HDU 4618 Palindrome Sub-Array(最大回文子矩阵)

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本文介绍了一个寻找二维数组中最大回文子矩阵的问题及其实现方法。通过遍历所有可能的子矩阵并检查其是否为回文序列来确定最大回文子矩阵的大小。

Palindrome Sub-Array

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 162 Accepted Submission(s): 72

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

  
   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

Source

2013 Multi-University Training Contest 2

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#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
int map[330][330];
int rec[330];
int n,m;
int ans;
int is_ok(int i,int j,int len)
{
   int k=0;
   len--;
   while(k<len)
   {
       if(map[i][j+k]!=map[i][j+len])
       return 0;
       k++,len--;
   }
   return 1;
}
int is_ok_ver(int i,int j,int len)
{
    int k=i+len-1;
    while(i<k)
    {
        if(map[i][j] !=map[k][j])
        return 0;
        i++,k--;
    }
    return 1;
}
int is_matrix(int i,int j,int len)
{
    int k,r;
    for(k=0;k<len;k++)
    {
        if(!is_ok(i+k,j,len))
        return 0;
        if(!is_ok_ver(i,j+k,len))
        return 0;
    }
    return 1;
}
int find_ans(int i,int j)
{
    int p=n-i,q=m-j,len;
    len= p < q ? p : q;
    if(ans >= len)
    return 0;
    while(len > ans)
    {
        if(is_matrix(i,j,len))
        {
            ans=len;
            return 0;
        }
        len--;
    }
    return 1;
}
int main()
{
    int t,i,j,k;
    scanf("%d",&t);
    while(t--)
    {
        ans=1;
        scanf("%d%d",&n,&m);
        for(i=0;i<n;i++)
        for(j=0;j<m;j++)
        scanf("%d",&map[i][j]);
        for(i=0;i<n;i++)
        for(j=0;j<m;j++)
        {
            find_ans(i,j);
        }
        printf("%d\n",ans);
    }
    return 0;
}