模板—LIS(最长上升子序列优化算法)
链接：HDU - 5532
https://vjudge.net/contest/161167#problem/F
复杂度 n*logn
二分查找+dp

输入n n个数
查找n个数的最长上升子序列
方法一：STL版

int LIS(void)
{
    int tail[100005],len = 0;
    //最长下降子序列 改为for（int i=n-1;i>=0;i--)
    //或 reverse（a，a+n）
    for(int i=0;i<n;i++) 
    {
        int pos = upper_bound(tail,tail+len,a[i])-tail;
        tail[pos] = a[i];
        if(pos==len)len++;
    }
    return len;
}
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>a[i];
    return 0;
}

方法2：
一个类似的但是长度大于等于n-1

#include <iostream>
#include<algorithm>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<queue>
#include<deque>
#include<map>
using namespace std;
int n,mmax;
int link[100002],along[100002];
  int a[100002],b[100002];
int find(int key)
{
    int l = 1,r = mmax,mid;
    while(l<=r)
    {
        mid = (l+r)/2;
        if(a[along[mid]]<=key)
        l = mid+1;
        else r = mid-1;
    }
    return l -1;
}
int find2(int key)
{
    int l = 1,r = mmax,mid;
    while(l<=r)
    {
        mid = (l+r)/2;
        if(b[along[mid]]<=key)
        l = mid+1;
        else r = mid-1;
    }
    return l -1;
}
int main()
{
    int t,l;
    cin>>t;
    while(t--)
    {
        //int n;
        cin>>n;
        //vector<int>o;
        int long1,long2;

        int dp1[100002],dp2[100002];
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            b[n-i+1] = a[i];
          // dp1[i] = dp2[i] =1;
        }

        mmax = 1;
        along[mmax] = 1;
        link[1] =1;
        for(int i=2;i<=n;i++)
        {
            if(a[i]>=a[along[mmax]])
            {
                mmax++;
                along[mmax] = i;
                link[i] = along[mmax-1];
            }
            else
            {
                l = find(a[i]);
                along[l+1] = i;
                link[i]= along[l];
            }
        }
    long1 = mmax;








mmax = 1;
        along[mmax] = 1;
        link[1] =1;
        for(int i=2;i<=n;i++)
        {
            if(b[i]>=b[along[mmax]])
            {
                mmax++;
                along[mmax] = i;
                link[i] = along[mmax-1];
            }
            else
            {
                l = find2(b[i]);
                along[l+1] = i;
                link[i]= along[l];
            }
        }

    long2 = mmax;
        //cout<<long1<<" "<<long2<<endl;

        if(long1>=n-1||long2>=n-1)
        {
        printf("YES\n");

            }
            else
            {
                printf("NO\n");
            }

    }


return 0;
}