB()
{int g = o;for (int i = n; i <= m; i++){for (int j = i; j <= m; j++){if (M[i][j] == 1 && M[j][i] != 1)//如果不关于对角线对称{R2[g][0] = j;//相对应的元素连接在R2后面R2[g][1] = i;g++;}if (M[i][j] != 1 && M[j][i] == 1)//如果不关于对角线对称{R2[g][0] = j;//相对应的元素连接在R2后面R2[g][1] = i;g++;}}}cout << "对称闭包是：" << endl;for (int j = 0; j < g ; j++){cout << '<' << R2[j][0] << ',' << R2[j][1] << '>';}cout << endl;
}//求传递闭包
void Relation::DeliveryB()
{int g = 0;for (int t = n; t <= m ; t++){for (int j = n; j <= m; j++){if (M[t][j] == 1){for (int k = n; k <= m; k++){M[j][k] = M[t][k] || M[j][k];//Warshall算法求解传递闭包}}}}for (int i = n; i <= m; i++)//将矩阵反转回二元数组{for (int j = n; j <= m; j++){if (M[i][j] == 1){R3[g][0] = i;R3[g][1] = j;g++;}}}cout << "传递闭包是：" << endl;for (int j = 0; j < g-1; j++){cout << '<' << R3[j][0] << ',' << R3[j][1] << '>';}cout << endl;
}int main()
{Relation real;real.Decide();
real.Give();real.ReflexB();real.SymmeB();real.DeliveryB();return 0;
}