HDU 5532 Almost Sorted Array

Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 5478    Accepted Submission(s): 1287

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

 注意  1 1 1 3 5 输出什么，严格递增递减排序

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int x,dp[100006],dps[100006],n,t;
int a[100006];
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int ans=1;
        int pos=1;
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        dp[1]=a[0];
        for(int i=1;i<n;i++)
        {
            if(a[i]>=dp[ans]) dp[++ans]=a[i];
            else
            {
                int l=upper_bound(dp+1,dp+ans,a[i])-dp;//重要
                dp[l]=a[i];
            }
        }
        dps[1]=a[n-1];
        for(int i=n-2;i>=0;i--)
        {
            if(a[i]>=dps[pos]) dps[++pos]=a[i];
            else
            {
                int l=upper_bound(dps+1,dps+pos,a[i])-dps;
                dps[l]=a[i];
            }
        }
        if(ans>=(n-1) || pos>=(n-1)) printf("YES
");
        else printf("NO
");
    }
    return 0;
}