poj-2533 longest Ordered subsequence "longest ascending subsequence"

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

Longest Ordered subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 35929		Accepted: 15778

Description

A Numeric sequence of AIis ordered if A1< A2< ... < an. Let the subsequence of the given numeric sequence ( A1, A2, ..., an) is any sequence ( Ai1, AI2, ..., AiK), where 1 <= I1< I2< ... < IK<= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) have ordered Subsequences, E. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences is of length 4, e. g., (1, 3, 5, 8).

Your program, when given the numeric sequence, must find the length of its longest ordered.

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

To find the longest ascending sub-sequence

#include <stdio.h> #include <iostream> #include <math.h> #include <stdlib.h> #include < ctype.h> #include <algorithm> #include <vector> #include <string> #include <queue> #include <stack> #include <set> #include <map>using namespace Std;int n,p[1010],dp[1010];int main () {while ( scanf ("%d", &n)! = EOF) {for (int i = 1; I <= n; i++) scanf ("%d", &p[i]), for (int i = 1; I <= n; i++) {Dp[i] = 1 ; for (int j = 1; j < I; J + +) {if (P[i] > P[j]) dp[i] = max (Dp[i],dp[j] + 1);}} int ans = -1;for (int i = 1; I <= n; i++) ans = max (ans,dp[i]);p rintf ("%d\n", ans);} return 0;}

poj-2533 longest Ordered subsequence "longest ascending subsequence"