[hdu 5532] [2015ACM/ICPC亚洲区长春站 ] Almost Sorted Array　最长不下降子序列

Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 1917 Accepted Submission(s): 494

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）

题目链接；http://acm.hdu.edu.cn/showproblem.php?pid=5532

题意；问给定序列去掉一个数后是否构成不下降｜不上升的序列

思路；正向｜反向作最长不下降子序列（nlogn的做法）；

如果len==n||len==n-1输出“Ｙｅｓ”；

代码；

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int a[100005];
int ans[100005];
int b[100005];
int inf=99999999;
int n;
int main()
{
int T;

scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
b[n-i+1]=a[i];
ans[i]=inf;
}
int len=0;

for(int i=1;i<=n;i++)
{
int pos=upper_bound(ans+1,ans+1+n,a[i])-ans;
ans[pos]=a[i];
}

int pos=lower_bound(ans+1,ans+1+n,inf)-ans;

if(pos==n||pos==n+1)
{
printf("YES\n");
continue;
}
for(int i=1;i<=n;i++) ans[i]=inf;
for(int i=1;i<=n;i++)
{
pos=upper_bound(ans+1,ans+1+n,b[i])-ans;
ans[pos]=b[i];
}
pos=lower_bound(ans+1,ans+1+n,inf)-ans;
if(pos==n||pos==n+1)
{
printf("YES\n");
continue;
}
printf("NO\n");

}

}