简单C程序，迷宫

 #include<stdio.h>
 #include <stdlib.h>
 int visit(int, int);
 int maze[][] =
 {
     {, , , , , , , ,  ,},
     {, , , , , , , ,  ,},
     {, , , , , , , ,  ,},
     {, , , , , , , ,  ,},
     {, , , , , , , ,  ,},
     {, , , , , , , ,  ,},
     {, , , , , , , ,  ,},
     {, , , , , , , ,  ,},
     {, , , , , , , ,  ,},
     {, , , , , , , , , },
     {, , , , , , , , , },
 };
 int startI = , startJ = ; // 入口
 int endI = , endJ = ; // 出口
 int success = ;

 int main(void) {
     int i, j;

     printf("显示迷宫：\n");
     for (i = ; i < ; i++) {
         for (j = ; j < ; j++)
             if (maze[i][j] == )
                 printf("■");
             else
                 printf("  ");
         printf("\n");
     }

     if (visit(startI, startJ) == )
         printf("\n沒有找到出口！\n");
     else {
         printf("\n显示路径：\n");
         for (i = ; i < ; i++) {
             for (j = ; j < ; j++) {
                 if (maze[i][j] == )
                     printf("■");
                 else if (maze[i][j] == )
                     printf("▲");
                 else
                     printf("  ");
             }
             printf("\n");
         }
     }

     return ;
 }

 int visit(int i, int j) {
     maze[i][j] = ;

     if (i == endI && j == endJ)
         success = ;

     if (success !=  && maze[i][j + ] == ) visit(i, j + );
     if (success !=  && maze[i + ][j] == ) visit(i + , j);
     if (success !=  && maze[i][j - ] == ) visit(i, j - );
     if (success !=  && maze[i - ][j] == ) visit(i - , j);

     if (success != )
         maze[i][j] = ;

     return success;
 }

 //参考.c
 #include<iostream>

 using namespace std;
 void strMatrix(int *a1, int **a2, int n);
 void output(int **a2, int n);
 int main()
 {
     int n, i, c, d;
     cout << "How many numbers do you want to input:" << endl;
     cin >> n;
     int *arr1 = new int[n];
     int **arr2 = new int*[n];
     for (i = ; i < n; i++)
         arr2[i] = new int[n];
     cout << "Please input the numbers:" << endl;
     for (i = ; i < n; i++)
         cin >> arr1[i];
     cout << "Please input the ralations:" << endl;
     cin >> c;
     cin >> d;
     strMatrix(arr1, arr2, n);
     output(arr2, n);
 }
 void strMatrix(int *a1, int **a2, int n)
 {
     for (int i = ; i < n; i++)
         for (int j = ; j < n; j++)
         {
             if (a1[i] % a1[j] == )
                 a2[i][j] = ;
             else
                 a2[i][j] = ;
         }
 }
 void output(int **a2, int n)
 {
     for (int i = ; i < n; i++)
     {
         for (int j = ; j < n; j++)
             cout << a2[i][j] << " ";
         cout << endl;
     }
 }

 void zfx(int a[M][M], int n)
 {
     int i = , vale1 = , vale2 = ;/*vale1确定关系的自反性,vale2
                                     判断关系的反自反性*/
     while (i < n)
     {
         if (a[i][i] == ) vale2 = ;
         else vale1 = ;
         i++;
     }
     if (vale1 == ) printf("zfx\n");
     if (vale2 == ) printf("fzfx\n");
     if (!(vale1) && !(vale2)) printf("wzfxywfzfx\n");
 }
 void dcx(int a[M][M], int n)
 {
     int i, j, vale3 = , vale4 = ;/*vale3确定关系的对称性,vale4
                                    判断关系的反对称性*/
     for (i = ; i < n; i++)
         for (j = ; j < i; j++)
             if (a[i][j] == a[j][i])
                 vale4 = ;
             else vale3 = ;
             if (vale3 == ) printf("dcx\n");
             if (vale4 == ) printf("fdcx\n");
             if (!(vale3) && !(vale4)) printf("wdcx wfdcx\n");
 } void cdx1(int a[M][M], int n)/*用MM ⊆2 算法来判断传递 性*/
 {
     int b[M][M], i, j, k, vale = ;
     for (i = ; i < n; i++)
         for (j = ; j < n; j++)
         {
             b[i][j] = ;
             for (k = ; k < n; k++)
                 b[i][j] += a[i][k] * a[k][j];/*计算 2M */
         }
         for (i = ; i < n; i++)
             for (j = ; j < n; j++)
                 if (b[i][j] != )
                     b[i][j] = ;
         for (i = ; i < n; i++)
             for (j = ; j < n; j++)
                 if (b[i][j] == )
                     if (a[i][j] != b[i][j])
                         vale = ;
         if (vale)
             printf("ycdx\n");
         else
             printf("wcdx\n");
 }
 void cdx2(int a[M][M], int n)/*用warshall算法来求传递闭
                              包数组b */
 {
     int b[M][M], i, j, k, vale = ;
     for (i = ; i < n; i++)
         for (j = ; j < n; j++)
             b{ [i][j] = a[i][j];
     for (k = ; k < n; k++)
         i{ (f a[i][j] && (a[i][k] || a[j][k]))
         b[i][k] = ;
     }
     }
     for (i = ; i < n; i++)     /* 判断数组a和b相等 */
         for (j = ; j < n; j++)
             (iaf[i][j] != b[i][j])
         {
             vale = ; break;
         }
         if (vale == ) printf("cdx\n");
         else printf("wcdx\n");
 }