SPOJ 题目694 Distinct Substrings（后缀数组，求不同的子串个数）

DISUBSTR - Distinct Substrings

no tags

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.



ac代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int s[2002];
char str[2002];
int sa[2002],t1[2002],t2[2002],c[2002];
int rank[2002],height[2002];
void build_sa(int s[],int n,int m)
{
int i,j,p,*x=t1,*y=t2;
for(i=0;i<m;i++)
c[i]=0;
for(i=0;i<n;i++)
c[x[i]=s[i]]++;
for(i=1;i<m;i++)
c[i]+=c[i-1];
for(i=n-1;i>=0;i--)
sa[--c[x[i]]]=i;
for(j=1;j<=n;j<<=1)
{
p=0;
for(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<m;i++)
c[i]=0;
for(i=0;i<n;i++)
c[x[y[i]]]++;
for(i=1;i<m;i++)
c[i]+=c[i-1];
for(i=n-1;i>=0;i--)
sa[--c[x[y[i]]]]=y[i];
swap(x,y);
p=1;
x[sa[0]]=0;
for(i=1;i<n;i++)
x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++;
if(p>=n)
break;
m=p;
}
}
void getHeight(int s[],int n)
{
int i,j,k=0;
for(i=0;i<=n;i++)
rank[sa[i]]=i;
for(i=0;i<n;i++)
{
if(k)
k--;
j=sa[rank[i]-1];
while(s[i+k]==s[j+k])
k++;
height[rank[i]]=k;
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int i;
scanf("%s",str);
int len=strlen(str);
for(i=0;i<len;i++)
{
s[i]=str[i];
}
s[len]=0;
build_sa(s,len+1,128);
getHeight(s,len);
long long ans=(len)*(len+1)/2;
for(i=1;i<=len;i++)
{
ans-=height[i];
}
printf("%lld\n",ans);
}
}