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?
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.
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
- 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
1 #include <iostream> 2 #include <string> 3 #include <cstdio> 4 #include <cstring> 5 #include <algorithm> 6 #include <climits> 7 #include <cmath> 8 #include <vector> 9 #include <queue> 10 #include <stack> 11 #include <set> 12 #include <map> 13 using namespace std; 14 typedef long long LL ; 15 typedef unsigned long long ULL ; 16 const int maxn = 1e5 + 10 ; 17 const int inf = 0x3f3f3f3f ; 18 const int npos = -1 ; 19 const int mod = 1e9 + 7 ; 20 const int mxx = 100 + 5 ; 21 const double eps = 1e-6 ; 22 const double PI = acos(-1.0) ; 23 24 int T, n, ans; 25 int a[maxn], c[maxn]; 26 int main(){ 27 // freopen("in.txt","r",stdin); 28 // freopen("out.txt","w",stdout); 29 while(~scanf("%d",&T)){ 30 while(T--){ 31 ans=1; 32 scanf("%d",&n); 33 scanf("%d",&a[1]); 34 c[0]=a[1]; 35 for(int i=2;i<=n;i++){ 36 scanf("%d",&a[i]); 37 if(a[i]>=c[ans-1]){ 38 c[ans++]=a[i]; 39 }else{ 40 c[upper_bound(c,c+ans,a[i])-c]=a[i]; 41 } 42 } 43 if(n-ans==1 || n==ans){ 44 puts("YES"); 45 }else{ 46 ans=1; 47 c[0]=a[n]; 48 for(int i=n-1;i>0;i--) 49 if(a[i]>=c[ans-1]){ 50 c[ans++]=a[i]; 51 }else{ 52 c[upper_bound(c,c+ans,a[i])-c]=a[i]; 53 } 54 if(n-ans==1 || n==ans) 55 puts("YES"); 56 else 57 puts("NO"); 58 } 59 } 60 } 61 return 0; 62 }
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