Phalanx （hdu 2859）

http://acm.hdu.edu.cn/showproblem.php?pid=2859  

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 996    Accepted Submission(s): 468

Problem DescriptionToday 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 InputThere 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. OutputEach test case output one line, the size of the maximum symmetrical sub- matrix. Sample Input3
abx
cyb
zca
4
zaba
cbab
abbc
cacq
0 Sample Output3
3 Source2009 Multi-University Training Contest 5 – Host by NUDT

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <string>
using namespace std;#define N 1100
#define MOD 1000000007
#define met(a, b) memset(a, b, sizeof(a))
#define INF 0x3f3f3f3fchar G[N][N];
int dp[N][N];int main()
{
    int n;    while(scanf("%d", &n), n)
    {
        int i, j, i1, j1, Max=;        met(G, );
        met(dp, );
        for(i=; i<n; i++)
            scanf("%s", G[i]);        for(i=; i<n; i++)
        for(j=; j<n; j++)
        {
            if(i== || j==n-)
            {
                dp[i][j] = ;
                continue;
            }
            i1=i, j1=j;
            while(j1<n && i1>= && G[i1][j]==G[i][j1] )
                 i1--, j1++;            if((i-i1)>=dp[i-][j+]+) dp[i][j] = dp[i-][j+] + ;
            else dp[i][j] = i-i1;            Max = max(Max, dp[i][j]);
        }        printf("%d\n", Max);
    }
    return ;
}