POJ-2533Longest Ordered Subsequence（LIS）

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#POJ #2533 #Longest Ordered Subs #LIS

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最长上升子序列

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本文介绍了一种寻找给定数列中最长上升子序列长度的高效算法，通过使用二分查找的方法来优化解决方案，适用于序列长度不超过1000的情况。

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 40736		Accepted: 17942

Description

A numeric sequence of a_i is ordered if a₁ < a₂ < ... < a_N. Let the subsequence of the given numeric sequence (a₁, a₂, ..., a_N) be any sequence (a_i₁, a_i₂, ..., a_{i_K}), where 1 <= i₁ < i₂ < ... < i_K <= 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

求最长上升子序列

调用函数

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
	int n,a[1010],i,t,r;
	while(scanf("%d",&n)!=EOF)
	{
		scanf("%d",&a[0]);
		int top=0;
		for(i=1;i<n;i++)
		{
			scanf("%d",&t);
			if(t>a[top])
			a[++top]=t;
			else
			{
				int r=upper_bound(a,a+top,t)-a;
				a[r]=t;
			}
		}
		printf("%d\n",top+1);
	}
	return 0;
}

二分法：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
	int n,i,j,r,l,a[1011],t;
	while(scanf("%d",&n)!=EOF)
	{
		scanf("%d",&a[0]);
		int top=0;
		for(i=1;i<n;i++)
		{
			scanf("%d",&t);
			if(t>a[top])
			a[++top]=t;
			else
			{
				int l=0,r=top;
				
				while(r>=l)
				{
					int mid=(l+r)/2;
					if(a[mid]<t)
					l=mid+1;
					else
					r=mid-1;
				}
				a[l]=t;
			}
		}
					printf("%d\n",top+1);

	}
return 0;
}