Phalanx

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3093 Accepted Submission(s): 1510

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

Source

2009 Multi-University Training Contest 5 - Host by NUDT

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gaojie

题意：找最大对称子矩阵（沿用上角到左下角的对角线对称）。

题解：dp遍历每一个点，比较这一个点所在的列上边和所在行的右边对称的数目，如果大于dp[i-1][j+1]，则dp[i][j] = dp[i-1][j+1] + 1,否则等于对称的数目。

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;const int maxn = 1050;char s[maxn][maxn];int dp[maxn][maxn];int main(){    int n,i,j,Max,a,b;    while(scanf("%d",&n)!=EOF&&n)    {        for(i=0;i
         
          =0&&b
          
           =dp[i-1][j+1] + 1) dp[i][j] = dp[i-1][j+1] + 1; else dp[i][j] = i - a; } Max = max(Max,dp[i][j]); } } printf("%d\n",Max); } return 0;}