Ashton appeared for a job interview and is asked the following question. Arrange all the distinct substrings of a given string in lexicographical order and concatenate them. Print the Kth character of the concatenated string. It is assured that given value of K will be valid i.e. there will be a Kth character. Can you help Ashton out with this?
Input Format
First line will contain a number T i.e. number of test cases.
First line of each test case will contain a string containing characters (a−z) and second line will contain a number K.Output Format
Print Kth character ( the string is 1 indexed )Constraints
1≤T≤5
1≤length≤105
K will be an appropriate integer.Sample Input #00
1 dbac 3Sample Output #00
cExplanation #00
The substrings when arranged in lexicographic order are as follows
a, ac, b, ba, bac, c, d, db, dba, dbacOn concatenating them, we get
aacbbabaccddbdbadbacThe third character in this string is
cand hence the answer.
题意：给出一个字符串，把这个字符串的所有字串按字典序升序拼接起来得到一个大的字符串。求该字符串的第ｋ个字符。
思路：该字符串的字串按字典序排序后的结果其实就是后缀数组中根据ｌｃｐ一个个数出来的字串。。。
Accepted Code:
1 #include <string>
2 #include <iostream>
3 #include <algorithm>
4 using namespace std;
5
6 const int MAX_N = 100005;
7 typedef long long LL;
8 int n, k, sa[MAX_N], rk[MAX_N], lcp[MAX_N], tmp[MAX_N];
9
10 bool compare_sa(int i, int j) {
11 if (rk[i] != rk[j]) return rk[i] < rk[j];
12 int ri = i + k <= n ? rk[i + k] : -1;
13 int rj = j + k <= n ? rk[j + k] : -1;
14 return ri < rj;
15 }
16
17 void construct_sa(const string &S, int *sa) {
18 n = S.length();
19 for (int i = 0; i <= n; i++) {
20 sa[i] = i;
21 rk[i] = (i < n ? S[i] : -1);
22 }
23
24 for (k = 1; k <= n; k *= 2) {
25 sort(sa, sa + n + 1, compare_sa);
26
27 tmp[sa[0]] = 0;
28 for (int i = 1; i <= n; i++) {
29 tmp[sa[i]] = tmp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0);
30 }
31 for (int i = 0; i <= n; i++) rk[i] = tmp[i];
32 }
33 }
34
35 void construct_lcp(const string &S, int *sa, int *lcp) {
36 n = S.length();
37 for (int i = 0; i <= n; i++) rk[sa[i]] = i;
38
39 int h = 0;
40 lcp[0] = 0;
41 for (int i = 0; i < n; i++) {
42 int j = sa[rk[i] - 1];
43
44 if (h > 0) h--;
45 for (; i + h < n && j + h < n; h++) if (S[i + h] != S[j + h]) break;
46
47 lcp[rk[i] - 1] = h;
48 }
49 }
50
51 string S;
52
53 void solve(LL k) {
54 n = S.length();
55 construct_sa(S, sa);
56 construct_lcp(S, sa, lcp);
57
58 for (int i = 0; i < n; i++) {
59 int L = lcp[i];
60 int left = n - sa[i + 1];
61 LL sum = (L + 1 + left) * (LL)(left - L) / 2;
62 if (k > sum) {k -= sum;}
63 else {
64 for (int j = L + 1; j <= left; j++) {
65 if (k <= j) {
66 int index = sa[i + 1] + k;
67 cout << S[index - 1] << endl;
68 return ;
69 } else {
70 k -= j;
71 }
72 }
73 }
74 }
75 }
76 int main(void) {
77 ios::sync_with_stdio(false);
78 int T;
79 cin >> T;
80 while (T--) {
81 LL k;
82 cin >> S >> k;
83 solve(k);
84 }
85 return 0;
86 }