后缀数组（笔记）

dalao的博客，讲的很详细

倍增法：设当前已知各后缀的前 (2^k) 个字符的相对大小关系（即排名数组 (rank)）， 于是只要用 ((rank[i], rank[i + 2 ^k])) 进行双关键字排序就可以得到前 (2^{k+1}) 个字符的相对大小关系

跟没有也差不多的upd：

注意 (tp[i]) 的意义：按 (rank[i + 2 ^ k]) 为关键字排序后的第 (i) 个后缀，然后由于基数排序的稳定性，再次按照 (rank[i]) 排序时相当于进行了双关键字排序

代码调了一个小时才知道原来还有指针引用这种玩意

#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){//丧病的复合类型
    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;
}