后缀数组：SPOJ SUBST1 - New 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 <= 50000

Output

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

Example

Input:
2
CCCCC
ABABA

Output:
5
9

　　题意：求一个字符串有多少不同子串
　　LCP的应用，会打后缀数组的模板后就十分简单了
　　后缀数组的理解：
　　http://www.cnblogs.com/staginner/archive/2012/02/02/2335600.html

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 const int maxn=;
 char S[maxn];
 int r[maxn],wa[maxn],wb[maxn],wv[maxn],ws[maxn],sa[maxn];

 bool cmp(int *p,int i,int j,int l)
 {return p[i]==p[j]&&p[i+l]==p[j+l];}

 void DA(int n,int m)
 {
     int i,j,p,*x=wa,*y=wb,*t;
     for(i=;i<m;i++)
         ws[i]=;
     for(i=;i<n;i++)
         ++ws[x[i]=r[i]];
     for(i=;i<m;i++)
         ws[i]+=ws[i-];
     for(i=n-;i>=;i--)
         sa[--ws[x[i]]]=i;

     for(j=,p=;p<n;j<<=,m=p)
     {
         for(p=,i=n-j;i<n;i++)
             y[p++]=i;
         for(i=;i<n;i++)
             if(sa[i]>=j)
                 y[p++]=sa[i]-j;
         for(i=;i<n;i++)
             wv[i]=x[y[i]];
         for(i=;i<m;i++)
             ws[i]=;
         for(i=;i<n;i++)
             ++ws[wv[i]];
         for(i=;i<m;i++)
             ws[i]+=ws[i-];
         for(i=n-;i>=;i--)
             sa[--ws[wv[i]]]=y[i];
         for(t=x,x=y,y=t,p=,x[sa[]]=,i=;i<n;i++)
             x[sa[i]]=cmp(y,sa[i-],sa[i],j)?p-:p++;
     }
 }

 int rank[maxn],lcp[maxn];
 void LCP(int n)
 {
     int i,j,k=;
     for(i=;i<=n;i++)
         rank[sa[i]]=i;
     for(i=;i<n;lcp[rank[i++]]=k)
         for(k?k--:k,j=sa[rank[i]-];r[i+k]==r[j+k];k++);
 }

 int main()
 {
     int Q;
     scanf("%d",&Q);
     while(~scanf("%s",S))
     {
         int i,n;
         long long ans=;
         for(i=;S[i];i++)
         r[i]=S[i];
         DA(i+,);
         LCP(i);
         n=i;
         for(i=;i<n;i++)
             ans+=n-i-lcp[rank[i]];
         printf("%d\n",ans);
     }
     return ;
 }