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

Longest Ordered Subsequence

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

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

Source

Northeastern Europe 2002, Far-Eastern Subregion

n*n算法

#include
#include
#include
#include
#include
#include

using namespace std;

int n;
int a[1001];
int dp[1010];

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int maxx = -1;
        for(int i=0;ia[j] && dp[j]+1>dp[i])
                {
                    dp[i] = dp[j] + 1;
                }
            }
            if(maxx 

n*logn算法：

#include
#include
#include
#include
#include

#define inf 999999

using namespace std;

int n;
int dp[1010];
int a[1010];

int res(int len,int num)
{
    int l = 0,r = len;
    while(l!=r)
    {
        int mid = (l+r)>>1;
        if(dp[mid] == num)
        {
            return mid;
        }
        else if(dp[mid]num)
        {
            r = mid;
        }
    }
    return l;
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i

服务器托管，北京服务器托管，服务器租用 http://www.fwqtg.net
机房租用，北京机房租用，IDC机房托管， http://www.e1idc.net