POJ 2533 Longest Ordered Subsequence

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 29540		Accepted: 12864

Description

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_{i_K}), 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 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

Source

Northeastern Europe 2002, Far-Eastern Subregion

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <vector>
 4 #include <set>
 5 #include <cstring>
 6 #include <string>
 7 #include <map>
 8 #include <cmath>
 9 #include <stack>
10 #include <ctime>
11 #include <algorithm>
12 #include <queue>
13 
14 using namespace std;
15 #define INF 0x7fffffff
16 #define maxm 1001
17 #define mod 1000000007
18 #define mp make_pair
19 #define pb push_back
20 #define rep(i,n) for(int i = 0; i < (n); i++)
21 #define re return
22 #define fi first
23 #define se second
24 #define sz(x) ((int) (x).size())
25 #define all(x) (x).begin(), (x).end()
26 #define sqr(x) ((x) * (x))
27 #define sqrt(x) sqrt(abs(x))
28 #define y0 y3487465
29 #define y1 y8687969
30 #define fill(x,y) memset(x,y,sizeof(x))
31 
32 typedef vector<int> vi;
33 typedef long long ll;
34 typedef long double ld;
35 typedef double D;
36 typedef pair<int, int> ii;
37 typedef vector<ii> vii;
38 typedef vector<string> vs;
39 typedef vector<vi> vvi;
40 
41 template<class T> T abs(T x) { re x > 0 ? x : -x; }
42 const int maxn = 100015;
43 int n, m, t, k, x, l, r, s, sk,y,top,temp;
44 int S[maxn];
45 int main(){
46     top = 0; S[top] = -1;
47     scanf("%d", &n);
48     for (int i = 0; i < n; i++){
49         scanf("%d", &temp);
50         if (temp>S[top])S[++top] = temp;
51         else {
52             int l = 1, r = top;
53             while (l < r){
54                 int mid = (l + r) >> 1;
55                 if (S[mid]<temp)l = mid + 1;
56                 else r = mid;
57             }
58             S[l] = temp;
59         }
60     }
61     printf("%d\n", top);
62 }

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posted @ 2013-12-01 18:44 HaibaraAi 阅读(139) 评论(0) 收藏举报

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