//时间复杂度

O(n*n)

C++ 代码

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int N = 1e3 + 1;
int n, m;
char v[N][N];
int visited[N][N];
int dx[] = { -1,0,1,0 };
int dy[] = { 0,1,0,-1 };
int ret[N][N];
struct node
{
    int x, y, step;
};
queue<node>q;
void bfs()
{
    while (!q.empty())
    {

        for (int a = 0; a < 4; a++)
        {
            int tx = q.front().x + dx[a];
            int ty = q.front().y + dy[a];
            if (tx<1 || tx>n || ty<1 || ty>m || v[tx][ty] == '1' || visited[tx][ty] == 1)continue;
            visited[tx][ty] = 1;
            node tmp;
            tmp.x = tx, tmp.y = ty,tmp.step = q.front().step + 1;
            ret[tx][ty] = tmp.step;
            //cout << tmp.step << " ";
            q.push(tmp);
        }
        q.pop();
    }

}

int main()
{
    cin >> n >> m;
    //1是源点，0是终点，求达到所有终点要多少步

    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            cin >> v[i][j];
            if (v[i][j] == '1')
                //源点步数为0
            {
                q.push({ i,j,0 });
                visited[i][j] = 1;
            }
        }
    }
    bfs();

    for (int i = 1; i <= n; i++)
    {
        for (int j=1; j <= m; j++)
        {
            cout << ret[i][j] << " ";
        }
        cout << endl;
    }

    return 0;
}