Almost Sorted Array

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

题意：给你一个n个数的序列，让你删除一个数后，这个序列是严格不下降或者不上升的序列

题解：最先想到就是LIS，可做

还有一种是O(N) 对于每一个数，要么呈现V形或者^形，或者单调，记录上升下降次数，我们对删除某个数所造成影响记录下来，假如删除它致使最后上升次数或者下降次数满足n-2次就行了

///1085422276
#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define inf 1000000007
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){
        if(ch==‘-‘)f=-1;ch=getchar();
    }
    while(ch>=‘0‘&&ch<=‘9‘){
        x=x*10+ch-‘0‘;ch=getchar();
    }return x*f;
}
//****************************************

#define maxn 100000+5
int a[maxn],n;
int main(){
    int T=read();
    while(T--){
        scanf("%d",&n);int flag=0;
        for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        }
        int res=0,fall=0;
        for(int i=1;i<=n;i++){
            if(i-1>=1&&a[i]>a[i-1])res++;
            else if(i-1>=1&&a[i]<a[i-1])fall++;
        }
       // cout<<fall<<" "<<res<<endl;
        for(int i=1;i<=n;i++){
                int t1=res,t2=fall;
            if(i-1>=1&&a[i]>a[i-1]){
                res--;
            }
            else if(i-1>=1&&a[i]<a[i-1]){
                fall--;
            }
            if(i+1<=n&&a[i]>a[i+1]){
                fall--;
            }
            else if(i+1<=n&&a[i]<a[i+1]){
                res--;
            }
            if(i-1>=1&&i+1<=n){
                if(a[i-1]<a[i+1]){
                    res++;
                }else if(a[i-1]>a[i+1])fall++;
            }
            if(fall==0||res==0){
                flag=1;break;
            }
        res=t1;fall=t2;
        }
        if(flag){
            cout<<"YES"<<endl;
        }
        else cout<<"NO"<<endl;

    }
  return 0;
}

代码