SPOJ 694 Distinct Substrings

后缀数组求不同的子串数。。

每一个子串都是某个后缀的前缀，每加入一个后缀都会增加 n-sa[ i ] 个子串，但是有h[ i ]个子串会是重复的，所以对每增加的一个后缀会产生 n-sa[ i ]-h[ i ]个不同的子串

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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.

Source

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#include 
  
   
#include 
   
     #include 
    
      #include 
     
       using namespace std; const int maxn=10000; int sa[maxn],rank[maxn],rank2[maxn],h[maxn],c[maxn],*x,*y,ans[maxn],n; char str[maxn]; bool cmp(int* r,int a,int b,int l,int n) { if(r[a]==r[b]&&a+l
      
       =0;i--) sa[--c[x[y[i]]]]=y[i]; } void get_sa(char c[],int n,int sz=128) { x=rank,y=rank2; for(int i=0;i
       
        =len) y[yid++]=sa[i]-len; radix_sort(n,sz); swap(x,y); x[sa[0]]=yid=0; for(int i=1;i
        
         =n) break; } for(int i=0;i