计蒜客 The Preliminary Contest for ICPC Asia Nanjing 2019

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:

s_i[1]=isi[1]=i;
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];
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.
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;
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.

题解：

输入 T组样例（T<=20）给定 n k，序列a是 1-n 乱序排列的一组数。

求有n个 s序列 s_i[0]= i . 从 a中选择数字， s序列是降序排列，满足最大字典序，且s中相邻的两个数在a中的下标绝对值的差小于k ∣pos[j]−pos[j−1]∣≤k (1≤ k ≤10^5)

输出n个s序列中非0的个数。

从 s₁={1,0,0,,,0} 答案为 ans=1.

s₂在s₁ 的基础上增加了 2 判断新加入的2是否满足k ，即ans[2] =ans[1]+1. 从i 到 1 满足的则加上，否则不加。

ans[i]+=ans[j]; 每个s序列的ans[i] 需要从 1计算到 i 由 j 控制。

#include <cstdio>

#include <algorithm>

#include <iostream>

#include <map>

#include <string>

#include <cstring>

#include <queue>

#include <stack>

#include <cmath>

#define int long long

#define Mod 1000000007

#define pi (acos(-1))

#define inf 0x3f3f3f3f3f

#define Maxn 100005

using namespace std;

int a[Maxn];

int pos[Maxn];

int ans[Maxn];

signed main(){

    int t;

    scanf("%lld",&t);

    while(t--){

        int n,k;

        scanf("%lld%lld",&n,&k);

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

        {

            scanf("%lld",&a[i]);

            pos[a[i]]=i;

        }

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

//        printf("%lld ",pos[i]);

//        ans[1]=1;

//        if(pos[2]-pos[1]<=k&&pos[2]-pos[1]>=-k)

//        ans[2]+=a[1];

//        printf("a2=%lld\n",a[2]);

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

        {

            ans[i]=;

            for(int j = i- ; j >=  ; j -- )

            {

                if(pos[i]-pos[j]>=-k&&pos[i]-pos[j]<=k)

                {

                    ans[i]+=ans[j];

//                    printf("%lld%lld%lld\n",i,i,i);

                    break;

                }

            }

        }

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

        printf("%lld ",ans[i]);

        printf("%lld\n",ans[n]);

    }

    return ;

}