Phalanx

Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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

/*
题意：给你一个n*n的字符矩阵，让你求出，最大的对称矩阵的边长，对角线是从左下角到右上角

初步思路：递推    ，从右上角向下递推，dp[i][j]表示是以(i,j)为左下角的最大对称矩阵的边长，这样就得到状态转移方程:
    从(i,j)开始向上，向右遍历，如果相等的长度>dp[i-1][j+1]的话，dp[i][j]=dp[i-1][j+1]+1,否则的话就是
    dp[i-1][j+1];

#错误：还在找错误原因
*/
#include <bits/stdc++.h>
using namespace std;
int n;
char s[1005][1005];
int dp[1005][1005];
int maxn=0;
void init(){
    memset(dp,0,sizeof dp);
    maxn=1;
}
int main(){
    // freopen("in.txt","r",stdin);
    while(scanf("%d",&n)!=EOF&&n){
        init();
        for(int i=1;i<=n;i++){
            scanf("%s",s[i]+1);
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                if(i==1||j==n){
                    dp[i][j]=1;
                    continue;
                }
                int a=i,b=j;//表示向上向右遍历的长度
                while(a>=1&&b<=n&&s[a][j]==s[i][b]){
                    a--;
                    b++;
                }
                // cout<<dp[i][j]<<" ";
                a=i-a;
                if(a>=dp[i-1][j+1]+1){
                    dp[i][j]=dp[i-1][j+1]+1;
                }else{
                    dp[i][j]=a;
                }
                maxn=max(dp[i][j],maxn);
            }
            // cout<<endl;
        }
        // for(int i=1;i<=n;i++){
        //     for(int j=1;j<=n;j++){
        //         cout<<dp[i][j]<<" ";
        //     }
        //     cout<<endl;
        // }
        printf("%d
",maxn);
    }
    return 0;
}