SPOJ694 -- DISUBSTR 后缀树组求不相同的子串的个数

DISUBSTR - Distinct Substrings

 

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.

题意：求不同的子串的个数。

首先要知道，任意子串必是某一后缀的前缀。对于suffix[sa[i]],  它有len-sa[i]个前缀，，其中lcp[i]个子串与suffix[sa[i-1]]的前缀重复。。

每次只需要加上 len-sa[i]-lcp[i]就行了。。

 #include <set>
 #include <map>
 #include <cmath>
 #include <ctime>
 #include <queue>
 #include <stack>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 typedef unsigned long long ull;
 typedef long long ll;
 const int inf = 0x3f3f3f3f;
 const double eps = 1e-;
 const int maxn = 2e4+;
 int sa[maxn], k, len, tmp[maxn], rank[maxn];
 string s;
 bool cmp(int i, int j)
 {
     if (rank[i] != rank[j])
         return rank[i] < rank[j];
     else
     {
         int x = (i+k <= len ? rank[i+k] : -);
         int y = (j+k <= len ? rank[j+k] : -);
         return x < y;
     }
 }
 void build_sa()
 {
     for (int i = ; i <= len; i++)
     {
         sa[i] = i;
         rank[i] = (i < len ? s[i] : -);
     }
     for (k = ; k <= len; k *= )
     {
         sort (sa,sa+len+,cmp);
         tmp[sa[]] = ;
         for (int i = ; i <= len; i++)
         {
             tmp[sa[i]] = tmp[sa[i-]] + (cmp(sa[i-],sa[i])?  : );
         }
         for (int i = ; i <= len; i++)
             rank[i] = tmp[i];
     }
 }
 int lcp[maxn];
 void get_lcp()
 {
     for (int i = ; i < len; i++)
         rank[sa[i]] = i;
     int h = ;
     lcp[] = ;
     for (int i = ; i < len; i++)
     {
         int j = sa[rank[i]-];
         if (h > )
             h--;
         for (; h+i < len && h+j < len; h++)
             if (s[i+h] != s[j+h])
                 break;
         lcp[rank[i]] = h;
     }
 }
 int main()
 {
     #ifndef ONLINE_JUDGE
         freopen("in.txt","r",stdin);
     #endif
     int T;
     scanf ("%d", &T);
     while (T--)
     {
         cin >> s;
         len = s.size();
         build_sa();
         get_lcp();
         int ans = ;
         for (int i = ; i <= len; i++)
         {
             ans += len - sa[i] - lcp[i];
         }
         printf ("%d\n",ans);
     }
     return ;
 }