HDU - 4618 H - Palindrome Sub-Array

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

1
5 10
1 2 3 3 2 4 5 6 7 8
1 2 3 3 2 4 5 6 7 8
1 2 3 3 2 4 5 6 7 8
1 2 3 3 2 4 5 6 7 8
1 2 3 9 10 4 5 6 7 8

Sample Output
4

题意：给你一个大的矩阵，然后让你计算出该矩阵中每行每列都是回文串的正方形子矩阵的个数

暴力水过不科学

ac代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int map[305][305],x,y;
int solve(int a,int b,int ans)
{
if(a+ans>x||b+ans>y)
return 0;
for(int i=a;i<a+ans;i++)
{
for(int j=b;j<b+ans;j++)
{
if(map[i][j]!=map[i][b+ans-1-j+b]||map[i][j]!=map[a+ans-1-i+a][j])
return 0;
}
}
return 1;
}
int main()
{
int n;
cin>>n;
while(n--)
{
cin>>x>>y;
for(int i=0;i<x;i++)
for(int j=0;j<y;j++)
scanf("%d",&map[i][j]);
int sum=min(x,y);
int flag=0;
for(int i=sum;i>=1;i--)
{
for(int j=0;j<x;j++)
{
for(int k=0;k<y;k++)
{
if(solve(j,k,i))
{
printf("%d\n",i);
flag=1;
break;
}
}
if(flag==1)
break;
}
if(flag==1)
break;
}
}
}