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
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 const int maxn=50010;
 7 char S[maxn];
 8 int r[maxn],wa[maxn],wb[maxn],wv[maxn],ws[maxn],sa[maxn];
 9 
10 bool cmp(int *p,int i,int j,int l)
11 {return p[i]==p[j]&&p[i+l]==p[j+l];}
12 
13 void DA(int n,int m)
14 {
15     int i,j,p,*x=wa,*y=wb,*t;
16     for(i=0;i<m;i++)
17         ws[i]=0;
18     for(i=0;i<n;i++)
19         ++ws[x[i]=r[i]];
20     for(i=1;i<m;i++)
21         ws[i]+=ws[i-1];
22     for(i=n-1;i>=0;i--)
23         sa[--ws[x[i]]]=i;
24         
25     for(j=1,p=1;p<n;j<<=1,m=p)
26     {
27         for(p=0,i=n-j;i<n;i++)
28             y[p++]=i;
29         for(i=0;i<n;i++)
30             if(sa[i]>=j)
31                 y[p++]=sa[i]-j;
32         for(i=0;i<n;i++)
33             wv[i]=x[y[i]];
34         for(i=0;i<m;i++)
35             ws[i]=0;    
36         for(i=0;i<n;i++)
37             ++ws[wv[i]];
38         for(i=1;i<m;i++)            
39             ws[i]+=ws[i-1];
40         for(i=n-1;i>=0;i--)
41             sa[--ws[wv[i]]]=y[i];
42         for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++)
43             x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;        
44     }
45 }
46 
47 int rank[maxn],lcp[maxn];
48 void LCP(int n)
49 {
50     int i,j,k=0;
51     for(i=1;i<=n;i++)
52         rank[sa[i]]=i;
53     for(i=0;i<n;lcp[rank[i++]]=k)
54         for(k?k--:k,j=sa[rank[i]-1];r[i+k]==r[j+k];k++);    
55 }
56 
57 int main()
58 {
59     int Q;
60     scanf("%d",&Q);
61     while(~scanf("%s",S))
62     {
63         int i,n;
64         long long ans=0;
65         for(i=0;S[i];i++)
66         r[i]=S[i];
67         DA(i+1,128);
68         LCP(i);
69         n=i;
70         for(i=0;i<n;i++)
71             ans+=n-i-lcp[rank[i]];
72         printf("%d\n",ans);    
73     }
74     return 0;
75 }

 

posted @ 2016-02-16 21:27  TenderRun  阅读(166)  评论(0)  编辑  收藏  举报