HDU - 2859 Phalanx（DP）

HDU - 2859 Phalanx

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<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

#include<iostream>
#include<cstring>
using namespace std;
const int N = 1010;
char s[N][N];
int dp[N][N];
int main()
{
    
	int n;
	while(cin>>n,n)
	{
    
		memset(dp,0,sizeof dp);
		for(int i=1;i<=n;i++) cin>>(s[i]+1);
		int ans=1;
		for(int i=1;i<=n;i++)
			for(int j=n;j>=1;j--)
			{
    
				dp[i][j]=1;
				if(i==1||j==n) continue;
				int l=dp[i-1][j+1];
				for(int k=1;k<=l;k++)
					if(s[i-k][j]==s[i][j+k]) dp[i][j]++;
					else break;
				ans=max(ans,dp[i][j]);	
			}
		cout<<ans<<endl;
	}
	return 0;
}