SPOJ #687 Repeats

题目描述：

给定字符串，求字符串中最大的连续重复子串出现次数是多少。

解题思路：

如果一个重复子串的长度是 l，那么它一定跨过s[0]、s[l]、s[l * 2]、s[l * ...] 中连续的两个，那我们就可以枚举 l，枚举起始位置 i * l，用 SA 求出s[i * l]、s[i * l + l]的lcp，那么 lcp / l + 1就是答案。对吗？不一定，有可能s[i * l]前面有段可以匹配，那我们就将位置前移多匹配，但不足l的多出来的那段。再求次lcp计算。

代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 using namespace std;
 5
 6 const int N = 5e4 + 10;
 7 int T, n, m, st[N][20], ans;
 8 int sa[N], height[N], rank[N], tax[N], tp[N];
 9 char s[N];
10
11 void rsort() {
12     for (int i = 0; i <= m; i ++) tax[i] = 0;
13     for (int i = 1; i <= n; i ++) tax[rank[tp[i]]] ++;
14     for (int i = 1; i <= m; i ++) tax[i] += tax[i - 1];
15     for (int i = n; i >= 1; i --) sa[tax[rank[tp[i]]] --] = tp[i];
16 }
17
18 int cmp(int *f, int x, int y, int w) {return f[x] == f[y] && f[x + w] == f[y + w];}
19
20 void SA() {
21     for (int i = 1; i <= n; i ++) rank[i] = s[i] - ‘a‘ + 1, tp[i] = i;
22     m = 2, rsort();
23     for (int w = 1, p = 1, i; p < n; w <<= 1, m = p) {
24         for (p = 0, i = n - w + 1; i <= n; i ++) tp[++ p] = i;
25         for (int i = 1; i <= n; i ++) if (sa[i] > w) tp[++ p] = sa[i] - w;
26         rsort(), swap(rank, tp), rank[sa[1]] = p = 1;
27         for (int i = 2; i <= n; i ++) rank[sa[i]] = cmp(tp, sa[i], sa[i - 1], w) ? p : ++ p;
28     }
29     int j, k = 0;
30     for (int i = 1; i <= n; height[rank[i ++]] = k)
31         for (k = k ? k - 1 : k, j = sa[rank[i] - 1]; s[i + k] == s[j + k]; k ++);
32 }
33
34 void st_init() {
35     for (int i = 1; i <= n; i ++) st[i][0] = height[i];
36     for (int j = 1; (1 << j) <= n; j ++)
37         for (int i = 1; i + (1 << j) - 1 <= n; i ++) st[i][j] = min(st[i][j - 1], st[i + (1 << j - 1)][j - 1]);
38 }
39
40 int rmq(int l, int r) {
41     if (l > r) swap(l, r);
42     l ++;
43     int k = 0;
44     while (1 << k + 1 <= r - l + 1) k ++;
45     return min(st[l][k], st[r - (1 << k) + 1][k]);
46 }
47
48 void work() {
49     ans = 0;
50     for (int l = 1; l <= n; l ++)
51         for (int i = 1; i + l <= n; i += l) {
52             int k, r;
53             k = rmq(rank[i], rank[i + l]);
54             r = k % l;
55             if (r) k = max(k, rmq(rank[i - l + r], rank[i + r]));
56             ans = max(ans, k / l + 1);
57         }
58     printf("%d\n", max(ans, 1));
59 }
60
61 int main() {
62     scanf("%d", &T);
63     while (T --) {
64         scanf("%d\n", &n);
65         scanf("%s", s + 1);
66         SA();
67         st_init();
68         work();
69     }
70     return 0;
71 }