poj-2533 Longest Ordered Subsequence（最长递增子序列）

Longest Ordered Subsequence

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 47572 Accepted: 21170

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

71 7 3 5 9 4 8

Sample Output

4

nlogn：

#include<cstdio>#include<iostream>#include<string>#include <algorithm>using namespace std;int zu[1000+10];int dp[1000+10];int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        for(int i=1;i<=n;i++)            scanf("%d",&zu[i]);//        fill(dp,dp+n,10*10000);///[dp[0],dp[n-1]]全部赋值10*10000//        for(int i=1;i<=n;i++)        *lower_bound(dp,dp+n,zu[i])=zu[i];        ///lower_bound()在[dp[0],dp[n-1]]返回大于或等于zu[i]的第一个元素位置        ///(找不到时返回的位置为n)        printf("%d\n",lower_bound(dp,dp+n,10*10000)-dp);    }    return 0;}