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Phalanx

** Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 452    Accepted Submission(s): 219
**

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 nn, 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 33 symmetrical matrix:
cbx
cpb
zcc

Input

There are several test cases in the input file. Each case starts with an integer n (0 <n<=1000), followed by n lines which has n character. There won’t be any blank spaces between characters or the end of line. The input file is ended with a 0.

Output

Each test case output one line, the size of the maximum symmetrical sub- matrix.

Sample Input

3 abx cyb zca 4 zaba cbab abbc cacq 0

Sample Output

3 3

求每个元素最上方的列和本行的最大对称长度，如果该元素的最大对称长度大于右上角的矩阵，则是右上角矩阵大小的加1，否则就是最大对称长度。

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<map>

using namespace std;

int n,ans;
char p[1010][1010];
int dp[1010][1010];

int main()
{
	while (cin >> n && n)
	{
		ans = 1;
		for (int i = 0; i < n; i++)
			cin >> p[i];
		memset(dp,0,sizeof(dp));
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < n; j++)
			{
				if (i == 0 || j == n - 1)
					dp[i][j] = 1;
				else
				{
					int t1 = i,t2 = j;
					while (t1 >= 0 && t2<=n-1 && p[t1][j] == p[i][t2])
					{
						t1--;
						t2++;
					}
					int t = i - t1;
					if (t>dp[i - 1][j + 1] + 1)
						dp[i][j] = dp[i - 1][j + 1] + 1;
					else
						dp[i][j] = t;
					ans = max(ans,dp[i][j]);
				}
			}
		}
		cout << ans << endl;
	}
	return 0;
}