dalao的博客，講的很詳細

倍增法：設當前已知各後綴的前 \(2^k\) 個字符的相對大小關系（即排名數組 \(rank\)）， 於是只要用 \((rank[i], rank[i + 2 ^k])\) 進行雙關鍵字排序就可以得到前 \(2^{k+1}\) 個字符的相對大小關系

代碼調了一個小時才知道原來還有指針引用這種玩意

#include<cstdio>
#include<cstring>
using namespace std;

const int MAXN = 1000010;

int t1[MAXN], t2[MAXN], sa[MAXN], tax[MAXN];
char str[MAXN];

inline void Swap(int* &a, int* &b){//喪病的復合類型
    int *t = a; a = b, b = t;
}

void getSA(char s[], int len){
    int p = 0, crd;
    int *rak = t1, *tp = t2;
    crd = 122; 
    for (int i = 1; i <= len; ++i) rak[i] = s[i], tp[i] = i;
    for (int i = 0; i <= crd; ++i) tax[i] = 0;
    for (int i = 1; i <= len; ++i) ++tax[rak[i]];
    for (int i = 1; i <= crd; ++i) tax[i] += tax[i - 1];
    for (int i = len; i >= 1; --i) sa[tax[rak[tp[i]]]--] = tp[i];
    for (int w = 1; p != len; w <<= 1, crd = p){
        p = 0;
        for (int i = len - w + 1; i <= len; ++i)
            tp[++p] = i;
        for (int i = 1; i <= len; ++i)
            if (sa[i] > w)
                tp[++p] = sa[i] - w;
        for (int i = 0; i <= crd; ++i) tax[i] = 0;
        for (int i = 1; i <= len; ++i) ++tax[rak[i]];
        for (int i = 1; i <= crd; ++i) tax[i] += tax[i - 1];
        for (int i = len; i >= 1; --i) sa[tax[rak[tp[i]]]--] = tp[i];
        Swap(rak, tp);
        rak[sa[1]] = p = 1;
        for (int i = 2; i <= len; ++i)
            rak[sa[i]] = (tp[sa[i]] == tp[sa[i - 1]] && tp[sa[i] + w] == tp[sa[i - 1] + w]) ? p : ++p;
    }
}

int main(){
    scanf("%s", str + 1);
    int slen = strlen(str + 1);
    getSA(str, slen);
    for (int i = 1; i <= slen; ++i)
        printf("%d ", sa[i]);
    return 0;
}
 

後綴數組（筆記）