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:

 #include <string>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 const int MAX_N = ;
 int sa[MAX_N], rk[MAX_N], lcp[MAX_N], tmp[MAX_N], n, k;

 bool compare_sa(int i, int j) {
     if (rk[i] != rk[j]) return rk[i] < rk[j];
     int ri = i + k <= n ? rk[i + k] : -;
     int rj = j + k <= n ? rk[j + k] : -;
     return ri < rj;
 }

 void construct_sa(const string &S, int *sa) {
     n = S.length();
     for (int i = ; i <= n; i++) {
         sa[i] = i;
         rk[i] = i < n ? S[i] : -;
     }

     for (k = ; k <= n; k *= ) {
         sort(sa, sa + n + , compare_sa);

         tmp[sa[]] = ;
         for (int i = ; i <= n; i++) {
             tmp[sa[i]] = tmp[sa[i - ]] + (compare_sa(sa[i - ], sa[i]) ?  : );
         }
         for (int i = ; i <= n; i++) rk[i] = tmp[i];
     }
 }

 void construct_lcp(const string &S, int *sa, int *lcp) {
     n = S.length();
     for (int i = ; i <= n; i++) rk[sa[i]] = i;

     int h = ;
     lcp[] = ;
     for (int i = ; i < n; i++) {
         int j = sa[rk[i] - ];

         if (h > ) h--;
         for (; i + h < n && j + h < n; h++) if (S[i + h] != S[j + h]) break;

         lcp[rk[i] - ] = h;
     }
 }

 string S;
 int lft[MAX_N], rgt[MAX_N], st[MAX_N], top;
 void solve() {
     construct_sa(S, sa);
     construct_lcp(S, sa, lcp);

     lcp[n] = n - sa[n];
    // for (int i = 1; i <= n; i++) cerr << lcp[i] << ' ';
    // cerr << endl;
     top = ;
     for (int i = ; i <= n; i++) {
         while (top && lcp[st[top-]] >= lcp[i]) top--;
         if (top) lft[i] = st[top - ] + ;
         else lft[i] = ;
         st[top++] = i;
     }
     top = ;
     for (int i = n; i > ; i--) {
         while (top && lcp[st[top-]] >= lcp[i]) top--;
         // attention: rgt[i] should be asigned to st[top - 1]
         // rather than st[top - 1] - 1 because lcp[i] is the
         // length of the longest common prefix of sa[i] and sa[i + 1].
         if (top) rgt[i] = st[top - ];
         else rgt[i] = n;
         st[top++] = i;
     }
     long long ans = n;
     for (int i = ; i <= n; i++) ans = max(ans, (long long)lcp[i] * (rgt[i] - lft[i] + ));
     cout << ans << endl;
 }

 int main(void) {
     //ios::sync_with_std(false);
     while (cin >> S) solve();
     return ;
 }