Hackerrank--String Function Calculation(后缀数组）

Jane loves string more than anything. She made a function related to the string some days ago and forgot about it. She is now confused about calculating the value of this function. She has a string

f(S)=|S|∗NumberoftimesSoccuringinT

Jane wants to know the maximum value of

Input Format
A single line containing string

Output Format
An integer containing the value of output.

Constraints

Sample Input #00
aaaaaa
Sample Output #00
12
Explanation #00
f('a') = 6
f('aa') = 10
f('aaa') = 12
f('aaaa') = 12
f('aaaaa') = 10
f('aaaaaa') = 6
Sample Input #01
abcabcddd
Sample Output #01
9
Explanation #01
f("a") = 2
f("b") = 2
f("c") = 2
f("ab") = 4
f("bc") = 4
f("ddd") = 3
f("abc") = 6
f("abcabcddd") = 9
Among the function values 9 is the maximum one.

题意：求字符串所有子串中，f(s)的最大值。这里ｓ是某个子串，f(s)　＝　ｓ的长度与ｓ在原字符串中出现次数的乘积。

求得lcp, 转化为求以lcp为高的最大子矩形。

Accepted Code:

 1 #include <string>
 2 #include <iostream>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 const int MAX_N = 100002;
 7 int sa[MAX_N], rk[MAX_N], lcp[MAX_N], tmp[MAX_N], n, k;
 8 
 9 bool compare_sa(int i, int j) {
10     if (rk[i] != rk[j]) return rk[i] < rk[j];
11     int ri = i + k <= n ? rk[i + k] : -1;
12     int rj = j + k <= n ? rk[j + k] : -1;
13     return ri < rj;
14 }
15 
16 void construct_sa(const string &S, int *sa) {
17     n = S.length();
18     for (int i = 0; i <= n; i++) {
19         sa[i] = i;
20         rk[i] = i < n ? S[i] : -1;
21     }
22     
23     for (k = 1; k <= n; k *= 2) {
24         sort(sa, sa + n + 1, compare_sa);
25         
26         tmp[sa[0]] = 0;
27         for (int i = 1; i <= n; i++) {
28             tmp[sa[i]] = tmp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0);
29         }
30         for (int i = 0; i <= n; i++) rk[i] = tmp[i];
31     }
32 }
33 
34 void construct_lcp(const string &S, int *sa, int *lcp) {
35     n = S.length();
36     for (int i = 0; i <= n; i++) rk[sa[i]] = i;
37     
38     int h = 0;
39     lcp[0] = 0;
40     for (int i = 0; i < n; i++) {
41         int j = sa[rk[i] - 1];
42         
43         if (h > 0) h--;
44         for (; i + h < n && j + h < n; h++) if (S[i + h] != S[j + h]) break;
45             
46         lcp[rk[i] - 1] = h;
47     }
48 }
49 
50 string S;
51 int lft[MAX_N], rgt[MAX_N], st[MAX_N], top;
52 void solve() {
53     construct_sa(S, sa);
54     construct_lcp(S, sa, lcp);
55     
56     lcp[n] = n - sa[n];
57    // for (int i = 1; i <= n; i++) cerr << lcp[i] << ' ';
58    // cerr << endl;
59     top = 0;
60     for (int i = 1; i <= n; i++) {
61         while (top && lcp[st[top-1]] >= lcp[i]) top--;
62         if (top) lft[i] = st[top - 1] + 1;
63         else lft[i] = 1;
64         st[top++] = i;
65     }
66     top = 0;
67     for (int i = n; i > 0; i--) {
68         while (top && lcp[st[top-1]] >= lcp[i]) top--;
69         // attention: rgt[i] should be asigned to st[top - 1]
70         // rather than st[top - 1] - 1 because lcp[i] is the
71         // length of the longest common prefix of sa[i] and sa[i + 1]. 
72         if (top) rgt[i] = st[top - 1];
73         else rgt[i] = n;
74         st[top++] = i;
75     }
76     long long ans = n;
77     for (int i = 1; i <= n; i++) ans = max(ans, (long long)lcp[i] * (rgt[i] - lft[i] + 1));
78     cout << ans << endl;
79 }
80 
81 int main(void) {
82     //ios::sync_with_std(false);
83     while (cin >> S) solve();
84     return 0;
85 }