第一种，用数组

#include <iostream>
#include <cstring>

using namespace std;

typedef pair<int, int> PII;

const int N = 200;

int n, m;
int g[N][N];
int d[N][N];
PII q[N * N], Prev[N][N];

int bfs()
{
    int hh = 0, tt = 0;
    q[0] = {0, 0};

    memset(d, -1, sizeof(d));

    d[0][0] = 0;
    //四个方向的向量
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

    while (hh <= tt)
    {
        PII t = q[hh++];
        for (int i = 0; i < 4; i++)
        {   int x = t.first + dx[i], y = t.second + dy[i];
            //if (x >= 0 && x < n && y >= 0 && y < n && g[x][y] == 0 && d[x][y] == -1)  打错字母了
            if (x >= 0 && x < n && y >= 0 && y < m && g[x][y] == 0 && d[x][y] == -1) 
            {
                d[x][y]= d[t.first][t.second] + 1;
                Prev[x][y] = t;
                q[++tt] = {x, y};
            }
        }
    }

/*    
    int x = n - 1, y = m - 1;
    while (x || y)
    {
        cout << x << ' ' << y << endl;
        PII t = Prev[x][y];
        x = t.first, y = t.second;
    }
*/  
    return d[n - 1][m - 1];
}

int main()
{
    cin >> n >> m;

    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            cin >> g[i][j];

    cout << bfs() << endl;

    /*
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
            printf("%3d", d[i][j]);
        cout << endl;
    }
    */

    return 0;
}

第二种，用queue< >

#include <iostream>
#include <cstring>
#include <queue>

using namespace std;

const int N = 110;

typedef pair<int, int> PII;

int n, m;
int g[N][N], d[N][N];

int bfs()
{
    queue< pair<int, int> > q;
    memset(d, -1, sizeof(d));
    d[0][0] = 0;
    q.push({0, 0});

    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, -1, 0, 1};
    while (q.size())
    {
        PII t = q.front();
        q.pop();

        for (int i = 0; i < 4; i++)
        {
            int x = t.first + dx[i], y = t.second + dy[i];
            if (x >= 0 && x < n && y >= 0 && y < m && g[x][y] == 0 && d[x][y] == -1)
            {
                d[x][y] = d[t.first][t.second] + 1;
                q.push({x, y});
            }
        }
    }

    return d[n - 1][m -1];
}

int main()
{
    cin >> n >> m;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            cin >> g[i][j];

    cout << bfs() << endl;

    return 0;
}

打印路径（维护路线）

1. bfs进入队列的时候，维护前一步的位置
2. 从最后一点倒序到起点，把每一步都放到一个队列或栈里面
3. 把队列从front()开始弹空（再逆序回来）
4. 要手写队列，否则要使用双端队列

#include <iostream>
#include <cstring>
#include <queue>

using namespace std;

const int N = 110;

typedef pair<int, int> PII;

int n, m;
int g[N][N], d[N][N];
PII Prev[N][N];
PII path[N * N];

int bfs()
{
    queue< pair<int, int> > q;
    memset(d, -1, sizeof(d));
    d[0][0] = 0;
    q.push({0, 0});

    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, -1, 0, 1};
    while (q.size())
    {
        PII t = q.front();
        q.pop();

        for (int i = 0; i < 4; i++)
        {
            int x = t.first + dx[i], y = t.second + dy[i];
            if (x >= 0 && x < n && y >= 0 && y < m && g[x][y] == 0 && d[x][y] == -1)
            {
                d[x][y] = d[t.first][t.second] + 1;
                q.push({x, y});
                Prev[x][y] = t;
            }
        }
    }

    return d[n - 1][m -1];
}

int main()
{
    cin >> n >> m;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            cin >> g[i][j];

    cout << bfs() << endl;

    int hh = 0, tt = 0;
    int x = n - 1, y = m - 1;
    while (x || y)
    {
        path[++tt] = {x, y};
        PII t = Prev[x][y];
        x = t.first, y = t.second;
    }
    path[++tt] = {0, 0};

    while (hh < tt)
    {
        PII t = path[tt--];
        printf("==>%d %d\n", t.first, t.second);
    }

    return 0;
}