HDU 5532 Almost Sorted Array(最长不上升||不下降子序列）_almost sorted array 答案错误-优快云博客

本文介绍了一种算法，用于判断给定的序列在移除一个元素后是否能变为有序。通过计算最长不上升和不下降子序列的长度，如果长度大于等于n-1，则序列满足条件。文章提供了一个C++实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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：给你一段长度为n的序列，问你去掉一个数字后，剩下的序列是否处于排好序的状态。

题解：显然，我们可以直接求最长不上升和不下降子序列，若答案大于等于n-1，则满足条件。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[100000+10];
int b[100000+10];
int dp[100000+10];
int find(int l,int r,int x)
{
    int ans=r;
    while(l<=r)
    {
        int mid=(l+r)/2;
        if(dp[mid]<=x)
        {
            l=mid+1;
        }
        else
        {
            r=mid-1;
            ans=min(ans,mid);
        }
    }
    return ans;
}
int n;
int solve()
{
    memset(dp,0,sizeof(dp));
    int len=1;
    dp[len]=a[1];
    for(int i=2; i<=n; i++)
    {
        if(a[i]>=dp[len])
        {
            len++;
            dp[len]=a[i];
        }
        else
        {
            int pos=find(1,len,a[i]);
            dp[pos]=a[i];
        }
    }
    //cout<<len<<endl;
    return len;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            b[i]=a[i];
        }
        bool flag=0;
        if(solve()>=n-1)
        {
            flag=1;
        }
        for(int i=1;i<=n;i++)
        {
            a[i]=b[n-i+1];
        }
        if(solve()>=n-1)
        {
            flag=1;
        }
        if(flag)
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }
    return 0;
}