思路： 循环每一格，判断是否上下左右都有为1的格子，如果满足，则变为2

#include<bits/stdc++.h>
using namespace std;
#define FO(i,a,b) for(int i=a;i<=b;i++)
int n,a[31][31],nx[31],nn[31],sx[31],sn[31];
bool ok(int a,int b){return a>=sn[b]&&a<=sx[b]&&b>=nn[a]&&b<=nx[a];}
int main(){
	memset(nn,0x3f,sizeof(nn));
	memset(sn,0x3f,sizeof(sn));
	cin>>n;
	FO(i,1,n){
		FO(j,1,n){
			cin>>a[i][j];
			if(a[i][j]==0)continue;
			nx[i]=max(nx[i],j);
			sx[j]=max(sx[j],i);
			nn[i]=min(nn[i],j);
			sn[j]=min(sn[j],i);
		}
	}
	FO(i,1,n){
		FO(j,1,n){
			if(a[i][j]==0&&ok(i,j)==1)a[i][j]=2;
			cout<<a[i][j]<<" ";
		}
		cout<<"\n";
	}
	return 0;
}

nn:每一行中最左边的1 nx:每一行中最右边的1 sn:每一列中最左边的1 sx:每一列中最右边的1