poj2533 Longest Ordered Subsequence

Longest Ordered Subsequence

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 41282 Accepted: 18182
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

#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;
/*
L存到当前为止最长有序子序列长度，即当前为止子问题最优解
	0 1 2 3 4 5 6
 a  1 7 3 5 9 4 8
 L  1 2 2 3 4 3 4   每次取比a[i]小的当中L最大的加一作为当前L最优方案
*/
int main()
{
	int N, a[1005], L[1005];
	scanf("%d", &N);
	int ans = 0;
	for (int i = 0; i < N; i++) {
		scanf("%d", &a[i]);
		int maxn = 0;
		for (int j = 0; j < i; j++) {
			if (a[j] < a[i] ) {
				maxn = max(maxn, L[j]);
			}
		}
		L[i] = maxn + 1;
		ans = max(ans, L[i]);
	}
	printf("%d\n", ans);
	return 0;
}