Longest Ordered Subsequence

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最新推荐文章于 2022-08-30 19:25:01 发布

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分类专栏： # dp动态规划 # 二分

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本文介绍了一种寻找最长递增子序列的有效算法，通过动态规划结合二分查找技术，解决给定序列中最长有序子序列的问题。适用于编程竞赛及算法优化场景。

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A numeric sequence of ai is ordered if a1 < a2 < … < aN. Let the subsequence of the given numeric sequence ( a1, a2, …, aN) be any sequence ( ai1, ai2, …, aiK), where 1 <= i1 < i2 < … < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).

Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.
Input
The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000
Output
Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.
Sample Input
7
1 7 3 5 9 4 8
Sample Output
4

解题思路：
找到最长的递增序列，可以用dp加二分的方法来筛选，把该序列存入数组b中，如果当前数比b中最大的数大，那就加入最后一个位置。如果比最后一个小的话，那就用二分找到b中比当前数大的最小数，这样既可以保证结果最优，又可以保证b的序列长度不变，一举两得。
AC代码如下：

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<stack>
using namespace std;
const int N = 1e5+10;
int a[N],b[N];
int check(int *a, int l, int r, int x)
{//二分
	int L = l;
	int R = r;
	while(L < R)
	{
		int M = (R + L) / 2;
		if(a[M] <= x)
		   L = M + 1;
		else R = M;
	}
	return R;
}
int main()
{
	int n;
	cin >> n;
	for(int i = 0; i < n; i++)
	{
		cin >> a[i];
	}
	int cur=0;
	b[0]=-99999999;
	for(int i = 0; i < n; i++)
	{
		if(a[i] > b[cur])
		{
			cur++;
			b[cur] = a[i];
			//cur++;
		}
		else if(a[i] < b[cur])
		{
			b[check(b, 1, cur, a[i])] = a[i];
		}
	}
	cout << cur << endl;
	return 0;
}