用C语言问题描述：以一个m*n的长方阵表示迷宫，0和1分别表示迷宫中的通路和障碍。设计一个程序，对任意设定的迷宫，求出一条从入口到出口的通路，或得出没有通路的结论。

以下是用C语言实现的迷宫求解程序：

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

#define MAX_ROW 20
#define MAX_COL 30

typedef struct {
    int row, col; // 行、列
} Position;

Position stack[MAX_ROW*MAX_COL]; // 用数组实现栈
int top = -1; // 栈顶指针

void push(Position p) {
    stack[++top] = p;
}

Position pop() {
    return stack[top--];
}

bool is_empty() {
    return top == -1;
}

int maze[MAX_ROW][MAX_COL] = {
    {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1},
    {1,0,0,0,1,0,0,0,1,0,0,0,1,1,1,1,1,1,1,0,1,0,1,0,1,0,1,0,1,1},
    {1,0,1,0,1,0,1,1,1,1,1,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,0,1},
    {1,0,1,0,1,1,1,0,0,0,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,0,1,0,1,1},
    {1,0,1,1,1,0,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,0,1},
    {1,0,1,0,0,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,1},
    {1,0,1,0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,1},
    {1,0,0,0,1,0,0,0,1,0,1,1,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,0,1},
    {1,0,1,0,1,1,1,1,1,0,1,0,0,0,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1},
    {1,0,1,0,0,0,1,0,0,0,1,0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,1,0,0,1},
    {1,0,1,1,1,0,1,0,1,1,1,0,1,0,0,0,1,0,1,0,1,1,1,0,1,0,1,0,1,1},
    {1,0,0,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,1,0,0,1},
    {1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1},
    {1,0,1,0,1,0,1,0,0,0,0,0,0,0,1,0,1,0,1,0,0,0,1,0,1,0,1,0,0,1},
    {1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,1},
    {1,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,1,0,0,1},
    {1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,1},
    {1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,0,1},
    {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1},
};

int visited[MAX_ROW][MAX_COL] = {0};

bool solve_maze(int row, int col) {
    push((Position){row, col});
    visited[row][col] = 1;
    
    while (!is_empty()) {
        Position p = pop();
        int r = p.row, c = p.col;
        
        if (r == MAX_ROW-1 && c == MAX_COL-1) { // 到达终点
            printf("到达终点！\n");
            return true;
        }
        
        // 向上探索
        if (r > 0 && maze[r-1][c] == 0 && visited[r-1][c] == 0) {
            push((Position){r-1, c});
            visited[r-1][c] = 1;
        }
        // 向下探索
        if (r < MAX_ROW-1 && maze[r+1][c] == 0 && visited[r+1][c] == 0) {
            push((Position){r+1, c});
            visited[r+1][c] = 1;
        }
        // 向左探索
        if (c > 0 && maze[r][c-1] == 0 && visited[r][c-1] == 0) {
            push((Position){r, c-1});
            visited[r][c-1] = 1;
        }
        // 向右探索
        if (c < MAX_COL-1 && maze[r][c+1] == 0 && visited[r][c+1] == 0) {
            push((Position){r, c+1});
            visited[r][c+1] = 1;
        }
    }
    
    printf("没有通路！\n");
    return false;
}

int main() {
    solve_maze(0, 0);
    return 0;
}

该程序使用深度优先搜索算法，在迷宫中搜索通路。将起点入栈，然后循环进行以下操作：从栈中取出一个位置，对其四个相邻位置进行探索，将可访问的位置入栈，并标记为已访问。如果到达终点，则输出“到达终点！”并返回true，否则继续循环。如果栈为空，则没有通路，输出“没有通路！”并返回false。

该程序的时间复杂度为O(mn)，其中m、n分别为迷宫的行数和列数。