后缀树组(SA)初探

还没有什么任意两个后缀的LCP这些玩意儿。

启蒙题：输入一个串S，求最长的串T使得T在S中出现过不止一次。输出T的长度。

 1 #include <algorithm>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #define N 100
 5 char s[N+1];
 6 int n, sa[N], rank[N], height[N];
 7 namespace SABuilder {
 8     int cnt[N], t1[N<<1], t2[N<<1];
 9     void work() {
10         int i, k, m = 26, p, *x = t1, *y = t2;
11         memset(cnt, 0, m * sizeof(int));
12         for(i = 0; i < n; ++i) ++cnt[x[i] = s[i] - ‘a‘];
13         for(i = 1; i < m; ++i) cnt[i] += cnt[i-1];
14         for(i = n - 1; i >= 0; --i) sa[--cnt[x[i]]] = i;
15         for(k = 1; k <= n; k <<= 1) {
16             for(i = n-k, p = 0; i < n; ++i) y[p++] = i;
17             for(i = 0; i < n; ++i) if(sa[i] >= k) y[p++] = sa[i] - k;
18             memset(cnt, 0, m * sizeof(int));
19             for(i = 0; i < n; ++i) ++cnt[x[y[i]]];
20             for(i = 1; i < m; ++i) cnt[i] += cnt[i-1];
21             for(i = n - 1; i >= 0; --i) sa[--cnt[x[y[i]]]] = y[i];
22             std::swap(x, y);
23             x[sa[0]] = 0;
24             for(i = m = 1; i < n; ++i) {
25                 x[sa[i]] = y[sa[i]]==y[sa[i-1]]&&sa[i]+k<n&&sa[i-1]+k<n&&y[sa[i]+k]==y[sa[i-1]+k] ? m-1 : m++;
26             }
27             if(m > n) break;
28         }
29         memcpy(rank, x, n * sizeof(int));
30         for(i = 0, p = 0; i < n; ++i) if(rank[i]) {
31             if(p) --p;
32             k = sa[rank[i]-1];
33             while(s[i+p] == s[k+p]) ++p;
34             height[rank[i]] = p;
35         }
36     }
37 }
38 int main() {
39     register int i, ans = 0;
40     gets(s);
41     n = strlen(s);
42     SABuilder::work();
43     printf("%d", ans);
44     return 0;
45 }