【题解】POJ-2533 Longest Ordered Subsequence

发表于 分类于 ， ， ， 本文字数： 734 阅读时长 ≈ 1 分钟
POJ-2533 Longest Ordered Subsequence 类型 LIS nlogn

Longest Ordered Subsequence (POJ - 2533)

题面

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

1
2
7
1 7 3 5 9 4 8

样例输出

1
4

提示

无

思路

n^2也能过

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int N = 1e3+5;

int a[N];
int n;

int dp[N], ans[N];

int main(void) {
    scanf("%d", &n);
    for(int i=0; i<n; i++)
        scanf("%d", &a[i]);

    ans[0] = a[0];
    int len = 0;
    for(int i=1; i<n; i++) {
        if(a[i]>ans[len]) {
            ans[++len] = a[i];
        } else {
            int pos = lower_bound(ans, ans+len, a[i])-dp;
            dp[pos] = a[i];
        }
    }
    printf("%d\n", len+1);

    return 0;
}