1.思路：

把二维矩阵转化成一维编号，之后将编号使用并查集，看最后是否在同一个集合中即可。

2.代码： 

#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10;
int n, m, cnt, root;
int fa[N * N];
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
char mp[N][N];
void init(int x)
{
    for (int i = 1; i <= x; i++)
    {
        fa[i] = i;
    }
}
int find(int x)
{ // 查找，带路径压缩
    return x == fa[x] ? x : (fa[x] = find(fa[x]));
}
void merge(int i, int j)
{
    int x = find(i);
    int y = find(j);
    if (x != y)
    {
        fa[x] = y;
    }
}
int convert(int x, int y)
{ // 将二维矩阵转化成一维编号
    return (x - 1) * m + y;
}
int main()
{
    cin >> n >> m;
    init(n * m);
    for (int i = 1; i <= n; i++)
    {
        cin >> (mp[i] + 1);
    }

    for (int i = 1; i <= m; i++)
    { // 判断下面至少有两层积木
        if (mp[n][i] == '1')
        {
            cnt++;
        }
    }
    if (cnt < 2)
    {
        cout << "No";
        return 0;
    }
    // solve
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            if (mp[i][j] == '0')
            {
                continue;
            }
            if (!root)
            {
                root = convert(i, j);
            }
            for (int k = 0; k < 4; k++)
            {
                int x = i + dx[k];
                int y = j + dy[k];
                if (x < 1 || x > n || y < 1 || y > m)
                {
                    continue;
                }
                if (mp[x][y] == '1')
                {
                    merge(convert(i, j), convert(x, y));
                }
            }
        }
    }
    // 判断是否全部联通
    root = find(root);
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            if (mp[i][j] == '1')
            {
                if (find(convert(i, j)) != root)
                {
                    cout << "No";
                    return 0;
                }
            }
        }
    }
    cout << "Yes";
    return 0;
}