1 //后缀数组模板，MANX为数组的大小
2 //支持的操作有计算后缀数组(sa数组)， 计算相邻两元素的最长公共前缀(height数组)，使用get_height();
3 //计算两个后缀a, 和b的最长公共前缀，请先使用lcp_init(),再调用get_lcp(a, b)得到
4 //下面的n是输入字符串的长度+1(n = strlen(s) + 1)， m是模板的范围 m=128表示在字母，数字范围内，可以扩大也可缩小
5 //s[len] 是插入的一个比输入字符都要小的字符
6 struct SufArray {
7 char s[MAXN];
8 int sa[MAXN], t[MAXN], t2[MAXN], c[MAXN], n, m;
9 int rnk[MAXN], height[MAXN];//rnk和height数组
10 int mi[MAXN][20], idxK[MAXN];//用于计算LCP
11
12 void init() {
13 mem0(s);
14 mem0(height);
15 }
16 //读入字符串作为输入
17 void read_str() {
18 gets(s);
19 m = 128;
20 n = strlen(s);
21 s[n++] = ' ';
22 }
23 void build_sa() {
24 int *x = t, *y = t2;
25 rep (i, 0, m - 1) c[i] = 0;
26 rep (i, 0, n - 1) c[x[i] = s[i]] ++;
27 rep (i, 1, m - 1) c[i] += c[i - 1];
28 dec (i, n - 1, 0) sa[--c[x[i]]] = i;
29 for(int k = 1; k <= n; k <<= 1) {
30 int p = 0;
31 rep (i, n - k, n - 1) y[p++] = i;
32 rep (i, 0, n - 1) if(sa[i] >= k) y[p++] = sa[i] - k;
33 rep (i, 0, m - 1) c[i] = 0;
34 rep (i, 0, n - 1) c[x[y[i]]] ++;
35 rep (i, 0, m - 1) c[i] += c[i - 1];
36 dec (i, n - 1, 0) sa[--c[x[y[i]]]] = y[i];
37 swap(x, y);
38 p = 1;
39 x[sa[0]] = 0;
40 rep (i, 1, n - 1) {
41 x[sa[i]] = y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k] ? p - 1 : p++;
42 }
43 if(p >= n) break;
44 m = p;
45 }
46 }
47 void get_height() {
48 int k = 0;
49 rep (i, 0, n - 1) rnk[sa[i]] = i;
50 rep (i, 0, n - 1) {
51 if(k) k --;
52 int j = sa[rnk[i] - 1];
53 while(s[i + k] == s[j + k]) k ++;
54 height[rnk[i]] = k;
55 }
56 }
57 void rmq_init(int *a, int n) {
58 rep (i, 0, n - 1) mi[i][0] = a[i];
59 for(int j = 1; (1 << j) <= n; j ++) {
60 for(int i = 0; i + (1<<j) - 1 < n; i ++) {
61 mi[i][j] = min(mi[i][j - 1], mi[i + (1 << (j - 1))][j - 1]);
62 }
63 }
64 rep (len, 1, n) {
65 idxK[len] = 0;
66 while((1 << (idxK[len] + 1)) <= len) idxK[len] ++;
67 }
68 }
69 int rmq_min(int l, int r) {
70 int len = r - l + 1, k = idxK[len];
71 return min(mi[l][k], mi[r - (1 << k) + 1][k]);
72 }
73 void lcp_init() {
74 get_height();
75 rmq_init(height, n);
76 }
77 int get_lcp(int a, int b) {
78 if(a == b) return n - a - 1;
79 return rmq_min(min(rnk[a], rnk[b]) + 1, max(rnk[a], rnk[b]));
80 }
81 void solve() {
82 }
83 };
