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;}// solvefor (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;
}