Longest Repeated Sequence

最新推荐文章于 2021-03-22 11:43:48 发布

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本文介绍了一种寻找给定整数序列中最长重复子序列的有效算法。通过使用二维数组记录局部结果，该算法能在多项式时间内解决问题。文章还提供了一个具体的实现示例，并通过样例输入输出展示了算法的应用。

题目：

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

You are given a sequence of integers, A = a1, a2, ... an. A consecutive subsequence of A (say ai, ai+1 ... aj) is called a "repeated sequence" if it appears more than once in A (there exists some positive k that ai+k = ai, ai+k+1 = ai+1, ... aj+k = aj) and its appearances are not intersected (i + k > j).

Can you find the longest repeated sequence in A?

输入

Line 1: n (1 <= n <= 300), the length of A.
Line 2: the sequence, a1 a2 ... an (0 <= ai <= 100).

输出

The length of the longest repeated sequence.

样例输入

5
2 3 2 3 2

样例输出

算法思想：

将运算中的局部结果保存在一个二维数组mark[][]中，mark[i][j]表示与子序列ai, ai+1 ... aj重复的下个子序列中首个数字的索引，

mark[i][j+1]由mark[i][j]很容易得到。

int  LongestRepeatedSequence(vector arr, int n)
{
	static int mark[300][300];

	int maxNum = 0;

	for (int i = 0; i < n; i++)
	{
		for (int j = i+1; j < n; j++)
		{
			if (arr[j] == arr[i])
			{
				mark[i][i] = j;
				maxNum = 1;

				break;
			}
		}
	}	

	for (int i = 1; i < n; i++)
	{
		for (int j = 0; j < n-i; j++)
		{
			int k = mark[j][j+i-1];

			while (k != 0)
			{
				if (k+i < n && arr[j+i] == arr[k+i])
				{
					mark[j][j+i] = k;
					break;
				}
				else if (k+i == n)
					break;
				else
					k = mark[k][k+i-1];
			}
		}

		for (int j = 0; j < n-i; j++)
		{
			int k = mark[j][j+i];

			while (k != 0)
			{
				if (j+i < k)
				{
					maxNum = i+1;
					break;
				}
				else
					k = mark[k][k+i];
			}
		}
	}

	return maxNum;
}