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

17
21 7 3 5 9 4 8

样例输出

14

提示

无

思路

n^2也能过

代码

 1using namespace std;
 2typedef long long ll;
 3const int inf = 0x3f3f3f3f;
 4const int N = 1e3+5;
 5
 6int a[N];
 7int n;
 8
 9int dp[N], ans[N];
10
11int main(void) {
12    scanf("%d", &n);
13    for(int i=0; i<n; i++)
14        scanf("%d", &a[i]);
15
16    ans[0] = a[0];
17    int len = 0;
18    for(int i=1; i<n; i++) {
19        if(a[i]>ans[len]) {
20            ans[++len] = a[i];
21        } else {
22            int pos = lower_bound(ans, ans+len, a[i])-dp;
23            dp[pos] = a[i];
24        }
25    }
26    printf("%d\n", len+1);
27
28    return 0;
29}