学步园

Problem Description
Today is army day, but the servicemen are busy with the phalanx for the celebration of the 60th anniversary of the PRC.
A phalanx is a matrix of size n*n, each element is a character (a~z or A~Z), standing for the military branch of the servicemen on that position.
For some special requirement it has to find out the size of the max symmetrical sub-array. And with no doubt, the Central Military Committee gave this task to ALPCs.
A symmetrical matrix is such a matrix that it is symmetrical by the “left-down to right-up” line. The element on the corresponding place should be the same. For example, here is a 3*3 symmetrical matrix:
cbx
cpb
zcc

Input
There are several test cases in the input file. Each case starts with an integer n (0

/*************************************************************************
    > File Name: hdu2859.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月14日 星期六 16时31分45秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

const int N = 1010;
int dp[N][N];
char mat[N][N];

int main ()
{
    int n;
    while (~scanf("%d", &n), n)
    {
        for (int i = 1; i <= n; ++i)
        {
            scanf("%s", mat[i] + 1);
            for (int j = 1; j <= n; ++j)
            {
                dp[i][j] = 1;
            }
        }
        for (int i = 1; i <= n; ++i)
        {
            for (int j = 1; j <= n; ++j)
            {
                int m = dp[i - 1][j + 1];
                bool flag = true;
                int x;
                for (int k = 1; k <= m; ++k)
                {
                    if (mat[i - k][j] == mat[i][j + k])
                    {
                        continue;
                    }
                    else
                    {
                        x = k;
                        flag = false;
                        break;
                    }
                }
                if (!flag)
                {
                    dp[i][j] = x;
                }
                else
                {
                    dp[i][j] = m + 1;
                }
            }
        }
        int ans = -1;
        for (int i = 1; i <= n; ++i)
        {
            for (int j = 1; j <= n; ++j)
            {
                ans = max (ans, dp[i][j]);
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}