Longest Ordered Subsequence
| Time Limit: 2000MS |
|
Memory Limit: 65536K |
| Total Submissions: 21310 |
|
Accepted: 9190 |
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
最长上升子序列
朴素的做法n^2的复杂度，n<=1000不会超时
有nlogn的做法
1
/**///////朴素做法 2
#include<stdio.h>
3#include<string.h>
4#include<math.h>
5#define MAX 1005
6int a[MAX],f[MAX];
7int n,i,j,ans;
8int max(int a,int b)
9
{
10
if (a>b) return a; else return b;
11
}
12int main()
13{
14 scanf("%d",&n);
15 for (i=1;i<=n ;i++) scanf("%d",&a[i]);
16 f[1]=1;
17 ans=f[1];
18 for (i=2;i<=n ;i++ )
19
{
20 f[i]=1;
21 for (j=1;j<=i-1 ;j++ )
22 {
23 if (a[j]<a[i])
24 {
25 f[i]=max(f[i],f[j]+1);
26
}
27 }
28 ans=max(ans,f[i]);
29 }
30 printf("%d\n",ans);
31 return 0;
32}
优化的解法 nlogn
1/**/////二分查找优化 nlogn 2#include<stdio.h>
3#include<string.h>
4#include<math.h>
5#define MAX 1005
6int f[MAX];
7int i,j,ans,n;
8int left,right,x,mid;
9int main()
10{
11 scanf("%d",&n);
12 ans=0;
13 for (i=1; i<=n ; i++)
14 {
15 scanf("%d",&x);
16 left=1;
17 right=ans;
18 while (left<right)
19 {
20 mid=(left+right)/2;
21 if (f[mid]<x) left=mid+1;
22 else right=mid;
23 }
24 if (left>=right&&x>f[ans]||ans==0)
25 {
26 ans++;
27 f[ans]=x;
28 }
29 else if (x<f[left])
30 {
31 f[left]=x;
32 }
33 }
34 printf("%d\n",ans);
35 return 0;
36}