POJ2533 Longest Ordered Subsequence (动态规划)

Longest Ordered Subsequence

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 49547 Accepted: 21994

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

Source

Northeastern Europe 2002, Far-Eastern Subregion

思路：求序列的最长递增子序列即可，状态转移方程：dp[i] = max(dp[i],dp[j]+1);

#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>using namespace std;int a[1005],dp[1005];int main(){int n;cin >> n;for(int i = 0; i < n; i ++)scanf("%d",&a[i]);for(int i = 0; i < n; i ++){dp[i] = 1;}for(int i = 1; i < n; i ++)for(int j = 0; j < i; j ++)if(a[i] > a[j])dp[i] = max(dp[i],dp[j] + 1);int Max = 0;for(int i = 0; i < n; i ++)Max = max(Max,dp[i]);printf("%d\n",Max);return 0;}