Greedy Sequence（主席树-区间小于每个数的最大值）

You're given a permutation aaa of length nnn (1≤n≤1051 le n le 10^51≤n≤105).

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

1. si[1]=is_i[1]=isi​[1]=i;
2. The length of sis_isi​ is nnn, and for each j∈[2,n]j in [2, n]j∈[2,n], si[j]≤si[j−1]s_i[j] le s_i[j-1]si​[j]≤si​[j−1];
3. First, we must choose all the possible elements of sis_isi​ from permutation aaa. If the index of si[j]s_i[j]si​[j] in permutation aaa is pos[j]pos[j]pos[j], for each j≥2j ge 2j≥2, ∣pos[j]−pos[j−1]∣≤k|pos[j]-pos[j-1]|le k∣pos[j]−pos[j−1]∣≤k (1≤k≤1051 le k le 10^51≤k≤105). And for each sis_isi​, every element of sis_isi​ must occur in aaa at most once.
4. After we choose all possible elements for sis_isi​, if the length of sis_isi​ is smaller than nnn, the value of every undetermined element of sis_isi​ is 000;
5. For each sis_isi​, we must make its weight high enough.

Consider two sequences C=[c1,c2,...cn]C = [c_1, c_2, ... c_n]C=[c1​,c2​,...cn​] and D=[d1,d2,...,dn]D=[d_1, d_2, ..., d_n]D=[d1​,d2​,...,dn​], we say the weight of CCC is higher than that of DDD if and only if there exists an integer kkk such that 1≤k≤n1 le k le n1≤k≤n, ci=dic_i=d_ici​=di​ for all 1≤i<k1 le i < k1≤i<k, and ck>dkc_k > d_kck​>dk​.

If for each i∈[1,n]i in [1,n]i∈[1,n], ci=dic_i=d_ici​=di​, the weight of CCC is equal to the weight of DDD.

For each i∈[1,n]i in [1,n]i∈[1,n], print the number of non-zero elements of sis_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≤T≤20)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 nnn and kkk (1≤n,k≤1051 le n,k le 10^51≤n,k≤105), the second line contains nnn distinct integers a1,a2,...,ana_1, a_2, ..., a_na1​,a2​,...,an​ (1≤ai≤n1 le a_i le n1≤ai​≤n) separated by a space, which is the permutation aaa.

Output

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

∣si∣|s_i|∣si​∣ is the number of non-zero elements of sequence sis_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
解题思路: 题目意思找[x-k,x+k]区间里小于其的最大值,然后在用那个数去更新 位置一个单调递减的数列！

#include <bits/stdc++.h>
using namespace std;
int tot=0,t,n,k;
const int maxn=1e5+5;
struct Node{
    int l,r,sum;
}A[maxn*32];
int arr[maxn],root[maxn];
int ma[maxn],res[maxn];

void updata(int left,int right,int &now,int pre,int value){
    A[++tot]=A[pre],now=tot,A[now].sum++;
    if(left==right) return;
    int mid=left+right>>1;
    if(mid>=value) updata(left,mid,A[now].l,A[pre].l,value);
    else updata(mid+1,right,A[now].r,A[pre].r,value);
}

int query(int left,int right,int x,int y,int value){   //返回的小于某个数的最大值
    if(A[y].sum-A[x].sum==0) return -1;   //当前区间的个数为0则减枝;
    if(left==right){      //递归到叶子节点
        return left<value?left:-1;
    }
    int mid=left+right>>1;
    if(value<=mid+1||A[A[y].r].sum-A[A[x].r].sum==0){  //查找的数在左边或者右边没有shu
        return query(left,mid,A[x].l,A[y].l,value);
    }
    int t=query(mid+1,right,A[x].r,A[y].r,value);  //先查找右边
    if(t==-1) query(left,mid,A[x].l,A[y].l,value);
    else return t;

}

int main(){
    ios::sync_with_stdio(false);
    cin>>t;
    while(t--){
        memset(root,0,sizeof(root)),tot=0,memset(A,0,sizeof(A));
        cin>>n>>k;
        for(int i=1;i<=n;i++) cin>>arr[i],ma[arr[i]]=i;
        for(int i=1;i<=n;i++) updata(1,n,root[i],root[i-1],arr[i]);
        res[1]=1;
        for(int i=2;i<=n;i++){
            int L=max(1,ma[i]-k);
            int R=min(n,ma[i]+k);
            int flag=query(1,n,root[L-1],root[R],i);
            if(flag==-1) res[i]=1;
            else res[i]=res[flag]+1;   //递推的公式
        }
        for(int i=1;i<=n;i++){
            printf("%d%c",res[i],i==n?'
':' ');
        }
    }
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
int t,n,m;
const int maxn=1e5+5;
int sum[maxn<<2];
int ma[maxn],res[maxn],arr[maxn];
void updata(int rt) { sum[rt]=max(sum[rt<<1],sum[rt<<1|1]); }  //区间存的是最大值
void build(int l,int r,int rt){
    sum[rt]=0;
    if(l==r) {return;}
    int mid=l+r>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    updata(rt);
}
void add(int l,int r,int rt,int pos,int value){
    if(l==r) {sum[rt]=value;return;}
    int mid=l+r>>1;
    if(mid>=pos) add(l,mid,rt<<1,pos,value);
    else add(mid+1,r,rt<<1|1,pos,value);
    updata(rt);
}
int query(int l,int r,int rt,int L,int R,int value){
    if(L<=l&&R>=r) return sum[rt];
    int mid=l+r>>1;
    int ans=0;
    if(mid>=L) ans=max(ans,query(l,mid,rt<<1,L,R,value));
    if(R>mid) ans=max(ans,query(mid+1,r,rt<<1|1,L,R,value));
    return ans;
}

int main(){
    ios::sync_with_stdio(false);
    cin>>t;
    while(t--){
        cin>>n>>m;
        for(int i=1;i<=n;i++){
            cin>>arr[i];
            ma[arr[i]]=i;
        }
        build(1,n,1);
        for(int i=1;i<=n;i++){
            int L=max(1,ma[i]-m);
            int R=min(n,ma[i]+m);
            int x=query(1,n,1,L,R,i);   //小于这个数的最大值
            add(1,n,1,ma[i],i);   //后添加必定保证是小于其的数
            res[i]=res[x]+1;
        }
        for(int i=1;i<=n;i++)
            printf("%d%c",res[i],i==n?'
':' ');
    }
    return 0;
}