HDU-5532 Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2865    Accepted Submission(s): 719

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：

在一串给定的数字中去掉一个数能否使剩余数字组成不上升或不下降数列。

好久不见 upper_bound();  关于upper_bound();函数介绍：http://www.cnblogs.com/Kiven5197/p/5767583.html

每次判断前一个是否>=/<=后一个 否时存在upper_bound();

判断剩余len是否为n-1;

附AC代码：

#include <bits/stdc++.h>

using namespace std;

const int N=100000+10;
 
int ans[N],a[N];
int main()
{
    int cas;
    scanf("%d",&cas);
    while(cas--){
       int n;scanf("%d",&n);
       for(int i=1;i<=n;i++) scanf("%d",&a[i]);
       int len=1,flag1=0,flag2=0;
       ans[1]=a[1];
       for(int i=2;i<=n;i++)
       {
             if(a[i]>=ans[len]) {len++;ans[len]=a[i];}
             else{
                int pos=upper_bound(ans+1,ans+len,a[i])-ans;
                ans[pos]=a[i];
             }
       }
       if(len>=n-1) flag1=1;
       len=1;ans[1]=a[n];
       for(int i=n-1;i>=1;i--)
       {
             if(a[i]>=ans[len]) {len++;ans[len]=a[i];}
             else{
                int pos=upper_bound(ans+1,ans+len,a[i])-ans;
                ans[pos]=a[i];
             }
       }
       if(len>=n-1) flag2=1;
       if(flag1||flag2) printf("YES
");
       else printf("NO
");
    }
    return 0;
}