【LOJ #129】【模板】 Lyndon 分解

传送门

由于$L$串是不增的
可以发现一定连续一段$L$串相同
于是考虑先分割成单减的，然后每一部分按循环节分割
考虑假设已经求出了$1...i-1$的$lyndon$分解且恰好分解完
考虑从$i$开始，设$j,k$两指针，初始$j=i,k=i+1$
$k$指向循环节$TTT{T}^{\prime }$的结尾，$j=k-|T|$
若$s[j]=s[k],j++,k++$即可
若$s[j]>s[k]$,那么$k$之后单独一定更优，而前面是$k-j$长度的循环节
若$s[j]<s[k],j=i,k++$重新形成了循环节

复杂度$O(n)$

#include
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair
#define ll long long
#define y1 shinkle
#define fi first
#define se second
#define bg begin
cs int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ib==ob)?EOF:*ib++;
}
inline int read(){
    char ch=gc();
    int res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
inline ll readll(){
    char ch=gc();
    ll res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
inline int readstring(char *s){
	int top=0;char ch=gc();
	while(isspace(ch))ch=gc();
	while(!isspace(ch)&&ch!=EOF)s[++top]=ch,ch=gc();
	s[top+1]='\0';return top;
}
template<typename tp>inline void chemx(tp &a,tp b){a=max(a,b);}
template<typename tp>inline void chemn(tp &a,tp b){a=min(a,b);}
cs int N=(1<<20)|1;
char s[N];
int n;
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	n=readstring(s);
	for(int i=1,j,k;i<=n;){
		j=i,k=i+1;
		while(s[j]<=s[k]){
			if(s[j]==s[k])j++,k++;
			else j=i,k++;
		}
		while(i<=j){
			cout<<i+k-j-1<<" ";
			i+=k-j;
		}
	}
}