POJ2533 Longest Ordered Subsequence —— DP 最长上升子序列（LIS）

题目链接：http://poj.org/problem?id=2533

 

Longest Ordered Subsequence
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 55459   Accepted: 24864

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

Source

Northeastern Europe 2002, Far-Eastern Subregion
 
 
O(n^2):
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <queue>
 8 #include <stack>
 9 #include <map>
10 #include <string>
11 #include <set>
12 #define ms(a,b) memset((a),(b),sizeof((a)))
13 using namespace std;
14 typedef long long LL;
15 const double EPS = 1e-8;
16 const int INF = 2e9;
17 const LL LNF = 2e18;
18 const int MAXN = 1e6+10;
19 
20 int dp[MAXN], a[MAXN];
21 
22 int main()
23 {
24     int n;
25     while(scanf("%d",&n)!=EOF)
26     {
27         for(int i = 1; i<=n; i++)
28             scanf("%d",&a[i]);
29 
30         ms(dp, 0);
31         for(int i = 1; i<=n; i++)
32         for(int j = 0; j<i; j++)
33             if(j==0 || a[i]>a[j])
34                 dp[i] = max(dp[i], dp[j]+1);
35 
36         int ans = -INF;
37         for(int i = 1; i<=n; i++)
38             ans = max(ans, dp[i]);
39         printf("%d\n",ans);
40     }
41  }
View Code
 
O(nlogn):
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <queue>
 8 #include <stack>
 9 #include <map>
10 #include <string>
11 #include <set>
12 #define ms(a,b) memset((a),(b),sizeof((a)))
13 using namespace std;
14 typedef long long LL;
15 const double EPS = 1e-8;
16 const int INF = 2e9;
17 const LL LNF = 2e18;
18 const int MAXN = 1e6+10;
19 
20 int dp[MAXN], a[MAXN];
21 
22 int main()
23 {
24     int n;
25     while(scanf("%d",&n)!=EOF)
26     {
27         for(int i = 1; i<=n; i++)
28             scanf("%d",&a[i]);
29 
30         int len = 0;
31         for(int i = 1; i<=n; i++)
32         {
33             if(i==1 || a[i]>dp[len])
34                 dp[++len] = a[i];
35 
36             else
37             {
38                 int pos = lower_bound(dp+1,dp+1+len,a[i]) - (dp+1);
39                 dp[pos+1] = a[i];
40             }
41         }
42         printf("%d\n",len);
43     }
44  }
View Code

 

posted on 2017-10-03 21:17  h_z_cong  阅读(202)  评论(0)  编辑  收藏  举报

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