POJ 2533 Longest Ordered Subsequence（最长上升子序列（NlogN）

传送门

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_iK), 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

4

思路

  模板题

#include<stdio.h>
const int maxn = 1005;

int binary(int pos,int len,int *ans,int *dp)
{
	int left = 1,right = len;
	while (left < right)
	{
		int mid = left + ((right - left) >> 1);
		if (dp[mid] < ans[pos] )	left = mid + 1;
		else right = mid;
	}
	return right;
}

int main()
{
	int N,len = 1;
	int ans[maxn],dp[maxn];
	scanf("%d",&N);
	for (int i = 1;i <= N;i++)	scanf("%d",&ans[i]);
	dp[1] = ans[1];
	for (int i = 2;i <= N;i++)
	{
		if (ans[i] > dp[len])	dp[++len] = ans[i];
		else
		{
			int pos = binary(i,len,ans,dp);dp[pos] = ans[i];  //找到第一个大于等于ans[i]的位置，更新dp[pos]的最小值
			//pos = low_bound(dp+1,dp+len,ans[i]) - dp;
		}
	}
	printf("%d\n",len);
	return 0;
}