杭电5532 Almost Sorted Array

Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1324    Accepted Submission(s): 396

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）

 

扣掉一个数能不能让数列成为非递减或者非递增，upper_bound正着一次，反着一次就行了：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[110000],i,j,k,l,m,n,help[110000],top;
int main()
{
	scanf("%d",&k);
	while(k--)
	{
		top=0;
		scanf("%d",&n);
		for(i=0;i<n;i++)
		scanf("%d",&a[i]);
		help[top]=a[0];
		
		for(i=1;i<n;i++)
		{
			if(a[i]>=help[top])
			help[++top]=a[i];
			else
			{
				int t=upper_bound(help,help+top,a[i])-help;
				help[t]=a[i];
			}
		}
		if(n-top<=2)
		{
			printf("YES\n");
			continue;
		}
			top=0;
			help[top]=a[n-1];
			for(i=n-2;i>=0;i--)
			{
			if(a[i]>=help[top])
			help[++top]=a[i];
			else
			{
				int t=upper_bound(help,help+top,a[i])-help;
				help[t]=a[i];
			}
		}
		if(n-top<=2)
		printf("YES\n");
		else 
		printf("NO\n");
		
	}
}