Almost Sorted Array Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 6019 Accepted Submission(s) ...
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
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亞洲區長春站-重現賽（感謝東北師大）
- 要求是去掉一個元素使得新數列為非嚴格單調數列
- 那就正逆跑一遍nlogn的LIS，把判定里的小於號改成小於等於號就可以做到對於非嚴格單調的要求
- 然後原來演算法裡面的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 }
