Longest Ordered Subsequence - POJ 2533 最长上升子序列

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本文介绍了一种解决最长有序子序列问题的高效算法，并通过一个示例详细展示了输入输出及核心实现逻辑，该算法利用二分查找优化了传统方法。

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 32647		Accepted: 14265

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

AC代码如下;

#include<cstdio>
#include<cstring>
using namespace std;
int n[1010],num[1010],len=0;
int solve(int k)
{ int l=1,r=len,mi;
  while(l<=r)
  { mi=(l+r)/2;
    if(num[mi]>=k)
     r=mi-1;
    else
     l=mi+1;
  }
  return l;
}
int main()
{ int n,i,j,k;
  scanf("%d",&n);
  num[0]=-1;
  for(i=1;i<=n;i++)
  { scanf("%d",&k);
    if(k>num[len])
     num[++len]=k;
    else
     num[solve(k)]=k;
  }
  printf("%d\n",len);
}