HDU 5532 Almost Sorted Array (LIS水题)

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该问题要求判断一个数组是否只需移除一个元素即可变为有序数组。输入包含测试用例数和数组信息，输出数组是否几乎有序的结果。样例输入和输出分别展示了符合条件的和不符合条件的数组情况。

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Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 7902 Accepted Submission(s): 1835

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）

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#pragma comment(linker, "/STACK:102400000,102400000")
#include<bits/stdc++.h>
using namespace std;

#define debug puts("YES");
#define rep(x,y,z) for(int (x)=(y);(x)<(z);(x)++)

#define lrt int l,int r,int rt
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define ll long long

const int  maxn =1e5+5;
const int mod=1e9+7;
/*
题目大意:给定一个序列，判断拿掉一个数字后是否是有序的（不增或不降都行）。

水题一枚。。。
只要两次看看最长不降子序列和不升子序列的长度即可。
WA了两发是在找界限的时候，应该用upper_bound,
这样相同的数字不会被覆盖。

*/

int n,a[maxn],b[maxn];
int len;

int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);for(int i=0;i<n;i++) scanf("%d",&a[i]);
        memset(b,0xf,sizeof(b)); len=0;
        for(int i=0;i<n;i++)
        {
            int pos=upper_bound(b,b+n+1,a[i])-b;
            b[pos]=a[i];
            if(pos>=len) len++;
        }
        if(len>=n-1) {puts("YES");continue;}

        memset(b,0xf,sizeof(b)); len=0;
        for(int i=0;i<n;i++)
        {
            int pos=upper_bound(b,b+n+1,-a[i])-b;
            b[pos]=-a[i];
            if(pos>=len) len++;
        }
        if(len>=n-1) {puts("YES");continue;}
        puts("NO");
    }
    return 0;
}