AGC001F

link

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;
}