dp解法:

如果（i，j）为1，那么只有在（i-1，j），（i，j-1），（i-1，j-1）都为1时，边长才能+1

所以可以得到：f[i][j]=min{(i-1,j),(i,j-1),(i-1,j-1)}

for example:

已知：

2 3

9 ?

那么“？”的值就是2+1=3

代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
int n, m;
int a[105][105], f[105][105];
int ans = 0;
int main()
{
	std::cin >> n >> m;
	for(int i = 0; i < 100; i ++)
	{
		for(int j = 0; j < 100; j ++)
		{
			f[i][j] = 0;
		}
	}//初始化
	for(int i = 1;i <= n; i ++)
	{
		for(int j = 1;j <= m;j ++)
		{
			scanf("%d", &a[i][j]);
			if(a[i][j] == 1)
			{
				f[i][j] = std::min(std::min(f[i - 1][j], f[i][j - 1]), f[i - 1][j - 1]) + 1;//状态转移公式	
			}
			ans = std::max(ans, f[i][j]);
		}
	}
	std::cout << ans;
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
bool check(bool map[101][101],short x1,short y1,short x2,short y2){
	for (short i=x1;i<=x2;i++){
		for (short i1=y1;i1<=y2;i1++){
			if (map[i][i1]==0){
				return 0;
			}
		}
	}
	return 1;
}
bool find(short a,bool map[101][101],short n,short m){
	for (short i=0;i<=n-a;i++){
		for (short i1=0;i1<=m-a;i1++){
			if (check(map,i,i1,i+a-1,i1+a-1)==1){
				return 1;
			}
		}
	}
	return 0;
}
int main(){
	short n,m;
	bool map[101][101];
	cin >> n >> m;
	for (short i=0;i<n;i++){
		for (short i1=0;i1<m;i1++){
			cin >> map[i][i1];
		}
	}
	for (short i=max(n,m);i>1;i--){
		if (find(i,map,n,m)==1){
			cout << i;
			return 0;
		}
	}
	cout << 1;
	return 0;
}

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
int n, m;
int a[105][105], f[105][105];
int ans = 0;
int main()
{
	std::cin >> n >> m;
	for(int i = 0; i < 100; i ++)
	{
		for(int j = 0; j < 100; j ++)
		{
			f[i][j] = 0;
		}
	}//初始化
	for(int i = 1;i <= n; i ++)
	{
		for(int j = 1;j <= m;j ++)
		{
			scanf("%d", &a[i][j]);
			if(a[i][j] == 1)
			{
				f[i][j] = std::min(std::min(f[i - 1][j], f[i][j - 1]), f[i - 1][j - 1]) + 1;//状态转移公式	
			}
			ans = std::max(ans, f[i][j]);
		}
	}
	std::cout << ans;
	return 0;
}