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: [email protected]
> 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;
}