如何把c++代码改成C语言代码，使其实现原有功能？

Question

本人在学习C语言，对c++一窍不通，希望帮忙
以下是代码（实现bfs迷宫自动寻路，希望改动尽量少）

#include <bits/stdc++.h>
using namespace std;
struct que
{
    int x, y, s, f;
} node[250];
int main()
{
    int fill[50][50], book[50][50];
    memset(book, 0, sizeof(book));
    int m, n; // m->x   n->y
    int start_x, start_y, finish_x, finish_y;
    int head = 1, tail = 1;
    int tx, ty;        //下一步的位置坐标
    bool flag = false; //用于结束双层循环
    //初始化迷宫地图
    cin >> m >> n;
    for (int y = 0; y < n; y++)
        for (int x = 0; x < m; x++)
            cin >> fill[y][x];
    cin >> start_x >> start_y >> finish_x >> finish_y;
    //初始化迷宫入口
    book[start_y][start_x] = 1;
    node[tail].x = start_x;
    node[tail].y = start_y;
    node[tail].f = 0;
    node[tail].s = 0;
    tail++;
    //寻路
    while (head < tail)
    {
        for (int ctrl = 0; ctrl < 4; ctrl++) //控制下一步行走方向
        {
            switch (ctrl)
            {
            case 0: //向右走
                tx = node[head].x + 1;
                ty = node[head].y;
                break;
            case 1: //向下走
                tx = node[head].x;
                ty = node[head].y + 1;
                break;
            case 2: //向左走
                tx = node[head].x - 1;
                ty = node[head].y;
                break;
            case 3: //向上走
                tx = node[head].x;
                ty = node[head].y - 1;
                break;
            }
            if (tx < 0 || tx > m - 1 || ty < 0 || ty > n - 1)
                continue; //如果撞了围墙就返回
            if (fill[ty][tx] == 0 && book[ty][tx] == 0)
            {
                node[tail].s = node[head].s + 1;
                node[tail].x = tx;
                node[tail].y = ty;
                node[tail].f = head;
                book[ty][tx] = 1;
                tail++;
            } //如果下一步没有撞墙，并且没有走过这个地方
            if (ty == finish_x && ty == finish_y)
            {
                flag = true;
                break; //如果走到终点，停止行走
            }
        }
        if (flag)
            break; //flag的妙用[手动滑稽]
        head++;
    }
    cout << node[tail - 1].s << endl; //打印出从起点走了多少步到终点
    //倒序输出所经过的位置的坐标
    cout << finish_x << finish_y << endl;
    int temp = node[tail - 1].f;
    while (temp != 0)
    {
        cout << node[temp].x << node[temp].y << endl;
        fill[node[temp].y][node[temp].x] = 2;
        temp = node[temp].f;
    }
    fill[finish_y][finish_x] = 2;
    for (int y = 0; y < n; y++)
    {
        for (int x = 0; x < m; x++)
            cout << fill[y][x] << " ";
        cout << endl;
    }
    system("pause");
    return 0;
}

Answer 1

TechWhizKid参考GPT回答：

• c语言中没有C++中那样的cout和cin对象，用printf和scanf函数就行。也没有命名空间的概念，也没有C++中bool类型，把bool改为int，并且用0和1代表false和true。C语言没有new关键字，也没有memset函数的直接替代品，所以用循环初始化数组。

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

struct que
{
    int x, y, s, f;
} node[250];

int main()
{
    int fill[50][50], book[50][50];
    int i, j;
    for(i=0; i<50; i++)
        for(j=0; j<50; j++)
            book[i][j] = 0;
    int m, n; // m->x   n->y
    int start_x, start_y, finish_x, finish_y;
    int head = 1, tail = 1;
    int tx, ty;        //下一步的位置坐标
    int flag = 0; //用于结束双层循环
    //初始化迷宫地图
    scanf("%d %d", &m, &n);
    for (int y = 0; y < n; y++)
        for (int x = 0; x < m; x++)
            scanf("%d", &fill[y][x]);
    scanf("%d %d %d %d", &start_x, &start_y, &finish_x, &finish_y);
    //初始化迷宫入口
    book[start_y][start_x] = 1;
    node[tail].x = start_x;
    node[tail].y = start_y;
    node[tail].f = 0;
    node[tail].s = 0;
    tail++;
    //寻路
    while (head < tail)
    {
        for (int ctrl = 0; ctrl < 4; ctrl++) //控制下一步行走方向
        {
            switch (ctrl)
            {
            case 0: //向右走
                tx = node[head].x + 1;
                ty = node[head].y;
                break;
            case 1: //向下走
                tx = node[head].x;
                ty = node[head].y + 1;
                break;
            case 2: //向左走
                tx = node[head].x - 1;
                ty = node[head].y;
                break;
            case 3: //向上走
                tx = node[head].x;
                ty = node[head].y - 1;
                break;
            }
            if (tx < 0 || tx > m - 1 || ty < 0 || ty > n - 1)
                continue; //如果撞了围墙就返回
            if (fill[ty][tx] == 0 && book[ty][tx] == 0)
            {
                node[tail].s = node[head].s + 1;
                node[tail].x = tx;
                node[tail].y = ty;
                node[tail].f = head;
                book[ty][tx] = 1;
                tail++;
            } //如果下一步没有撞墙，并且没有走过这个地方
            if (ty == finish_x && ty == finish_y)
            {
                flag = 1;
                break; //如果走到终点，停止行走
            }
        }
        if (flag)
            break; //flag的妙用[手动滑稽]
        head++;
    }
    printf("%d\n", node[tail - 1].s); //打印出从起点走了多少步到终点
    //倒序输出所经过的位置的坐标
    printf("%d %d\n", finish_x, finish_y);
    int temp = node[tail - 1].f;
    while (temp != 0)
    {
        printf("%d %d\n", node[temp].x, node[temp].y);
        fill[node[temp].y][node[temp].x] = 2;
        temp = node[temp].f;
    }
    fill[finish_y][finish_x] = 2;
    for (int y = 0; y < n; y++)
    {
        for (int x = 0; x < m; x++)
            printf("%d ", fill[y][x]);
        printf("\n");
    }
    system("pause");
    return 0;
}