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): 2136    Accepted Submission(s): 537

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亞洲區長春站-重現賽（感謝東北師大）

 

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方法一：

解：當時首先想到的是用O(n)的思路答，每次遇到逆序的就選擇一個刪掉，具體選擇思路：比如求上升時，若a[i]<a[i-1]，那麼肯定要刪除a[i]或者a[i-1]。那麼如果a[i]>a[i-2]則刪除a[i-1]，否則刪除a[i]；看最終刪除的次數來判斷。

方法2：

直接用nlogn的方法求最長上升子序列和最長下降子序列的長度，如果大於等於n-1則YES。

關於nlogn求LIS的方法。用一個D[]的數組來輔助，D[i]表示LIS爲i時結尾數字最小時多少。可知D[]數組是一個遞增數組，所以LIS的循環中可以用一個二分來查找比a[j]小且LIS最大是多少，求出之後維護D數組，最後的時間複雜度就是nlogn。

<pre name="code" class="cpp">#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 100005
using namespace std;
int a[maxn];
int d[maxn];
int dp[maxn];
int find(int l,int r,int val)
{
    int ans;
    int mid;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(d[mid]<=val)
        {
            ans=mid;
            l=mid+1;
        }
        else {
            r=mid-1;
        }
    }
    return ans;
}
int LIS(int n)
{
    int now;
    int maxx=0;    memset(d,0,sizeof(d));
    d[0]=-1000000000;
    for(int i=1;i<=n;i++)
    {
       // if(a[i]>d[maxx])
        //{
          //  maxx++;
            //d[maxx]=a[i];
        //}
        //else {
            now=find(0,maxx,a[i]);
            d[now+1]=a[i];
            maxx=max(maxx,now+1);
        //}
    }
    return maxx;
}
int main()
{
    int flag1,flag2;
    int n,T;
    scanf("%d",&T);
    while(T--)
    {
        flag1=0;flag2=0;

        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        flag1=LIS(n);
        for(int i=1;i<=n/2;i++)
        {
            swap(a[i],a[n-i+1]);
        }
        flag2=LIS(n);
        if(flag1>=n-1||flag2>=n-1)
            printf("YES\n");
        else printf("NO\n");

    }
}