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 = 100002;
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] : -1;
    int rj = j + k <= n ? rk[j + k] : -1;
    return ri < rj;
}

void construct_sa(const string &S, int *sa) {
    n = S.length();
    for (int i = 0; i <= n; i++) {
        sa[i] = i;
        rk[i] = i < n ? S[i] : -1;
    }
    
    for (k = 1; k <= n; k *= 2) {
        sort(sa, sa + n + 1, compare_sa);
        
        tmp[sa[0]] = 0;
        for (int i = 1; i <= n; i++) {
            tmp[sa[i]] = tmp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0);
        }
        for (int i = 0; i <= n; i++) rk[i] = tmp[i];
    }
}

void construct_lcp(const string &S, int *sa, int *lcp) {
    n = S.length();
    for (int i = 0; i <= n; i++) rk[sa[i]] = i;
    
    int h = 0;
    lcp[0] = 0;
    for (int i = 0; i < n; i++) {
        int j = sa[rk[i] - 1];
        
        if (h > 0) h--;
        for (; i + h < n && j + h < n; h++) if (S[i + h] != S[j + h]) break;
            
        lcp[rk[i] - 1] = 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 = 0;
    for (int i = 1; i <= n; i++) {
        while (top && lcp[st[top-1]] >= lcp[i]) top--;
        if (top) lft[i] = st[top - 1] + 1;
        else lft[i] = 1;
        st[top++] = i;
    }
    top = 0;
    for (int i = n; i > 0; i--) {
        while (top && lcp[st[top-1]] >= 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 - 1];
        else rgt[i] = n;
        st[top++] = i;
    }
    long long ans = n;
    for (int i = 1; i <= n; i++) ans = max(ans, (long long)lcp[i] * (rgt[i] - lft[i] + 1));
    cout << ans << endl;
}

int main(void) {
    //ios::sync_with_std(false);
    while (cin >> S) solve();
    return 0;
}