hdu 4618 Palindrome Sub-Array(暴力啊，3级)

Palindrome Sub-Array

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 249    Accepted Submission(s): 129

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

4

 

Source

2013 Multi-University Training Contest 2

 

Recommend

zhuyuanchen520

思路：直接暴力N^5，诸多的break优化就能过，多校最水的一题了，居然没敢敲，泪奔啊。

#include<iostream>
#include<cstring>
#include<cstdio>
#define FOR(ii,nn) for(int ii=1;ii<=nn;++ii)
using namespace std;
const int mm=309;
int n,m,cas;
int f[mm][mm];
bool judge(int x,int y,int zk)
{
  int a,b;
  for(int i=x;i<x+zk;++i)
  { a=y;b=y+zk-1;
    while(a<b)
    {
      if(f[i][a++]!=f[i][b--])return 0;
    }
  }
  for(int i=y;i<y+zk;++i)
  {
    a=x;b=x+zk-1;
    while(a<b)
    {
      if(f[a++][i]!=f[b--][i])return 0;
    }
  }
  return 1;
}
int main()
{
  while(~scanf("%d",&cas))
  {while(cas--)
    {scanf("%d%d",&n,&m);
    FOR(i,n)FOR(j,m)
    scanf("%d",&f[i][j]);
    int z,ans;ans=0;
    FOR(i,n)FOR(j,m)
    {
      z=min(n-i,m-j)+1;
      if(z<=ans)break;
      for(int k=z;k>=1;--k)
      if(judge(i,j,k))
      { ans=max(ans,k);
        break;
      }
    }
     printf("%d\n",ans);
    }
  }
  return 0;
}