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): 6019    Accepted Submission(s): 1446

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）

 

• 要求是去掉一个元素使得新数列为非严格单调数列
• 那就正逆跑一遍nlogn的LIS，把判定里的小于号改成小于等于号就可以做到对于非严格单调的要求
• 然后原来算法里面的lower_bound改成upper_bound就行了

 #include <iostream>
 #include <string>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <climits>
 #include <cmath>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <set>
 #include <map>
 using namespace std;
 typedef long long           LL ;
 typedef unsigned long long ULL ;
 const int    maxn = 1e5 +    ;
 const int    inf  = 0x3f3f3f3f ;
 const int    npos = -         ;
 const int    mod  = 1e9 +     ;
 const int    mxx  =  +     ;
 const double eps  = 1e-       ;
 const double PI   = acos(-1.0) ;

 int T, n, ans;
 int a[maxn], c[maxn];
 int main(){
     // freopen("in.txt","r",stdin);
     // freopen("out.txt","w",stdout);
     while(~scanf("%d",&T)){
         while(T--){
             ans=;
             scanf("%d",&n);
             scanf("%d",&a[]);
             c[]=a[];
             for(int i=;i<=n;i++){
                 scanf("%d",&a[i]);
                 if(a[i]>=c[ans-]){
                     c[ans++]=a[i];
                 }else{
                     c[upper_bound(c,c+ans,a[i])-c]=a[i];
                 }
             }
             if(n-ans== || n==ans){
                 puts("YES");
             }else{
                 ans=;
                 c[]=a[n];
                 for(int i=n-;i>;i--)
                     if(a[i]>=c[ans-]){
                         c[ans++]=a[i];
                     }else{
                         c[upper_bound(c,c+ans,a[i])-c]=a[i];
                     }
                 if(n-ans== || n==ans)
                     puts("YES");
                 else
                     puts("NO");
             }
         }
     }
     return ;
 }