Lazy wording on Cardinal sort in SA

Because I have not learned the standard SA writing, has been written in accordance with their own blind yy n*log^2 to do

But it's been stuck a few times.

The practice is to record rk[] and rk2[] as the first and second keywords for sort//to do so the first keyword ordered cardinal sort constant huge

Luogu Template n*log^2 70 min

Change to cardinality (great constant edition) sort 9s AC

So I yy out an unknown complexity of the algorithm. Complexity is probably n*log*log (N/log)

Measured 5s AC

Algorithm record Sa[i] indicates where the position of rank i is

There is no difference between an SA sequence and an ordinary base row.

Next we consider

Each rk[sa[i]] is ordered, and Rk2[sa[i]] is unordered

So according to the RK segment, each time only the RK equal to a section of the order, so that the length of each order can be considered n/log.

#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <ctime > #include <cstdlib> #include <algorithm> #include <queue> #define for (i,j,k) for (int i=j;i<=k;+
+i) #define DOW (I,J,K) for (int i=k;i>=j;--i) #define LL Long long #define INF 1e9 using namespace std;
	inline int read () {int T=0,f=1;char c=getchar ();
	while (!isdigit (c)) {if (c== '-') F=-1;c=getchar ();}
	while (IsDigit (c)) t=t*10+c-' 0 ', C=getchar ();
return t*f;
} inline void write (int x) {if (x>=10) write (X/10);p Utchar (x%10+ ' 0 ');} inline void Writeln (int x) {write (x);p UTS ("");
inline void write_p (int x) {write (x);p Utchar (");} int sa[2000001],n,rk[2000001],rk2[2000001],tmp[2000001];
Char s[2000001]; inline BOOL cmp (int x,int y) {return rk[x]==rk[y]?rk2[x]<rk2[y]:rk[x]<rk[y];} inline void Resort (int p[]) {i
    NT Las=1;
            for (I,2,n) if (rk[p[i]]!=rk[p[i-1])//processing las~i-1 {sort (p+las,p+i,cmp);
        Las=i;} sort (p+las,p+n+1,cmp);
	} int main () {scanf ("%s", s+1); N=strlen (s+1);
	for (i,1,n) rk[i]=s[i],sa[i]=i;
	Sort (sa+1,sa+n+1,cmp);
		for (int. len=1;len<=n;len*=2) {for (i,1,n) rk2[i]=i+len<=n?rk[i+len]:0;
		Resort (SA);
		int tot=1;
		Tmp[sa[1]]=1;
		for (I,2,n) if (Rk[sa[i]]==rk[sa[i-1]]&&rk2[sa[i]]==rk2[sa[i-1]]) Tmp[sa[i]]=tot;else Tmp[sa[i]]=++tot;
	Swap (RK,TMP);
} for (I,1,n) write_p (Sa[i]); }