2019 ICPC 南京网络赛 F Greedy Sequence

You're given a permutation aa of length nn (1 \le n \le 10^51≤n≤105).

For each i \in [1,n]i∈[1,n], construct a sequence s_isi​ by the following rules:

1. s_i[1]=isi​[1]=i;
2. The length of s_isi​ is nn, and for each j \in [2, n]j∈[2,n], s_i[j] \le s_i[j-1]si​[j]≤si​[j−1];
3. First, we must choose all the possible elements of s_isi​ from permutation aa. If the index of s_i[j]si​[j] in permutation aa is pos[j]pos[j], for each j \ge 2j≥2, |pos[j]-pos[j-1]|\le k∣pos[j]−pos[j−1]∣≤k (1 \le k \le 10^51≤k≤105). And for each s_isi​, every element of s_isi​ must occur in aa at most once.
4. After we choose all possible elements for s_isi​, if the length of s_isi​ is smaller than nn, the value of every undetermined element of s_isi​ is 00;
5. For each s_isi​, we must make its weight high enough.

Consider two sequences C = [c_1, c_2, ... c_n]C=[c1​,c2​,...cn​] and D=[d_1, d_2, ..., d_n]D=[d1​,d2​,...,dn​], we say the weight of CC is higher thanthat of DD if and only if there exists an integer kk such that 1 \le k \le n1≤k≤n, c_i=d_ici​=di​ for all 1 \le i < k1≤i<k, and c_k > d_kck​>dk​.

If for each i \in [1,n]i∈[1,n], c_i=d_ici​=di​, the weight of CC is equal to the weight of DD.

For each i \in [1,n]i∈[1,n], print the number of non-zero elements of s_isi​ separated by a space.

It's guaranteed that there is only one possible answer.

Input

There are multiple test cases.

The first line contains one integer T(1 \le T \le 20)T(1≤T≤20), denoting the number of test cases.

Each test case contains two lines, the first line contains two integers nn and kk (1 \le n,k \le 10^51≤n,k≤105), the second line contains nn distinct integers a_1, a_2, ..., a_na1​,a2​,...,an​ (1 \le a_i \le n1≤ai​≤n) separated by a space, which is the permutation aa.

Output

For each test case, print one line consists of nn integers |s_1|, |s_2|, ..., |s_n|∣s1​∣,∣s2​∣,...,∣sn​∣ separated by a space.

|s_i|∣si​∣ is the number of non-zero elements of sequence s_isi​.

There is no space at the end of the line.

样例输入复制

2
3 1
3 2 1
7 2
3 1 4 6 2 5 7

样例输出复制

1 2 3
1 1 2 3 2 3 3

这题是队友出的，这个题好像线段树能做，但是我们是暴力加贪心做的，前几遍是枚举的区间位置纯暴力，后来是枚举左右区间长度，对于每个元素的下一个元素都是唯一固定的。之后记忆化搜索就可以求出每个序列。

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<math.h>
#include<vector>
#include<queue>
#include<map>
using namespace std;
#define LL long long
#define MAXN 100100
int a[MAXN],pos[MAXN];
int s[MAXN],oo[MAXN];
int main()
{
    int t;
    scanf("%d",&t);
    int n,k,l,r,maxx;
    while(t--){
        scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%d",a+i);

        pos[a[i]]=i;
        s[i]=1;
    }


    s[0]=0;
    for(int i=2;i<=n;i++){
        int j=i-1;
        while(j>0){
            if(abs(pos[i]-pos[j])<=k){
                s[i]+=s[j];
                break;
            }
            j--;
        }
    }
    for(int i=1;i<=n;i++){printf("%d",s[i]);if(i<n)printf(" ");}
    
    printf("\n");
    }
}