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2013
11-11

POJ 2533 Longest Ordered Subsequence [解题报告] Java

Longest Ordered Subsequence

问题描述 :

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.

输入:

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

输出:

Output file must contain a single integer – the length of the longest ordered subsequence of the given sequence.

样例输入:

7
1 7 3 5 9 4 8

样例输出:

4

解题代码:

import java.io.BufferedInputStream;   
import java.util.Scanner;   
public class Main {   
  
    public static void main(String[] args) {   
        Scanner scan = new Scanner(new BufferedInputStream(System.in));   
        if (scan.hasNext()) {   
            int n = scan.nextInt();   
            int[] data = new int[n];   
            int[] count = new int[n];   
            for (int i = 0; i < n; i++) {   
                int a = scan.nextInt();   
                data[i] = a;   
                count[i] = 1;   
                int max = count[i];   
                int flag = 0;   
                for (int j = i - 1; j >= 0; j--) {   
                    if (data[i] > data[j]) {   
                        if (count[j] > max) {   
                            max = count[j];   
                        }  
                        flag = 1;   
                    }   
                }   
                if (flag == 1) {   
                    count[i] = max + 1;   
                }   
            }   
            int max = count[0];   
            for (int i = 1; i < n; i++) {   
                if (count[i] > max) {   
                    max = count[i];   
                }   
            }   
            System.out.println(max);   
        }   
    }   
}