hdu4618(KMP+动态规划)

Palindrome Sub-Array

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 317    Accepted Submission(s): 166

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
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    本题在比赛的时候没有想到好的方法，索性没做，赛后相结合KMP算法运用动态规划思想，不了能力太弱，也没能动手，结果就写了一个5重循环，竟能过
   
   
   
   
   
   
   
   

    
    
    
    后来索性提交了几次，得到了一些无关技巧的小技巧，裸函数、宏定义、内联函数对运行时间的影响
   
   
   
   
   
   
   
   

    
    
    
    
#include<iostream>
#include<cstdio>
using namespace std;
//#define Min(a,b) a<b?a:b
//#define Max(a,b) a>b?a:b

const int MAX=300+10;
int da[MAX][MAX];

//宏定义Max(a,b),Min(a,b)，运行时间为512Ms
//直接写裸函数int Min(int a,int b),int Max(int a,int b)，运行时间为656Ms
//写内联函数inline int Min(int a,int b)，inline int Max(int a,int b)，运行时间为525Ms
int Min(int a,int b)
{
	return a<b?a:b;
}
int Max(int a,int b)
{
	return a<b?b:a;
}

inline int Judge(int x,int y,int len)
{
	int i,j,k;
	for(k=y;k<=y+len-1;k++)
	{
		for(i=x,j=x+len-1;i<=j;i++,j--)
		{
			if(da[i][k]!=da[j][k])
				return 0;
		}
	}
	
	for(k=x;k<=x+len-1;k++)
	{
		for(i=y,j=y+len-1;i<=j;i++,j--)
		{
			if(da[k][i]!=da[k][j])
				return 0;
		}
	}
	return len;
}


int main()
{
	int cas,i,j,n,m,ans,len,tmp;
	cin>>cas;
	while(cas--)
	{
		scanf("%d%d",&n,&m);
		for(i=0;i<n;i++)
		{
			for(j=0;j<m;j++)
				scanf("%d",&da[i][j]);
		}
		
		ans=0;
		for(i=0;i<n;i++)
		{
			for(j=0;j<m;j++)
			{
				len=Min(n-i,m-j)+1;
				do
				{
					tmp=Judge(i,j,len);
					ans=Max(ans,tmp);
				}while(--len);
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}