首页 > ACM题库 > HDU-杭电 > HDU 4618-Palindrome Sub-Array-枚举-[解题报告]HOJ
2015
09-17

HDU 4618-Palindrome Sub-Array-枚举-[解题报告]HOJ

Palindrome Sub-Array

问题描述 :

  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.

输入:

  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.

输出:

  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.

样例输入:

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

样例输出:

4

/*
分析:
    各种暴力各种过。。

                             2013-07-26
*/

#include"iostream"
#include"cstdio"
#include"cstring"
using namespace std;
const int N=306;

int n,m,ans,map[N][N];
int max(int a,int b){
	return a>b?a:b;
}
int solve(int x1,int y1,int x2,int y2)
{
	int i,l,a,b;
	for(i=x1;i<=x2;i++)
	{
		a=y1;b=y2;
		while(a<b && map[i][a]==map[i][b])	{a++;b--;}
		if(a<b)	return 0;
	}
	for(l=y1;l<=y2;l++)
	{
		a=x1;b=x2;
		while(a<b && map[a][l]==map[b][l])	{a++;b--;}
		if(a<b)	return 0;
	}
	return 1;
}
int main()
{
	int T;
	int i,l,a,b,t;
	cin>>T;
	while(T--)
	{
		cin>>n>>m;
		for(i=0;i<n;i++)
		for(l=0;l<m;l++)
			scanf("%d",&map[i][l]);
		ans=1;
		for(i=0;i<n;i++)
		{
			if(i+ans>=n)	break;
			for(l=0;l<m;l++)
			{
				if(l+ans>=m)break;
				t=(n-i-1)>(m-l-1)?(m-l-1):(n-i-1);
				a=i+t;
				b=l+t;
				while(i+ans<=a)
				{
					if(solve(i,l,a,b))
					{
						ans=max(ans,a-i+1);
						break;
					}
					a--;b--;
				}
			}
		}
		cout<<ans<<endl;
	}
	return 0;
}

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参考:http://blog.csdn.net/ice_crazy/article/details/9494457