N行M列的01矩阵
其中1为起点
求出从起点到0的最短距离

1 <= N,M <= 1000

思路：将所有起点入队，即为1的点，然后进行BFS

#include <iostream>
#include <cstdio>
#include <queue>

using namespace std;

const int N = 1010;

int n,m;
char g[N][N];
int dis[N][N];
int dx[10] = {0,0,-1,1},dy[10] = {-1,1,0,0};

void bfs()
{
    queue<int> x,y;
    for(int i = 0;i < n;i ++)
        for(int j = 0;j < m;j ++)
            if(g[i][j] == '1')
            {
                x.push(i),y.push(j);
                dis[i][j] = 0;
            }

    while(x.size())
    {
        int hx = x.front(),hy = y.front();
        for(int i = 0;i < 4;i ++)
        {
            int nx = hx + dx[i],ny = hy + dy[i],ns = dis[hx][hy] + 1;
            if(nx >= 0 && nx < n && ny >= 0 && ny < m && g[nx][ny] == '0' && !dis[nx][ny])
            {
                x.push(nx),y.push(ny);
                dis[nx][ny] = ns;
            }
        }

        x.pop(),y.pop();
    }
}

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

    for(int i = 0;i < n;i ++)
        scanf("%s",g[i]);

    bfs();

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

    return 0;
}