POJ 2533 Longest Ordered Subsequence 典型LIS

最长递增子序列

最新推荐文章于 2018-11-16 13:58:45 发布

最新推荐文章于 2018-11-16 13:58:45 发布 · 69 阅读

本文介绍了一种寻找给定数列中最长递增子序列长度的有效算法。通过使用二分查找的方法来优化解决方案，该算法能在O(n log n)的时间复杂度内找到最长递增子序列的长度。

http://poj.org/problem?id=2533

Longest Ordered Subsequence

Time Limit:2000MS		Memory Limit:65536K

Description

A numeric sequence of a_iis 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

/* Author : yan
 * Question : POJ 2533 Longest Ordered Subsequence
 * Date && Time : Monday, January 24 2011 10:21 AM
 * Compiler : gcc (Ubuntu 4.4.3-4ubuntu5) 4.4.3
*/
#include<stdio.h>
#define MAX 10001
int n;
int value[MAX];//原始数组
int opt[MAX+1];//记录长度为k的LIS的最后元素

int binary_search(const int opt[],const int size,const int value)
{
	int left=1;
	int right=size;
	int mid;
	while(left<=right)
	{
		mid=(left+right)>>1;
		if(value>opt[mid] && value<=opt[mid+1]) return  mid+1;
		else if(value<opt[mid+1]) right=mid-1;
		else left=mid+1;
	}
}

int LIS(const int value[],const int n)
{
	int i,j;
	int size=1;
	opt[1]=value[0];
	for(i=1;i<n;i++)
	{
		if(value[i]<=opt[1]) j=1;
		else if(value[i]>opt[size]) j=++size;
		else j=binary_search(opt,size,value[i]);
		opt[j]=value[i];
	}
	return size;
}
int main()
{
	//freopen("input","r",stdin);
	int i;
	scanf("%d",&n);
	for(i=0;i<n;i++) scanf("%d",&value[i]);
	printf("%d",LIS(value,n));
	return 0;
}