Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 447    Accepted Submission(s): 201

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）

题意：如果一组数去掉一个是不降或不升的，那么这组数是满足性质的，问给出的这组数是否满足性质。

分析：显然如果这组数本来就是不降或不升的，显然满足性质。

然后发现。。。不就是求max（最长不降子序列的长度，最长不升子序列的长度）是否大于等于n-1吗。。。

求最长不升子序列可以将数组反过来，再做一次最长不降子序列

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cstdlib>
  4 #include <cmath>
  5 #include <ctime>
  6 #include <iostream>
  7 #include <map>
  8 #include <set>
  9 #include <algorithm>
 10 #include <vector>
 11 #include <deque>
 12 #include <queue>
 13 #include <stack>
 14 using namespace std;
 15 typedef long long LL;
 16 typedef double DB;
 17 #define MIT (2147483647)
 18 #define MLL (1000000000000000001LL)
 19 #define INF (1000000001)
 20 #define For(i, s, t) for(int i = (s); i <= (t); i ++)
 21 #define Ford(i, s, t) for(int i = (s); i >= (t); i --)
 22 #define Rep(i, n) for(int i = (0); i < (n); i ++)
 23 #define Repn(i, n) for(int i = (n)-1; i >= (0); i --)
 24 #define mk make_pair
 25 #define ft first
 26 #define sd second
 27 #define puf push_front
 28 #define pub push_back
 29 #define pof pop_front
 30 #define pob pop_back
 31 #define sz(x) ((int) (x).size())
 32 inline void SetIO(string Name)
 33 {
 34     string Input = Name + ".in";
 35     string Output = Name + ".out";
 36     freopen(Input.c_str(), "r", stdin);
 37     freopen(Output.c_str(), "w", stdout);
 38 }
 39
 40 inline int Getint()
 41 {
 42     char ch = ‘ ‘;
 43     int Ret = 0;
 44     bool Flag = 0;
 45     while(!(ch >= ‘0‘ && ch <= ‘9‘))
 46     {
 47         if(ch == ‘-‘) Flag ^= 1;
 48         ch = getchar();
 49     }
 50     while(ch >= ‘0‘ && ch <= ‘9‘)
 51     {
 52         Ret = Ret * 10 + ch - ‘0‘;
 53         ch = getchar();
 54     }
 55     return Ret;
 56 }
 57
 58 const int N = 100010;
 59 int n, Arr[N];
 60 int Cnt[N], Len, Ans;
 61
 62 inline void Solve();
 63
 64 inline void Input()
 65 {
 66     int TestNumber = Getint();
 67     while(TestNumber--)
 68     {
 69         n = Getint();
 70         For(i, 1, n) Arr[i] = Getint();
 71         Solve();
 72     }
 73 }
 74
 75 inline int Find(int x)
 76 {
 77     int Left = 1, Right = Len, Mid, Ret = Len + 1;
 78     Cnt[Len + 1] = INF;
 79     while(Left <= Right)
 80     {
 81         Mid = (Left + Right) >> 1;
 82         if(Cnt[Mid] > x) Ret = Mid, Right = Mid - 1;
 83         else Left = Mid + 1;
 84     }
 85     return Ret;
 86 }
 87
 88 inline void Work()
 89 {
 90     Cnt[1] = Arr[1], Len = 1;
 91     For(i, 2, n)
 92     {
 93         int x = Find(Arr[i]);
 94         Cnt[x] = Arr[i];
 95         Len = max(Len, x);
 96     }
 97 }
 98
 99 inline void Solve()
100 {
101     Ans = 0;
102     Work();
103     Ans = max(Ans, Len);
104     For(i, 1, n / 2) swap(Arr[i], Arr[n - i + 1]);
105     Work();
106     Ans = max(Ans, Len);
107
108     if(Ans >= n-1) puts("YES");
109     else puts("NO");
110 }
111
112 int main()
113 {
114     Input();
115     //Solve();
116     return 0;
117 }