Sequence in the Pocket ZOJ - 4104 （待解决

https://vjudge.net/problem/ZOJ-4104/origin

https://vjudge.net/contest/395635#problem/E

DreamGrid has just found an integer sequence a_1, a_2, \dots, a_na1,a2,…,an in his right pocket. As DreamGrid is bored, he decides to play with the sequence. He can perform the following operation any number of times (including zero time): select an element and move it to the beginning of the sequence.

What's the minimum number of operations needed to make the sequence non-decreasing?

Input

There are multiple test cases. The first line of the input contains an integer TT, indicating the number of test cases. For each test case:

The first line contains an integer nn (1 \le n \le 10^51≤n≤105), indicating the length of the sequence.

The second line contains nn integers a_1, a_2, \dots, a_na1,a2,…,an (1 \le a_i \le 10^91≤ai≤109), indicating the given sequence.

It's guaranteed that the sum of nn of all test cases will not exceed 10^6106.

Output

For each test case output one line containing one integer, indicating the answer.

Sample Input

Sample Output

2

0

Hint

For the first sample test case, move the 3rd element to the front (so the sequence become {2, 1, 3, 4}), then move the 2nd element to the front (so the sequence become {1, 2, 3, 4}). Now the sequence is non-decreasing.

For the second sample test case, as the sequence is already sorted, no operation is needed.

附未通过代码方便日后补题：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <bitset>

#include <cassert>

#include <cctype>

#include <cmath>

#include <cstdlib>

#include <ctime>

#include <deque>

#include <iomanip>

#include <list>

#include <map>

#include <queue>

#include <set>

#include <stack>

#include <vector>

#include <iterator>

#include <utility>

#include <sstream>

#include <limits>

#include <numeric>

#include <functional>

using namespace std;

#define gc getchar()

#define mem(a) memset(a,0,sizeof(a))

#define debug(x) cout<<"debug:"<<#x<<" = "<<x<<endl;

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef long long ll;

typedef unsigned long long ull;

typedef long double ld;

typedef pair<int,int> pii;

typedef char ch;

typedef double db;

const double PI=acos(-1.0);

const double eps=1e-6;

const int inf=0x3f3f3f3f;

const int maxn=1e5+10;

const int maxm=100+10;

const int N=1e6+10;

const int mod=1e9+7;

ll a[100005] = {0};

int main()

{

    int t = 0;

    cin >> t;

    while(t--)

    {

        int n = 0;

        cin >> n;

        int counter = 0;

        int flag = 0;

        for(int i = 0;i<n;i++)

        {

            cin >> a[i];

        }

        if(a[1]>a[0] && a[1]>a[2])

        {

            flag = 1;

            //cout <<1<< " !A\n";

        }

        for(int i = 2;i<n-2;i++)

        {

            if(a[i]>a[i-1] && a[i]>a[i+1])

            {

                if(flag < 2)

                {

                    if(a[i] == a[i-2])

                    {

                        counter -= 1;

                        if(flag == 1)counter += 1;

                        a[i-1] = a[i];

                        flag = 2;

                        //cout <<i<< " !1\n";

                        continue;

                    }

                    if(a[i] == a[i+2])

                    {

                        if(flag == 1)counter += 1;

                        a[i+1] = a[i];

                        flag = 2;

                        //cout <<i<< " !2\n";

                        continue;

                    }

                }

                if(flag < 1)

                {

                    if(a[i-1]<a[i-2] && a[i-1]<a[i+1] && a[i+1]<a[i+2])

                    {

                        counter -= 1;

                        a[i] = a[i-1];

                        flag = 1;

                        //cout <<i<< " !3\n";

                        continue;

                    }

                    if(a[i+1]<a[i+2] && a[i+1]<a[i-1] && a[i-1]<a[i-2])

                    {

                        counter -= 1;

                        a[i] = a[i+1];

                        flag = 1;

                        //cout <<i<< " !4\n";

                        continue;

                    }

                    if(a[i-1] == a[i-2] || a[i-1] == a[i+1] || a[i+1] == a[i+2])

                    {

                        a[i] = a[i-1];

                        flag = 1;

                        //cout <<i<< " !5\n";

                        continue;

                    }

                }

                counter += 1;

                //cout <<i<< " !++\n";

            }

        }

        if(a[n-2]>a[n-1] && a[n-2]>a[n-3] && n!=3)

        {

            counter += 1;

            //cout <<n-2<< " !B\n";

        }

        cout << counter << endl;

    }

	return 0;

}