每日三题-Day5-A（POJ 2533 Longest Ordered Subsequence 最长上升子序列O(nlogn)解法）

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 51451		Accepted: 22885

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

相似的题已经发过很多，裸的LIS也不难。不再累述。
可以参考这个题的题解： http://blog.csdn.net/lulu11235813/article/details/70312914
AC代码：
#include<cstdio>
#include<cstring>

using namespace std;
int num[1005];
int min_end[1005];

int Bsea(int l,int r,int a)
{
if(l>=r) return l;
int mid=(l+r)/2;
if(min_end[mid]>=a) return Bsea(l,mid,a);
return Bsea(mid+1,r,a);
}
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=0;i<n;i++)
{
scanf("%d",&num[i]);
}
int ans=1;
min_end[1]=num[0];
for(int i=1;i<n;i++)
{
if(num[i]>min_end[ans])
{
ans++;
min_end[ans]=num[i];
}
else
{
min_end[Bsea(1,ans,num[i])]=num[i];
}
}
printf("%d\n",ans);
}
return 0;
}