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 si [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 = 1 ; 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 = 1 ; i <= n ; i ++ )
{
ans[i]=1;
for(int j = i-1 ; j >= 1 ; 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 = 1 ; i < n ; i ++ )
printf("%lld ",ans[i]);
printf("%lld\n",ans[n]);
}
return 0;
}