PKU2533 最长上升子序列 DP

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15397		Accepted: 6552

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

 

 

毫无疑问
,

最长上升子序列
…

当作练习
nlogn

的算法把它水掉了
…

代码如下
:

 
 

#include<stdio.h>
#include<string.h>
#define MAXN 1005
int i,n,a[MAXN],c[MAXN],len[MAXN],pos,max,total;
int find(int a,int l,int r)
{
int mid=(l+r)/2;
while (l<=r)
{
if (c[mid]==a) return mid;
else
{
if (a>c[mid]) l=mid+1;
else r=mid-1;
mid=(l+r)/2;
}
}
return l;
}
int main()
{
while (scanf("%d",&n)!=EOF)
{
for (i=1;i<=n;++i) scanf("%d",&a[i]);
for (i=1;i<=n;++i) c[i]=20000;
c[1]=a[1];
len[1]=1;
total=1;
max=1;
for (i=1;i<=n;++i)
{
pos=find(a[i],1,total+1);
c[pos]=a[i];
len[i]=pos;
if (pos>max) {max=pos;total++;}
}
printf("%d/n",max);
}
return 0;
}