SPOJ 694 Distinct Substrings

Description

Given a string, we need to find the total number of its distinct substrings.

Input

T- number of test cases. T<=20;
Each test case consists of one string, whose length is <= 1000

Output

For each test case output one number saying the number of distinct substrings.

Example

Sample Input:
2
CCCCC
ABABA

Sample Output:
5
9

Explanation for the testcase with string ABABA: 
len=1 : A,B
len=2 : AB,BA
len=3 : ABA,BAB
len=4 : ABAB,BABA
len=5 : ABABA

Thus, total number of distinct substrings is 9.

后缀数组

#include<iostream>#include<cmath>#include<map>#include<vector>#include<queue>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int Min = 0;const int Max = 0x7FFFFFFF;const int maxn = 20005;const int bit = 1000005;class suffix{private:char s[maxn];int r[maxn], w[bit], ss[maxn], h[maxn];int sa[maxn], rk[maxn + maxn], size;int limit;public:bool get(){if (scanf("%s", s + 1) != 1) return false;size = strlen(s + 1);return true;}void pre(){memset(rk, 0, sizeof(rk));for (int i = 1; i <= bit; i++) w[i] = 0;for (int i = 1; i <= size; i++) w[(int)s[i]]++;for (int i = 1; i <= bit; i++) w[i] += w[i - 1];for (int i = size; i; i--) sa[w[(int)s[i]]--] = i;for (int i = 1, j = 1; i <= size; i++)rk[sa[i]] = (s[sa[i]] == s[sa[i + 1]] ? j : j++);for (int j = 1; j < size; j += j){for (int i = 1; i <= size; i++) w[i] = 0;for (int i = 1; i <= size; i++) w[rk[i + j]]++;for (int i = 1; i <= size; i++) w[i] += w[i - 1];for (int i = size; i; i--) ss[w[rk[i + j]]--] = i;for (int i = 1; i <= size; i++) w[i] = 0;for (int i = 1; i <= size; i++) w[rk[ss[i]]]++;for (int i = 1; i <= size; i++) w[i] += w[i - 1];for (int i = size; i; i--) sa[w[rk[ss[i]]]--] = ss[i];for (int i = 1, k = 1; i <= size; i++)r[sa[i]] = (rk[sa[i]] == rk[sa[i + 1]] && rk[sa[i] + j] == rk[sa[i + 1] + j]) ? k : k++;for (int i = 1; i <= size; i++) rk[i] = r[i];}for (int i = 1, k = 0, j; i <= size; h[rk[i++]] = k)for (k ? k-- : 0, j = sa[rk[i] - 1]; s[i + k] == s[j + k]; k++);}void work(){int ans = 0;for (int i = 1; i <= size; i++)ans += size - sa[i] + 1 - h[i];printf("%d\n", ans);}}f;int main(){int T;scanf("%d", &T);while (T--){f.get();f.pre();f.work();}return 0;}