简单C程序，迷宫

#include<stdio.h>
#include <stdlib.h> 
int visit(int, int);
int maze[11][10] =
{
    {1, 1, 1, 1, 1, 1, 1, 1, 1 ,1},
    {1, 0, 0, 1, 0, 0, 0, 1, 0 ,1},
    {1, 0, 0, 1, 0, 0, 0, 1, 0 ,1},
    {1, 0, 0, 0, 0, 1, 1, 0, 1 ,1},
    {1, 0, 1, 1, 1, 0, 0, 1, 0 ,1},
    {1, 0, 0, 0, 1, 0, 0, 0, 0 ,1},
    {1, 0, 1, 0, 0, 0, 1, 0, 1 ,1},
    {1, 0, 1, 1, 1, 1, 0, 0, 1 ,1},
    {1, 1, 1, 0, 0, 0, 1, 0, 1 ,1},
    {1, 1, 1, 0, 0, 0, 0, 0, 0, 1},
    {1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
};
int startI = 1, startJ = 1; // 入口
int endI = 9, endJ = 8; // 出口
int success = 0;

int main(void) {
    int i, j;

    printf("显示迷宫：\n");
    for (i = 0; i < 9; i++) {
        for (j = 0; j < 8; j++)
            if (maze[i][j] == 1)
                printf("■");
            else
                printf("  ");
        printf("\n");
    }

    if (visit(startI, startJ) == 0)
        printf("\n沒有找到出口！\n");
    else {
        printf("\n显示路径：\n");
        for (i = 0; i < 9; i++) {
            for (j = 0; j < 8; j++) {
                if (maze[i][j] == 1)
                    printf("■");
                else if (maze[i][j] == 2)
                    printf("▲");
                else
                    printf("  ");
            }
            printf("\n");
        }
    }

    return 0;
}

int visit(int i, int j) {
    maze[i][j] = 2;

    if (i == endI && j == endJ)
        success = 2;

    if (success != 2 && maze[i][j + 1] == 0) visit(i, j + 1);
    if (success != 2 && maze[i + 1][j] == 0) visit(i + 1, j);
    if (success != 2 && maze[i][j - 1] == 0) visit(i, j - 1);
    if (success != 2 && maze[i - 1][j] == 0) visit(i - 1, j);

    if (success != 2)
        maze[i][j] = 0;

    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 = 0; i < n; i++)
        arr2[i] = new int[n];
    cout << "Please input the numbers:" << endl;
    for (i = 0; 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 = 0; i < n; i++)
        for (int j = 0; j < n; j++)
        {
            if (a1[i] % a1[j] == 0)
                a2[i][j] = 1;
            else
                a2[i][j] = 0;
        }
}
void output(int **a2, int n)
{
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
            cout << a2[i][j] << " ";
        cout << endl;
    }
}

void zfx(int a[M][M], int n)
{
    int i = 0, vale1 = 1, vale2 = 1;/*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] == 1) vale2 = 0;
        else vale1 = 0;
        i++;
    }
    if (vale1 == 1) printf("zfx\n");
    if (vale2 == 1) printf("fzfx\n");
    if (!(vale1) && !(vale2)) printf("wzfxywfzfx\n");
}
void dcx(int a[M][M], int n)
{
    int i, j, vale3 = 1, vale4 = 1;/*vale3确定关系的对称性,vale4
                                   判断关系的反对称性*/
    for (i = 0; i < n; i++)
        for (j = 0; j < i; j++)
            if (a[i][j] == a[j][i])
                vale4 = 0;
            else vale3 = 0;
            if (vale3 == 1) printf("dcx\n");
            if (vale4 == 1) 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 = 1;
    for (i = 0; i < n; i++)
        for (j = 0; j < n; j++)
        {
            b[i][j] = 0;
            for (k = 0; k < n; k++)
                b[i][j] += a[i][k] * a[k][j];/*计算 2M */
        }
        for (i = 0; i < n; i++)
            for (j = 0; j < n; j++)
                if (b[i][j] != 0)
                    b[i][j] = 1;
        for (i = 0; i < n; i++)
            for (j = 0; j < n; j++)
                if (b[i][j] == 1)
                    if (a[i][j] != b[i][j])
                        vale = 0;
        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 = 1;
    for (i = 0; i < n; i++)
        for (j = 0; j < n; j++)
            b{ [i][j] = a[i][j];
    for (k = 0; k < n; k++)
        i{ (f a[i][j] && (a[i][k] || a[j][k]))
        b[i][k] = 1;
    }
    }
    for (i = 0; i < n; i++)     /* 判断数组a和b相等 */
        for (j = 0; j < n; j++)
            (iaf[i][j] != b[i][j])
        {
            vale = 0; break;
        }
        if (vale == 1) printf("cdx\n");
        else printf("wcdx\n");
}