Phalanx （二维矩阵dp）

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

思路：

对每个点对它上面和右边的点进行匹配，如果匹配量大于右上角记录的矩阵最大的值，那么这个值就+1，否则就等于匹配量

代码：

#include<iostream>
using namespace std;
const int maxn=1005;
int dp[maxn][maxn];
char a[maxn][maxn];
int main()
{
	int n,ans;
	while(cin>>n,n)
	{
		ans=1;
		for(int i=0;i<n;i++)
		  for(int j=0;j<n;j++)
		     cin>>a[i][j];
		for(int i=0;i<n;i++)
		  for(int j=n-1;j>=0;j--)
		  {
		  	  dp[i][j]=1;
		  	  if(i==0||j==n-1) continue;
		  	  int k=dp[i-1][j+1];
		  	  for(int s=1;s<=k;s++)
		  	  {
		  	  	 if(a[i-s][j]==a[i][j+s]) dp[i][j]++;
		  	  	 else break;
			  }
			  ans=max(ans,dp[i][j]);
		  }
	    cout<<ans<<endl;
	}
}