705. New Distinct Substrings spoj（后缀数组求所有不同子串）

705. New Distinct Substrings Problem code: SUBST1

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

---

Added by:	Hoang Hong Quan
Date:	2006-01-18
Time limit:	2s
Source limit:	50000B
Memory limit:	256MB
Cluster:	Pyramid (Intel Pentium III 733 MHz)
Languages:	All except: NODEJS PERL 6
Resource:	Base on a problem in ByteCode06

 

 

 

 

 

 

 

 

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 using namespace std;
 #define N 51000
 char a[N];
 int c[N],d[N],e[N],sa[N],height[N],n,b[N],m;
 int cmp(int *r,int a,int b,int l)
 {
     return r[a]==r[b]&&r[a+l]==r[b+l];
 }
 void da()
 {
     int i,j,p,*x=c,*y=d,*t;
     for(i=;i<m;i++)b[i]=;
     for(i=; i<n; i++)b[x[i]=a[i]]++;
     for(i=; i<m; i++)b[i]+=b[i-];
     for(i=n-; i>=; i--)sa[--b[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++)e[i]=x[y[i]];
         for(i=;i<m;i++)b[i]=;
         for(i=; i<n; i++)b[e[i]]++;
         for(i=; i<m; i++)b[i]+=b[i-];
         for(i=n-; i>=; i--)sa[--b[e[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++;
     }
 }
 void callheight()
 {
     int i,j,k=;
     for(i=; i<n; i++)b[sa[i]]=i;
     n--;
     for(i=; i<n; height[b[i++]]=k)
         for(k?k--:,j=sa[b[i]-]; a[i+k]==a[j+k]; k++);
 }

 int main()
 {
     int t,i;
     cin>>t;
     for(i=;i<t;i++)
     {
         scanf("%s",a);
         n=strlen(a);
         n++;
         m=;
         da();
         callheight();
         int sum=;
         for(int j=;j<=n;j++)sum+=n-sa[j]-height[j];
         cout<<sum<<endl;
     }
 }