[POJ]2533 Longest Ordered Subsequence

问题

Description

A numeric sequence of ai is ordered if a1 < a2 < … < aN. Let the subsequence of the given numeric sequence (a1, a2, …, aN) be any sequence (ai1, ai2, …, aiK), where 1 <= i1 < i2 < … < iK <= 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

分析

题目要求找最长非降序子序列，可以采用动态规划的思想做。

f(i)=⎧⎩⎨1,f(j)+1,1,i = 1seq[i] > seq[j](1 < j < i)不存在seq[i] < seq[j]

源代码

#include <iostream>#include <vector>#include <algorithm>using namespace std;int main() {    int N, ans = 1;    cin >> N;    vector<int> seq(N, 0), dp(N, 0);    for (int i = 0; i < N; ++i)        cin >> seq[i];    dp[0] = 1;    for (int i = 1; i < N; ++i) {        int tmp = 0;        for (int j = 0; j < i; ++j)            if (seq[i] > seq[j])                tmp = max(tmp, dp[j]);        dp[i] = tmp + 1;        ans = max(dp[i], ans);    }    cout << ans << endl;    return 0;}

程序结果

Result Memory Time Language Code Length Accepted 244K 16MS C++ 422B

不使用系统库函数max，自己编写后，速度会提升，结果如下：

Result Memory Time Language Code Length Accepted 244K 0MS C++ 462B