Algorithm

USACO Section5.1 theme

Problem Summary

Given an array of N integers, find the maximal length of its “repeating” subsequence.

“A repeats B” means that subsequence A and B meet the following conditions:

  1. A and B are not overlapping
  2. A and B have the same length N, where N>=5
  3. There is a constant c, s.t. A[i]+c=B[i], 1<=i<=N

Solution

This is a pretty staight forward dynamic programming problem.

Let diff[i]=seq[i]-seq[i-1],2<i<=n. Then the problem is equivalent to finding the maximal length of repeating subsequence of diff.

Let f[i][j] be the maximal length of repeating subsequence ending at i and j in the sequence diff. Apparently, f[i][j] has the same value with f[j][i]. Then the answer to the original problem is max{f[i][j]}+1, 2<=i<n, i+1<j<=n

The dynamic programming equations are:

f[i][j]=f[i-1][j-1]+1, if diff[i]==diff[j] && f[i-1][j-1]<=j-i-2
f[i][j]=f[i-1][j-1], if diff[i]==diff[j] && f[i-1][j-1]==j-i-2
f[i][j]=0, if diff[i]!=diff[j]

Well, for any i, f[i][j] only depends on f[i-1][j], so we can rewrite the eqations:

f[j]=f[j-1]+1, if diff[i]==diff[j] && f[j-1]<=j-i-2
f[j]=f[j-1], if diff[i]==diff[j] && f[j-1]==j-i-2
f[j]=0, if diff[i]!=diff[j]

Using the new equations, we need to do the inner loop backwards. And we can save a lot of space.

Code

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#include <iostream>
#include <fstream>
using namespace std;
const int maxn = 5002;
int n,diff[maxn];
int f[maxn];
int main()
{
	fstream fin("theme.in",ios::in);
	fstream fout("theme.out",ios::out);
	fin>>n;
	int pre=0,cur=0;
	for(int i=1;i<=n;i++)
	{
		fin>>cur;
		diff[i]=cur-pre;
		pre=cur;
	}
	int ans=0;
	for(int j=3;j<=n;j++)
		f[j]=(diff[j]==diff[2]);
	for(int i=3;i<=n;i++)
	{
		for(int j=n;j>=i+2;j--)
		{
			if(diff[i]==diff[j])
			{
				if(f[j-1]<=j-i-2)
					f[j]=f[j-1]+1;
				else
					f[j]=f[j-1];
			}
			else
				f[j]=0;
			ans=ans>f[j] ? ans:f[j];
		}
	}
	
	ans++;
	if(ans<5) ans=0;
	fout<<ans<<endl;
	fin.close();
	fout.close();
	return 0;
}