2533. Longest Ordered Subsequence

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 contains the length of sequence n (1 ≤ n ≤ 1000). The second line contains the elements of sequence – n integers in the range from 0 to 10000 each, separated by spaces.

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

.

Sample input	Sample output
7 1 7 3 5 9 4 8	4

РЕШЕНИЕ

наибольшая возрастающая подпоследовательность

Анализ алгоритма

В задаче следует вычислить длину наибольшей возрастающей подпоследовательности.

Реализация алгоритма

#include <cstdio>

#include <algorithm>

#define MAX 1001

using namespace std;

int x[MAX], lis[MAX];

int i, n, len, pos;

int main(void)

{

scanf("%d",&n);

for(i = 0; i < n; i++) scanf("%d",&x[i]);

for (len = i = 0; i < n; i++)

{

pos = lower_bound(lis,lis+len,x[i]) - lis;

if (pos < len) lis[pos] = x[i]; else lis[len++] = x[i];

}

printf("%d\n",len);

return 0;

}