[开发测试]B. Moamen and k-subarrays

B. Moamen and k-subarrays

B.莫阿门和k-子阵

time limit per test: 2 seconds
每次测试的时间限制：2秒

memory limit per test: 256 megabytes
每次测试的内存限制：256兆字节

input: standard input
输入：标准输入

output: standard output
产出：标准产出

Moamen has an array of n distinct integers. He wants to sort that array in non-decreasing order by doing the fllwing operations in order
莫阿门有一个n个不同整数的数组。他希望通过按顺序执行fllwing操作，按非递减顺序对数组进行排序。

exactly once:
准确地说是一次：

1. Split the array into exactly k non-empty subarrays such that each element belongs to exactly one subarray.
   1.将数组拆分为完全k个非空子数组，使每个元素恰好属于一个子数组。

2. Reorder these subarrays arbitrary.
   2.任意重新排序这些子数组。

3. Merge the subarrays in their new order.
   3.将子数组按新顺序合并。

A sequence a is a subarray of a sequence b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the
序列a是序列b的子数组，如果可以从b中删除几个(可能是零或全部)元素来获得a。

beginning and several (possibly, zero or all) elements from the end.
开始和从末尾开始的几个(可能是零或全部)元素。

Can you tell Moamen if there is a way to sort the array in non-decreasing order using the operations written above?
你能告诉莫阿门，如果有一种方法来排序数组的非递减顺序使用上面写的操作？

Input
输入

The first line contains a single integert(1 < t < 103)- the number of test cases. The description of the test cases fllws.
第一行包含一个整数(1<t<103)–测试用例的数量。测试用例的描述。

The first line of each test case contains two integers nandk(1≤k≤n≤105).
每个测试用例的第一行包含两个整数nandk(1≤k≤n≤105)。

The second line contains n integers a1, a…,an(0≤|a|≤10"). It is guaranteed that all numbers are distinct.
第二行包含n个整数A1，a.，a，≤，≤10“，保证所有数字都是不同的。

It is guaranteed that the sum of n over all test cases does not exceed 3.105.
保证所有测试用例上n的和不超过3.105。

模拟
看有多少组符合顺序即可

int a[100007],b[100007];
map<ll ,ll>book;
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
    ll n,m;
     ll sum=0;
      cin>>n>>m;
        for (int i = 0; i < n; ++i) {
            cin>>a[i];
          b[i]=a[i];
        }
        sort(b,b+n);
        for (int i = 0; i < n-1; ++i) {
            book[b[i]]=b[i+1];
        }
        book[b[n-1]]=-1e10;
        for (int i = 0; i < n; ++i) {
            sum++;
            while(a[i+1]==book[a[i]])
            {
                ++i;
            }
        }
        //cout<<sum<<endl;///
        if(sum<=m)
            cout<<"Yes"<<endl;
        else
            cout<<"No\n";
        }
 
    return 0;
}