We are all familiar with sorting algorithms: quick sort, merge sort, heap sort, insertion sort, selection sort, bubble sort, etc. But sometimes it is an overkill to use these algorithms for an almost sorted array.

We say an array is sorted if its elements are in non-decreasing order or non-increasing order. We say an array is almost sorted if we can remove exactly one element from it, and the remaining array is sorted. Now you are given an array a1,a2,…,an, is it almost sorted?

The first line contains an integer T indicating the total number of test cases. Each test case starts with an integer n in one line, then one line with n integers a1,a2,…,an.

1≤T≤2000
2≤n≤105
1≤ai≤105
There are at most 20 test cases with n>1000.

For each test case, please output "`YES`" if it is almost sorted. Otherwise, output "`NO`" (both without quotes).

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大） 

思路：

题意就是问你一串数去掉一个数之后还是不是非递减或者非递增的。

转化为求数字串正反两面的最大上升子序列长度。只要LIS长度大于等于n-1，那么他一定可以去掉一个数之后还是单调的。

我这里用了kuangbi大神的板子求的LIS的长度。正反扫两次就好。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=500010; 
int a[MAXN],b[MAXN],c[MAXN],d[MAXN]; 
 
int Search1(int num,int low,int high) 
{     
	int mid;     
	while(low<=high)     
	{ 
		mid=(low+high)/2;         
		if(num>=b[mid])  
		low=mid+1;         
		else   
		high=mid-1;     
	}     
	return low; 
} 
int DP1(int n) 
{     
	int i,len,pos;     
	b[1]=a[1];     
	len=1;     
	for(i=2;i<=n;i++)     
	{         
		if(a[i]>=b[len])       
		{             
			len=len+1;             
			b[len]=a[i];         
		}         
		else      
		{           
			pos=Search1(a[i],1,len);           
			b[pos]=a[i];        
		}    
	}    
	return len;
} 

int Search2(int num,int low,int high) 
{     
	int mid;     
	while(low<=high)     
	{ 
		mid=(low+high)/2;         
		if(num>=d[mid])  
		low=mid+1;         
		else   
		high=mid-1;     
	}     
	return low; 
} 
int DP2(int n) 
{     
	int i,len,pos;     
	d[1]=c[1];     
	len=1;     
	for(i=2;i<=n;i++)     
	{         
		if(c[i]>=d[len])      
		{             
			len=len+1;             
			d[len]=c[i];         
		}         
		else         
		{           
			pos=Search2(c[i],1,len);           
			d[pos]=c[i];        
		}    
	}    
	return len;
} 
		 
int main()
{
	int num;
    while(scanf("%d",&num)!=EOF)
    {
    	while(num--)
    	{
    		int n;
    		cin>>n;
    		for(int i=1;i<=n;i++)
	        {
	            scanf("%d",&a[i]);
	            c[n-i+1]=a[i];
	        }    
	        int res1=DP1(n);
	        int res2=DP2(n);
	        //cout<<res1<<" "<<res2<<endl; 
	        if(res1>=n-1||res2>=n-1)printf("YES\n");
			else printf("NO\n");    
    	}  
    }
    return 0;    	
}