题干：
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.
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题目大意：
给你一个1~n的排列
解题报告：
首先贪心的思想，每个元素的下一个元素一定是与他距离不超过 k 的所有元素中，权值最大的元素，所以每个元素的下一个元素是固定的，我们可以通过滑动窗口 + set 上二分的预处理计算出每个元素的下一个元素，之后通过一个记忆化搜索即可 O(n) 求出每一个贪心序列的长度。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int a[MAX];
int n,q;
set<int> ss;
int db[MAX];//cun index
int ans[MAX];
int k;
int main()
{int T;cin>>T;while(T--) {scanf("%d%d",&n,&k);ss.clear();for(int i = 1; i<=n; i++) scanf("%d",a+i);int l=1-k,r=1+k;for(int i = 1; i<=r; i++) ss.insert(a[i]);for(int i = 1; i<=n; i++) { l = i-k,r = i+k;auto it = ss.lower_bound(a[i]);if(it != ss.begin()) {it--;db[a[i]] = (*it);}else db[a[i]] = 0; if(l >= 1) {auto it = ss.lower_bound(a[l]);ss.erase(it);}if(r+1<=n) {ss.insert(a[r+1]);}}for(int i = 1; i<=n; i++) ans[i] = ans[db[i]]+1;for(int i = 1; i<=n; i++) {printf("%d%c",ans[i],i == n ? '\n' :' ');}}return 0 ;
}
