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=0;i<m;i++)b[i]=0;
    for(i=0; i<n; i++)b[x[i]=a[i]]++;
    for(i=1; i<m; i++)b[i]+=b[i-1];
    for(i=n-1; i>=0; i--)sa[--b[x[i]]]=i;
    for(j=1,p=1; p<n; j*=2,m=p)
    {
        for(p=0,i=n-j; i<n; i++)y[p++]=i;
        for(i=0; i<n; i++)if(sa[i]>=j)y[p++]=sa[i]-j;
        for(i=0; i<n; i++)e[i]=x[y[i]];
        for(i=0;i<m;i++)b[i]=0;
        for(i=0; i<n; i++)b[e[i]]++;
        for(i=1; i<m; i++)b[i]+=b[i-1];
        for(i=n-1; i>=0; i--)sa[--b[e[i]]]=y[i];
        for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; i<n; i++)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
    }
}
void callheight()
{
    int i,j,k=0;
    for(i=1; i<n; i++)b[sa[i]]=i;
    n--;
    for(i=0; i<n; height[b[i++]]=k)
        for(k?k--:0,j=sa[b[i]-1]; a[i+k]==a[j+k]; k++);
}

int main()
{
    int t,i;
    cin>>t;
    for(i=0;i<t;i++)
    {
        scanf("%s",a);
        n=strlen(a);
        n++;
        m=200;
        da();
        callheight();
        int sum=0;
        for(int j=1;j<=n;j++)sum+=n-sa[j]-height[j];
        cout<<sum<<endl;
    }
}