hdoj--5532--Almost Sorted Array(正反LIS)

Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 1459    Accepted Submission(s): 413

Problem Description

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?

 

Input

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.

 

Output

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

 

Sample Input

3
3
2 1 7
3
3 2 1
5
3 1 4 1 5

 

Sample Output

YES
YES
NO

 

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

 

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给你一个数列，找到一个非递减数列，正反两次LIS就行，好久不用upper_bound写了不对，

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int a[100100],b[100100],s[100100];
int Find(int l,int r,int val)
{
	int mid;
	while(l<=r)
	{
		mid=(l+r)>>1;
		if(s[mid]>val)
		r=mid-1;
		else
		l=mid+1;
	}
	return l;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		memset(s,0,sizeof(s));
		int n;
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		scanf("%d",&a[i]);
		for(int i=0;i<n;i++)
		b[n-1-i]=a[i];
		int len=1;
		s[0]=-1;
		for(int i=0;i<n;i++)
		{
			if(a[i]>=s[len-1])
			s[len++]=a[i];
			else
			{
				s[len]=10000000;
//				int q=lower_bound(s,s+len,a[i])-s;
				int q=Find(0,len,a[i]);
				if(q==len)
				len++;
				s[q]=a[i];
			}
		}
		if(len-1>=n-1)
		{
			printf("YES
");
			continue;
		}
		else
		{
			int len=1;
			memset(s,0,sizeof(s));
			s[0]=-1;
			for(int i=0;i<n;i++)
			{
				if(b[i]>s[len-1])
				s[len++]=b[i];
				else
				{
					s[len]=10000000;
					int q=Find(0,len,b[i]);
					if(q==len)
					len++;
					s[q]=b[i];
				}
			}
			if(len-1>=n-1)
			printf("YES
");
			else printf("NO
");
		}
	}
	return 0;
}