AGC001F

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令\(a_i\)为\(p\)中\(i\)的出现位置\(j\)，也就是\(a_{p_j}=j\)，那么原操作就可以转化为若\(|a_i-a_{i+1}|\ge K\)可以交换\(a_i,a_{i+1}\)。

原问题（\(p\)字典序最小）就可以转化为\(a\)中1位置尽量小，同时2位置尽量小\(\dots\)

实际上这就相当于\(a\)翻转后字典序最大。

如何理解呢？大概的感性理解下：

令\(a_n\)尽量大，那么\(<a_n\)的数出现位置都会往前。

或者说要使得一个序列字典序尽量大，首先最小的必须尽量靠后，然后是第二小的....

由于\(|a_i-a_{i+1}|\ge K\)才可以交换，那么若\(|a_i-a_j|\lt K\)，它们相对顺序就不会变了 。

于是可以连边然后拓扑排序。

但是\(O(n^2)\)的边数有点大。

事实上，\(i\)只需要向最大满足\(j\lt i,0\lt a_j-a_i\lt k\)的\(j\)和最大满足\(j\lt i,0\lt a_i-a_j\lt k\)

的\(j\)连边就好了。

其它满足\(j\lt i,|a_j-a_j|\lt k\)的\(j\)都会被这两个\(j\)中至少一个直接或间接的连边，那么\(i\)也间接向\(j\)连了边。

时间复杂度\(O(n\log n)\)。

/*
Author: CNYALI_LK
LANG: C++
PROG: f.cpp
Mail: cnyalilk@vip.qq.com
*/
#include<bits/stdc++.h>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const signed inf=0x3f3f3f3f;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
	const int SIZE = (1 << 21) + 1;
	char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
	// getchar
	#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
	// print the remaining part
	inline void flush () {
		fwrite (obuf, 1, oS - obuf, stdout);
		oS = obuf;
	}
	// putchar
	inline void putc (char x) {
		*oS ++ = x;
		if (oS == oT) flush ();
	}
	// input a signed integer
	inline void read (signed &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
	}

	inline void read (long long &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
	}
	inline void read (char &x) {
		x=gc();
	}
	inline void read(char *x){
		while((*x=gc())=='\n' || *x==' '||*x=='\r');
		while(!(*x=='\n'||*x==' '||*x=='\r'))*(++x)=gc();
	}
	template<typename A,typename ...B>
	inline void read(A &x,B &...y){
		read(x);read(y...);
	}
	// print a signed integer
	inline void write (signed x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
	}

	inline void write (long long x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
	}
	inline void write (char x) {
		putc(x);
	}
	inline void write(const char *x){
		while(*x){putc(*x);++x;}
	}
	inline void write(char *x){
		while(*x){putc(*x);++x;}
	}
	template<typename A,typename ...B>
	inline void write(A x,B ...y){
		write(x);write(y...);
	}
	//no need to call flush at the end manually!
	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
int a[500005],p[500005];
struct smt{
	int ls,rs,mx;
	smt *l,*r;	
	smt(int la,int ra){
		ls=la;rs=ra;	
		mx=0;
		if(la==ra){
			l=r=0;
		}
		else{
			int mid=(ls+rs)>>1;
			l=new smt(ls,mid);
			r=new smt(mid+1,rs);
		}
	}
	int query(int la,int ra){
		if(la<=ls && rs<=ra)return mx;
		int mx=0;	
		if(la<=l->rs)chkmax(mx,l->query(la,ra));
		if(r->ls<=ra)chkmax(mx,r->query(la,ra));
		return mx;
	}
	void upd(int x,int y){
		mx=y;
		if(ls==rs)return;
		if(x<=l->rs)l->upd(x,y);
		else r->upd(x,y);
	}
};
smt *rt;
int ind[500005],c[500005];
vector<int> to[500005];
priority_queue<pii> pq;
int main(){
#ifdef cnyali_lk
	freopen("f.in","r",stdin);
	freopen("f.out","w",stdout);
#endif
	int n,k,x;	
	read(n,k);
	for(int i=1;i<=n;++i){
		read(a[i]);
		p[a[i]]=i;
	}
	rt=new smt(1,n);
	for(int i=1;i<=n;++i){
		to[i].push_back(x=rt->query(p[i]-k+1,p[i]));++ind[x];
		to[i].push_back(x=rt->query(p[i],p[i]+k-1));++ind[x];
		rt->upd(p[i],i);	
	}
	for(int i=1;i<=n;++i)if(!ind[i])pq.push(make_pair(p[i],i));
	int t=n+1;
	while(!pq.empty()){
		c[--t]=pq.top().x;
		int x=pq.top().y;	
		pq.pop();
		for(auto i:to[x]){
			if(!--ind[i]){
				pq.push(make_pair(p[i],i));	
			}
		}
	}
	for(int i=1;i<=n;++i)a[i]=i;
	sort(a+1,a+n+1,cmp_a<c>);
	for(int i=1;i<=n;++i)printf("%d\n",a[i]);
	return 0;
}