POJ 2533 Longest Ordered Subsequence-优快云博客

本文链接：https://blog.youkuaiyun.com/hellohelloC/article/details/47699655

本文探讨了如何通过维护一个单调队列来解决寻找给定数列中最长严格单调上升子序列的问题，提供了时间复杂度为O(n*logn)的高效算法。详细解释了算法逻辑和实现步骤。

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Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 39104		Accepted: 17188

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

题意：求一个严格单调上升子序列的最大长度

解题思路：维护一个单调的队列
新的元素，如果大于队尾元素，即插入队尾
否则二分查找比它大的最小元素，替换掉
最后队列长度即为LIS的解

复杂度分析：时间复杂度：o(n*logn)
空间复杂度：o(n)

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=10010;
int a[maxn],b[maxn];
int n;
int w(int *a,int x,int y,int t)
{
  int L=x,R=y,M=0;
  while(L<R)
  {
      M=((R-L)>>1)+L;
      if(a[M]<t) L=M+1;
      else R=M;
  }
  return L;
}
int main()
{
    while(cin>>n && n)
    {
        int t,c=1;
        cin>>b[0];
        for(int i=1;i<n;i++)
        {
            cin>>t;
            if(t>b[c-1]) b[c++]=t;
            else
                b[w(b,0,c,t)]=t;
        }
        cout << c << endl;
    }
    return 0;
}