Hackerrank--String Function Calculation(后缀数组）

题目链接

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 T with her, and value of string S over function f can be calculated as given below:

f(S)=|S|∗NumberoftimesSoccuringinT

Jane wants to know the maximum value of f(S) among all the substrings (S) of string T. Can you help her?

Input Format
A single line containing string T in small letter('a' - 'z').

Output Format
An integer containing the value of output.

Constraints
1 ≤|T|≤ 105

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 }