Problem Description

　　A palindrome sequence is a sequence which is as same as its reversed order. For example, 1 2 3 2 1 is a palindrome sequence, but 1 2 3 2 2 is not. Given a 2-D array of N rows and M columns, your task is to find a maximum sub-array of P rows and P columns, of which each row and each column is a palindrome sequence.

 

Input

　　The first line of input contains only one integer, T, the number of test cases. Following T blocks, each block describe one test case.
　　There is two integers N, M (1<=N, M<=300) separated by one white space in the first line of each block, representing the size of the 2-D array.
　　Then N lines follow, each line contains M integers separated by white spaces, representing the elements of the 2-D array. All the elements in the 2-D array will be larger than 0 and no more than 31415926.

 

Output

　　For each test case, output P only, the size of the maximum sub-array that you need to find.

 

Sample Input

 

Sample Output

 

Source

 

Recommend

zhuyuanchen520

 

#include <stdio.h>

#include <algorithm>

#include <iostream>

#include <string.h>

#include <set>

#include <map>

#include <vector>

#include <queue>

#include <string>

#include <math.h>

using namespace std;

const int MAXN = ;

int a[MAXN][MAXN];

int n,m;

int calc(int x,int y,int tt)//tt=0奇数，tt=1偶数

{

    int ans;

    int x1,y1,x2,y2;//两个角坐标

    if(tt == )

    {

        ans = ;

        x1 = x; y1 = y;

        x2 = x; y2 = y;

    }

    else

    {

        if(x+>=n || y+>=m)return ;

        ans = ;

        x1 = x;y1 = y;

        x2 = x+;y2 = y+;

        if(a[x1][y1]!=a[x2][y1]||a[x1][y1]!=a[x2][y1])

        return ;

        if(a[x2][y2]!=a[x2][y1]||a[x2][y2]!=a[x2][y1])

        return ;

    }

    while()

    {

        x1--;y1--;

        x2++;y2++;

        if(x1 <  || y1 <  || x2 >= n || y2 >= m)

            return ans;

        for(int i = x1;i <= x2;i++)

            if(a[i][y1]!=a[x2-i+x1][y1])

              return ans;

        for(int i = x1;i <= x2;i++)

            if(a[i][y2]!=a[x2-i+x1][y2])

              return ans;

        for(int i = y1;i <= y2;i++)

            if(a[x1][i]!=a[x1][y2-i+y1])

              return ans;

        for(int i = y1;i <= y2;i++)

            if(a[x2][i]!=a[x2][y2-i+y1])

              return ans;

        for(int i = x1;i <= x2;i++)

            if(a[i][y1]!=a[i][y2])

              return ans;

        for(int i = y1;i <= y2;i++)

            if(a[x1][i]!=a[x2][i])

               return ans;

        ans += ;

    }

    return ans;

}

int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    int T;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d%d",&n,&m);

        for(int i = ; i < n;i++)

            for(int j = ;j < m;j++)

                scanf("%d",&a[i][j]);

        int ans = ;

        for(int i = ;i < n;i++)

            for(int j = ;j < m;j++)

        {

            ans = max(ans,calc(i,j,));

            ans = max(ans,calc(i,j,));

        }

        printf("%d\n",ans);

    }

    return ;

    return ;

}